Chapter 4 – Quadratic Equations Exercise Ex. 4.1
Question 1 (i)
Solution 1 (i)
Question 1 (ii)
Solution 1 (ii)
Question 1 (iii)
Solution 1 (iii)
Question 1 (iv)
Solution 1 (iv)
Question 1 (v)
Solution 1 (v)
Question 1(vi)
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/102591_image123.png?w=1200&ssl=1)
Solution 1(vi)
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/102591_image124.png?w=1200&ssl=1)
Question 1 (vii)
Solution 1 (vii)
Question 1 (viii)
Solution 1 (viii)
Question 1 (ix)
Solution 1 (ix)
Question 1 (x)
Solution 1 (x)
Question 1 (xi)
Solution 1 (xi)
Question 1 (xii)
Solution 1 (xii)
Question 1 (xiii)
Solution 1 (xiii)
Question 1 (xiv)
Solution 1 (xiv)
Question 1 (xv)
Is X(x+1)+8=(x+2)(x-2) a quadratic equation?Solution 1 (xv)
Question 2 (i)
Solution 2 (i)
Question 2 (ii)
Solution 2 (ii)
Question 2 (iii)
Solution 2 (iii)
Question 2 (iv)
Solution 2 (iv)
Question 2 (v)
Are x = 2 and x = 3, solutions of the equation 2x2 – x + 9 = x2 + 4x + 3 , Solution 2 (v)
= 2x2 – x + 9 – x2 + 4x + 3
= x2 – 5x + 6 = 0
Here, LHS = x2 – 5x + 6 and RHS = 0
Substituting x = 2 and x = 3
= x2 – 5x + 6
= (2)2 – 5(2) + 6
=10-10
=0
= RHS
= x2 – 5x + 6
= (3)2 – 5(3) + 6
= 9 – 15 + 6
=15 – 15
=0
= RHS
x = 2 and x = 3 both are the solutions of the given quadratic equation.Question 2 (vi)
Solution 2 (vi)
Question 2 (vii)
Solution 2 (vii)
Question 3 (i)
Solution 3 (i)
Question 3 (ii)
Solution 3 (ii)
Question 3 (iii)
Solution 3 (iii)
Question 3 (iv)
Solution 3 (iv)
Question 4
Solution 4
Question 5
Solution 5
Chapter 4 – Quadratic Equations Exercise Ex. 4.2
Question 1
The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.Solution 1
Question 2
John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John had x marbles.Solution 2
Question 3
A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation to find x.Solution 3
Question 4
The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.Solution 4
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/96411_PICC440.jpg?w=1200&ssl=1)
Question 5
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.Solution 5
Question 6
A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.Solution 6
Chapter 4 – Quadratic Equations Exercise Ex. 4.3
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/102601_image125.png?w=1200&ssl=1)
Solution 9
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/102601_image126.png?w=1200&ssl=1)
Question 10
Solution 10
Question 11
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image006.gif?w=1200&ssl=1)
Solution 11
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image008.gif?w=1200&ssl=1)
Question 12
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/98271_E_94.gif?w=1200&ssl=1)
Solution 12
Question 13
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/98381_E_98.gif?w=1200&ssl=1)
Solution 13
Question 14
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/98401_E_100.gif?w=1200&ssl=1)
Solution 14
Question 15
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image016.gif?w=1200&ssl=1)
Solution 15
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image018.gif?w=1200&ssl=1)
Question 16
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/110531_image053.gif?w=1200&ssl=1)
Solution 16
Question 17
9x2 – 6b2x – (a4 – b4) = 0Solution 17
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image006.png?w=1200&ssl=1)
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solve the following quadratic equation by factorisation:
2x2 + ax – a2 = 0Solution 20
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image014.gif?w=1200&ssl=1)
Question 21
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image032.gif?w=1200&ssl=1)
Solution 21
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image034.gif?w=1200&ssl=1)
Question 22
Solve the following quadratic equations by factorization:
Solution 22
Question 23
Solution 23
Question 24
Solution 24
Question 25
Solution 25
Question 26
Solution 26
Question 27
Solution 27
Question 28
Solution 28
Question 29
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image024.gif?w=1200&ssl=1)
Solution 29
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image026.gif?w=1200&ssl=1)
Question 30
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image028.gif?w=1200&ssl=1)
Solution 30
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image030.gif?w=1200&ssl=1)
Question 31
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image036.gif?w=1200&ssl=1)
Solution 31
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image038.gif?w=1200&ssl=1)
Question 32
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image040.gif?w=1200&ssl=1)
Solution 32
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image042.gif?w=1200&ssl=1)
Question 33
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image044.gif?w=1200&ssl=1)
Solution 33
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image046.gif?w=1200&ssl=1)
Question 34
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image008.png?w=1200&ssl=1)
Solution 34
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image010.png?w=1200&ssl=1)
Question 35
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image012.png?w=1200&ssl=1)
Solution 35
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image014.png?w=1200&ssl=1)
Question 36
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/110541_image055.gif?w=1200&ssl=1)
Solution 36
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/110541_image056.gif?w=1200&ssl=1)
Question 37
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image002.png?w=1200&ssl=1)
Solution 37
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image004.png?w=1200&ssl=1)
Question 38
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image010.gif?w=1200&ssl=1)
Solution 38
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image012.gif?w=1200&ssl=1)
Question 39
Solution 39
Question 40
Solution 40
Question 41
Solution 41
Question 42
Solution 42
Question 43
Solve the following quadratic equations by factorization:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_Sharma_X_Math_Ch4_TBS_Rwd_files/20180719105705_image002.png?w=1200&ssl=1)
Solution 43
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_Sharma_X_Math_Ch4_TBS_Rwd_files/20180719105705_image004.png?w=1200&ssl=1)
Question 44
Solution 44
Question 45
Solution 45
Question 46
Solution 46
Question 47
Solve the following quadratic equations by factorization:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_3_files/20170608155448_image002.png?w=1200&ssl=1)
Solution 47
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_3_files/20170608155448_image004.png?w=1200&ssl=1)
Question 48
Solve the following quadratic equations by factorization:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_3_files/20170608155448_image006.png?w=1200&ssl=1)
Solution 48
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_3_files/20170608155448_image008.png?w=1200&ssl=1)
Question 49
Solution 49
Question 50
Solution 50
Question 51
Solution 51
Question 52
Solution 52
Question 53
Solution 53
Question 54
Solution 54
Question 55
Solution 55
Question 56
Solution 56
Question 57
Solution 57
Question 58
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image020.gif?w=1200&ssl=1)
Solution 58
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image022.gif?w=1200&ssl=1)
Question 59
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image048.gif?w=1200&ssl=1)
Solution 59
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image050.gif?w=1200&ssl=1)
Question 60
Solve the following quadratic equation by factorisation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image052.gif?w=1200&ssl=1)
Solution 60
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_3_ASH_files/20150713160457_image054.gif?w=1200&ssl=1)
Question 61
Solution 61
Question 62
Solution 62
Chapter 4 – Quadratic Equations Exercise Ex. 4.4
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Find the roots of the following quadratic equations (if they exist) by the method of completing the square
x2 – 8x + 18 = 0Solution 6
Given equation is x2 – 8x + 18 = 0
x2 – 2 × x × 4 + + 42 – 42 + 18 = 0
(x – 4)2 – 16 + 18 = 0
(x – 4)2 = 16 – 18
(x – 4)2 = -2
Taking square root on both the sides, we get
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/QuadraticEquations_Addition_files/20210518115340_image002.png?w=1200&ssl=1)
Therefore, real roots does not exist.Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Chapter 4 – Quadratic Equations Exercise Ex. 4.5
Question 1 (i)
Solution 1 (i)
Question 1 (ii)
Solution 1 (ii)
Question 1(iii)
Solution 1(iii)
Question 1 (iv)
Solution 1 (iv)
Question 1(v)
Solution 1(v)
Question 1(vi)
Solution 1(vi)
Question 1 (vii)
Write the discriminant of the following quadratic equations:
(x + 5)2 = 2(5x – 3)Solution 1 (vii)
x2 + 2 × x × 5 + 52 = 10x – 6
x2 + 10x + 25 = 10x – 6
x2 + 31 = 0
Here, a = 1, b = 0 and c = 31
Therefore, the discriminant is
D = b2 – 4ac
= 0 – 4 × 1 × 31
= -124Question 2 (i)
Solution 2 (i)
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/13121_PICC441.jpg?w=1200&ssl=1)
Question 2 (ii)
Solution 2 (ii)
Question 2 (iii)
Solution 2 (iii)
Question 2 (iv)
Solution 2 (iv)
Question 2 (v)
Solution 2 (v)
Question 2 (vi)
Solution 2 (vi)
Question 2 (vii)
Solution 2 (vii)
Question 2 (viii)
Solution 2 (viii)
Question 2 (ix)
Solution 2 (ix)
Question 2(x)
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/96431_swsw.gif?w=1200&ssl=1)
Solution 2(x)
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/96431_sws.gif?w=1200&ssl=1)
Question 2(xi)
Solution 2(xi)
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/98711_E_108.gif?w=1200&ssl=1)
Question 2(xii)
3x2 – 5x + 2 = 0Solution 2(xii)
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/98721_E_109.gif?w=1200&ssl=1)
Question 3(i)
Solution 3(i)
Question 3(ii)
Solve for x:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_5_files/20170608160617_image002.png?w=1200&ssl=1)
Solution 3(ii)
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_5_files/20170608160617_image004.png?w=1200&ssl=1)
Question 3(iii)
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/98431_E_103.gif?w=1200&ssl=1)
Solution 3(iii)
Question 3(iv)
Solve for x:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_5_ASH_files/20150714142846_image002.gif?w=1200&ssl=1)
Solution 3(iv)
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_5_ASH_files/20150714142846_image004.gif?w=1200&ssl=1)
Question 3(v)
Solve for x:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_5_files/20170608160617_image006.png?w=1200&ssl=1)
Solution 3(v)
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_5_files/20170608160617_image008.png?w=1200&ssl=1)
Chapter 4 – Quadratic Equations Exercise Ex. 4.6
Question 1(i)
Determine the nature of the roots of the following quadratic equations:
2x2 – 3x + 5 = 0Solution 1(i)
Question 1(ii)
Determine the nature of the roots of the following quadratic equations:
2x2 – 6x + 3 = 0Solution 1(ii)
Question 1(iii)
Solution 1(iii)
Question 1(iv)
Solution 1(iv)
Question 1(v)
Solution 1(v)
Question 1 (vi)
Determine the nature of the roots of the following quadratic equations:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/QuadraticEquations_Addition_files/20210518115340_image004.png?w=1200&ssl=1)
Solution 1 (vi)
Given quadratic equation is
Here,
Therefore, we have
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/QuadraticEquations_Addition_files/20210518115340_image008.png?w=1200&ssl=1)
As D = 0, roots of the given equation are real and equal.Question 2(i)
Solution 2(i)
Question 2(ii)
Solution 2(ii)
Question 2(iii)
Solution 2(iii)
Question 2(iv)
Solution 2(iv)
Question 2(v)
Solution 2(v)
Question 2(vi)
Solution 2(vi)
Question 2(vii)
Solution 2(vii)
Question 2(viii)
Solution 2(viii)
Question 2(ix)
Solution 2(ix)
Question 2(x)
Solution 2(x)
Question 2(xi)
Solution 2(xi)
Question 2(xii)
Solution 2(xii)
Question 2(xiii)
Solution 2(xiii)
Question 2(xiv)
Solution 2(xiv)
Question 2(xv)
Solution 2(xv)
Question 2(xvi)
Solution 2(xvi)
Question 2(xvii)
Solution 2(xvii)
Question 2(xviii)
Find the values of k for which the roots are real and equal in each of the following equations:
4x2 – 2 (k + 1)x + (k + 1) = 0 Solution 2(xviii)
4x2 – 2 (k + 1)x + (k + 1) = 0 Comparing with ax2 + bx + c = 0, we get a = 4, b = -2(k + 1), c = k + 1 According to the question, roots are real and equal. Hence, b2 – 4ac = 0Question 3(i)
Solution 3(i)
Question 3(ii)
In the following, determine the set of values of k for which the given quadratic equation has real roots:
2x2 + x + k = 0Solution 3(ii)
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_6_files/20170608161042_image002.png?w=1200&ssl=1)
Question 3(iii)
Solution 3(iii)
Question 3(iv)
Solution 3(iv)
Question 3(v)
Solution 3(v)
Question 4(i)
Solution 4(i)
Question 4(ii)
Solution 4(ii)
Question 4(iii)
Solution 4(iii)
Question 4(iv)
Find the values of k for which the following equations have real and equal roots:
x2 + k(2x + k – 1) + 2 = 0 Solution 4(iv)
x2 + k(2x + k – 1) + 2 = 0 x2 + 2kx + k(k – 1) + 2 = 0 Comparing with ax2 + bx + c = 0, we get a = 1, b = 2k, c = k(k – 1) + 2 According to the question, roots are real and equal. Hence, b2 – 4ac = 0Question 5(i)
Find the values of k for which the roots are real and equal in each of the following equations:
2x2 + kx + 3 = 0Solution 5(i)
Question 5(ii)
Find the values of k for which the roots are real and equal in each of the following equations:
kx (x – 2) + 6 = 0Solution 5(ii)
Question 5(iii)
Find the values of k for which the roots are real and equal in each of the following equations:
x2 – 4kx + k = 0Solution 5(iii)
Question 5(iv)
Find the value of k for which the roots are real and equal in the following equation:
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_6_ASH_files/20150713180419_image002.gif?w=1200&ssl=1)
Solution 5(iv)
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_6_ASH_files/20150713180419_image004.gif?w=1200&ssl=1)
Question 5(v)
Find the value of p for which the roots are real and equal in the following equation:
px(x – 3) + 9 = 0Solution 5(v)
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_6_ASH_files/20150713180419_image006.gif?w=1200&ssl=1)
Question 5(vi)
Find the values of k for which the following equations have real roots.
4x2 + kx + 3 = 0 Solution 5(vi)
4x2 + kx + 3 = 0 Comparing with ax2 + bx + c = 0, we get a = 4, b = k, c = 3 According to the question, roots are real and equal. Hence, b2 – 4ac = 0Question 6 (i)
Solution 6 (i)
Question 6 (ii)
Solution 6 (ii)
Question 7
Solution 7
Question 8
Solution 8
Question 9(i)
Find the values of k for which the quadratic equation (3k + 1)x2 + 2(k + 1)x + 1 = 0 has equal roots. Also, find the roots.Solution 9(i)
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_6_ASH_files/20150713180419_image010.gif?w=1200&ssl=1)
Question 9 (ii)
Write all the values of k for which the quadratic equation x2 + kx + 16 = 0 has equal roots. Find the roots of the equation so obtained.Solution 9 (ii)
Given quadratic equation is x2 + kx + 16 = 0
As it has equal roots, the discriminant will be 0.
Here, a = 1, b = k, c = 16
Therefore, D = k2 – 4(1)(16) = 0
i.e. k2 – 64 = 0
i.e. k = ± 8
When k = 8, the equation becomes x2 + 8x + 16 = 0
or x2 – 8x + 16 = 0
As D = 0, roots of the given equation are real and equal.Question 10
Find the values of p for which the quadratic equation (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0 has equal roots. Also, find these roots.Solution 10
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_6_ASH_files/20150713180419_image012.gif?w=1200&ssl=1)
Question 11
If -5 is a root of the quadratic equation, 2x2 + px – 15 = 0, and the quadratic equation p(x2 + x) + k = 0 has equal roots, find the value of k.Solution 11
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_6_ASH_files/20150713180419_image014.gif?w=1200&ssl=1)
Question 12
If 2 is a root of the quadratic equation 3x2 + px – 8 = 0 and the quadratic equation 4x2 – 2px + k = 0 has equal roots, find the value of k.Solution 12
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_6_ASH_files/20150713180419_image016.gif?w=1200&ssl=1)
Question 13
If 1 is root of the quadratic equation 3x2 + ax – 2 = 0 and the quadratic equation a(x2 + 6x) – b = 0 has equal roots, find the value of b.Solution 13
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_6_ASH_files/20150713180419_image018.gif?w=1200&ssl=1)
Question 14
Find the value of p for which the quadratic equation (p + 1)x2 – 6(p + 1)x + 3(p + 9) = 0, p ≠ -1 has equal roots. Hence, find the roots of equation.Solution 14
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image016.png?w=1200&ssl=1)
Question 15(i)
Solution 15(i)
Question 15(ii)
Solution 15(ii)
Question 15(iii)
Solution 15(iii)
Question 15(iv)
Solution 15(iv)
Question 16(i)
Solution 16(i)
Question 16(ii)
Solution 16(ii)
Question 16(iii)
Solution 16(iii)
Question 16(iv)
Solution 16(iv)
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24
Solution 24
Question 25
Solution 25
Chapter 4 – Quadratic Equations Exercise Ex. 4.7
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
Question 9
Solution 9
Question 10
Solution 10
Question 11
The sum of a number and its square is 63/4. Find the numbers.Solution 11
Question 12
Solution 12
Question 13
Solution 13
Question 14
Solution 14
Question 15
Solution 15
Question 16
Solution 16
Question 17
Solution 17
Question 18
Solution 18
Question 19
Solution 19
Question 20
Solution 20
Question 21
Solution 21
Question 22
Solution 22
Question 23
Solution 23
Question 24
A natural number when increased by 12 equals 160 times its reciprocal. Find the number.Solution 24
Let x be the natural number.
As per the question, we have
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/QuadraticEquations_Addition_files/20210518115340_image010.png?w=1200&ssl=1)
Therefore, x = 8 as x is a natural number.
Hence, the required natural number is 8.Question 25
Solution 25
Question 26
Solution 26
Question 27
Solution 27
Question 28
The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers.Solution 28
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_7_ASH_files/20150713181419_image002.gif?w=1200&ssl=1)
Question 29
Solution 29
Question 30
Solution 30
Question 31
The sum of two numbers is 9. The sum of their reciprocals is 1/2. Find the numbers.Solution 31
Question 32
Solution 32
Question 33
Solution 33
Question 34
Find two consecutives odd positive integers, sum of whose squares is 970.Solution 34
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_7_ASH_files/20150713181419_image004.gif?w=1200&ssl=1)
Question 35
The difference of two natural numbers is 3 and the difference of their reciprocal is . Find the numbers.Solution 35
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_7_ASH_files/20150713181419_image008.gif?w=1200&ssl=1)
Question 36
The sum of the squares of two consecutive odd numbers is 394. Find the numbers.Solution 36
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_7_ASH_files/20150713181419_image010.gif?w=1200&ssl=1)
Question 37
The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.Solution 37
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_7_ASH_files/20150713181419_image012.gif?w=1200&ssl=1)
Question 38
The sum of the squares of two consecutive even numbers is 340. Find the numbers.Solution 38
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_7_ASH_files/20150713181419_image014.gif?w=1200&ssl=1)
Question 39
The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is. Find the original fraction.Solution 39
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image020.png?w=1200&ssl=1)
Question 40
Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.Solution 40
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_7_files/20170608161420_image002.png?w=1200&ssl=1)
Chapter 4 – Quadratic Equations Exercise Ex. 4.8
Question 1
Solution 1
Question 2
A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train.Solution 2
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_8_files/20170608161957_image002.png?w=1200&ssl=1)
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/110551_image057.gif?w=1200&ssl=1)
Solution 6
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/110551_image058.gif?w=1200&ssl=1)
Question 7
Solution 7
Question 8
A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed, If it takes 3 hours to complete total journey, what is its original average speed?Solution 8
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_8_files/20170608161957_image004.png?w=1200&ssl=1)
Question 9
Solution 9
Question 10
Solution 10
Concept Insight: Use the relation s =d/t to crack this question and remember here distance is constant so speed and time will vary inversely.Question 11
Solution 11
Question 12
An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.Solution 12
Question 13
While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.Solution 13
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_8_ASH_files/20150713181846_image002.gif?w=1200&ssl=1)
Question 14
A motor boat whose speed in still water is 18 km/hr takes 1 hour more to go 24 km up stream than to return downstream to the same spot. Find the speed of the stream.Solution 14
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_8_ASH_files/20150713181846_image004.gif?w=1200&ssl=1)
Question 15
A car moves a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.Solution 15
Let the speed of a car be x km/hr. According to the question, time is hr. Distance = Speed × Time 2592 =
x = 72 km/hr Hence, the time taken by a car to cover a distance of 2592 km is 36 hrs.Question 16
A motor boat whose speed instill water is 9 km/hr, goes 15 km downstream and comes back to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.Solution 16
Let x km/hr be the speed of the stream.
Therefore, we have
Downstream speed = (9 + x) km/hr
Upstream speed = (9 – x) km/hr
Distance covered downstream = distance covered upstream
Total time taken = 3 hours 45 minutes = hours
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/QuadraticEquations_Addition_files/20210518115340_image014.png?w=1200&ssl=1)
Therefore, x = 3 as the speed can’t be negative.
Hence, the speed of the motor boat is 3 km/hr.
Chapter 4 – Quadratic Equations Exercise Ex. 4.9
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now?Solution 8
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_9_files/20170608162355_image002.png?w=1200&ssl=1)
Question 9
At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.Solution 9
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_9_files/20170608162355_image004.png?w=1200&ssl=1)
Chapter 4 – Quadratic Equations Exercise Ex. 4.10
Question 1
Solution 1
Question 2
Solution 2
![](https://i0.wp.com/images.topperlearning.com/topper/tinymce/imagemanager/files/1238683047a2b953bfd894b5f7c836b0591bef495e2c94.19716768chp810q2.png?w=1200&ssl=1)
Question 3
Solution 3
Question 4
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?Solution 4
Chapter 4 – Quadratic Equations Exercise Ex. 4.11
Question 1
Solution 1
Question 2
Solution 2
Question 3
Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm2. Find the sides of the squares.Solution 3
Question 4
Solution 4
Question 5
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.Solution 5
Question 6
Solution 6
Question 7
Sum of the areas of two squares is 640 m2. If the difference of their perimeter is 64 m, find the sides of the two squares.Solution 7
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_11_ASH_files/20150713182334_image002.gif?w=1200&ssl=1)
Question 8
Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of two squares.Solution 8
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_11_ASH_files/20150713182334_image004.gif?w=1200&ssl=1)
Question 9
The area of a rectangular plot is 528 m2. The length of the plot (in meters) is one meter more than twice its breadth. Find the length and the breadth of the plot.Solution 9
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_Ex8_11_ASH_files/20150713182334_image006.gif?w=1200&ssl=1)
Question 10
In the centre of a rectangular lawn of dimension 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2. Find the length and breadth of the pond.Solution 10
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_11_files/20170608162833_image002.png?w=1200&ssl=1)
Chapter 4 – Quadratic Equations Exercise Ex. 4.12
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?Solution 5
Let us assume that the larger pipe takes ‘x’ hours to fill the pool.
So, as per the question, the smaller pipe takes ‘x + 10’ hours to fill the same pool.
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/RD_X_Math_QuadraticEq_TBS_files/20160701120602_image022.png?w=1200&ssl=1)
Question 6
Two water taps together can fill a tank in hours. The tap with longer diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately. Solution 6
Let the tap with smaller diameter takes x hours to completely fill the tank.
So, the other tap takes (x – 2) hours to fill the tank completely.
Total time taken to fill the tank hours
As per the question, we have
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/QuadraticEquations_Addition_files/20210518115340_image020.png?w=1200&ssl=1)
When x = 5, then (x – 2) = 3
When which can’t be possible as the time becomes negative.
Hence, the smaller diameter tap fills in 5 hours and the larger diameter tap fills in 3 hours.
Chapter 4 – Quadratic Equations Exercise Ex. 4.13
Question 1
Solution 1
Question 2
Solution 2
Question 3
Solution 3
Question 4
Solution 4
Question 5
Solution 5
Question 6
Solution 6
Question 7
Solution 7
Question 8
Solution 8
![](https://i0.wp.com/images.topperlearning.com/topper/bookquestions/14521_RDS_C10_CH8.13_Q8.gif?w=1200&ssl=1)
Question 9
In a class test, the sum of the marks obtained by P in Mathematics and Science is 28. Had he got 3 marks more in Mathematics and 4 marks less in Science. The product of his marks, would have been 180. Find his marks in the two subjects.Solution 9
Question 10
Solution 10
Question 11
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.Solution 11
Question 12
At t minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than minutes. Find t.Solution 12
![](https://i0.wp.com/images.topperlearning.com/topper/question_uploads/CBSE_RDSharma_X_chap08_Ex8_13_files/20170608164116_image004.png?w=1200&ssl=1)
Chapter 4 – Quadratic Equations Exercise 4.82
Question 1
If the equation x2 + 4x + k = 0 has real and distinct root, then
(a) k < 4
(b) k > 4
(c) k ≥ 4
(d) k ≤ 4Solution 1
We know for the quadratic equation
ax2 + bx + c = 0
condition for roots to be real and distinct is
D = b2 – 4ac > 0 ……….(1)
for the given question
x2 + 4x + k = 0
a = 1, b = 4, c = k
from (1)
16 – 4k > 0
k < 4
So, the correct option is (a).
Chapter 4 – Quadratic Equations Exercise 4.83
Question 2
If the equation x2 – ax + 1 = 0 has two distinct roots, then
(a) |a| = 2
(b) |a| < 2
(c) |a| > 2
(d) None of theseSolution 2
For the equation x2 – ax + 1 = 0 has two distinct roots, condition is
(-a)2 – 4 (1) (1) > 0
a2 – 4 > 0
a2 > 4
|a| > 2
So, the correct option is (c).Question 3
![begin mathsize 12px style If space the space equation space 9 x squared space plus space 6 kx space plus space 4 space equals space 0 space has space equal space roots comma space then space the space roots space are space both space equal space to
left parenthesis straight a right parenthesis space plus-or-minus space 2 over 3
left parenthesis straight b right parenthesis space plus-or-minus space 3 over 2
open parentheses straight c close parentheses space 0
open parentheses straight d close parentheses space plus-or-minus space 3 end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=27ea04cf2b95c7e75dd40ab425444022.png)
Solution 3
![begin mathsize 12px style For space quadratic space equation
ax squared space plus space bx space plus space straight c space equals space 0
condition space for space equal space roots space is
straight D space equals space straight b squared space minus space 4 ac space equals space 0 space space space space space space space space space space space.......... open parentheses 1 close parentheses
For space the space given space question
9 straight x squared space plus space 6 kx space plus space 4 space equals space 0
straight a space equals space 9 comma space space space straight b space equals space 6 straight k comma space space space space straight c space equals space 4
from space open parentheses 1 close parentheses
open parentheses 6 straight k close parentheses squared space minus space 4 space open parentheses 9 close parentheses space open parentheses 4 close parentheses space equals space 0
36 straight k squared space minus space 36 space cross times space 4 space equals space 0
36 space open parentheses straight k to the power of 2 space end exponent minus space 4 close parentheses space equals space 0
straight k squared space equals space 4
straight k space equals space plus-or-minus 2
Now comma space equation space is space 9 straight x squared space plus space 6 kx space plus space 4 space equals space 0
rightwards double arrow space 9 straight x squared space plus-or-minus space 12 straight x space plus space 4 space equals 0
rightwards double arrow open parentheses 3 straight x space plus-or-minus space 2 close parentheses squared space equals space 0
rightwards double arrow straight x space equals space plus-or-minus 2 over 3
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=4d8fbaa3cdf1387c90ea58696e9d9b82.png)
Question 4
![begin mathsize 12px style If space ax squared space plus space bx space plus space straight c space equals 0 space has space equal space roots comma space then space straight c space equals
left parenthesis straight a right parenthesis space fraction numerator negative straight b over denominator 2 straight a end fraction
left parenthesis straight b right parenthesis space fraction numerator straight b over denominator 2 straight a end fraction
open parentheses straight c close parentheses space fraction numerator negative straight b squared over denominator 4 straight a end fraction
open parentheses straight d close parentheses space fraction numerator straight b squared over denominator 4 straight a end fraction end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=64b1ab862e3f0b4a35aac088c1f2ecbb.png)
Solution 4
![begin mathsize 12px style If space ax squared space plus space bx space plus space straight c space equals space 0 space has space equal space roots
then
straight b squared space minus space 4 ac space equals space 0
straight b squared space equals space 4 ac
straight c space equals space fraction numerator straight b squared over denominator 4 straight a end fraction
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=ac33c27c52a3582b897dad4f962fa98b.png)
Question 5
If the equation ax2 + 2x + a = 0 has two distinct roots, if
(a) a = ± 1
(b) a = 0
(c) a = 0, 1
(d) a = -1, 0Solution 5
For any quadratic equation
ax2 + bx + c = 0
having two distinct roots, condition is
b2 – 4ac > 0
For the equation ax2 + 2x + a = 0 to have two distinct roots,
(2)2 – 4 (a) (a) > 0
4 – 4a2 > 0
4(1 – a2) > 0
1 – a2 > 0 since 4 > 0
that is, a2 – 1 < 0
Hence -1 < a < 1, only integral solution possible is a = 0
So, the correct option is (b).Question 6
The positive value of k for which the equation x2 + kx + 64 = 0 and x2 – 8x + k = 0 will both have real roots, is
(a) 4
(b) 8
(c) 12
(d) 16Solution 6
For any quadratic equation
ax2 + bx + c = 0
having real roots, condition is
b2 – 4ac ≥ 0 …….(1)
According to question
x2 + kx + 64 = 0 have real root if
k2 – 4 × 64 ≥ 0
k2 ≥ 256
|k|≥ 16 ……..(2)
Also, x2 – 8x + k = 0 has real roots if
64 – 4k ≥ 0
k ≤ 16 ………(3)
from (2), (3) the only positive solution for k is
k = 16
So, the correct option is (d).Question 7
![begin mathsize 12px style The space value space of space square root of 6 space plus square root of 6 space plus square root of 6 space plus space...... end root end root end root is
left parenthesis straight a right parenthesis space 4
left parenthesis straight b right parenthesis space 3
left parenthesis straight c right parenthesis space minus 2
left parenthesis straight d right parenthesis space 3.5 end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=07771db0d7913301d3f6783ca9c5f0f8.png)
Solution 7
![begin mathsize 12px style Let space straight x space equals space square root of 6 space plus square root of 6 space plus square root of 6 space plus space...... end root end root end root space space space space space space space space space space space space space space space space....... left parenthesis 1 right parenthesis
On space squaring space both space sides
straight x squared space equals space 6 space plus space square root of 6 space plus square root of 6 space plus space..... end root end root space space space space space space space space space space space space space space space space space space space space space space space space space space....... left parenthesis 2 right parenthesis
From space left parenthesis 1 right parenthesis space & space left parenthesis 2 right parenthesis
straight x squared space equals space 6 space plus space straight x
straight x squared space minus space straight x space minus space 6 space equals space 0
straight x squared space minus space 3 straight x space plus space 2 straight x space minus space 6 space equals space 0
straight x space open parentheses straight x space minus space 3 close parentheses space plus space 2 left parenthesis straight x space minus space 3 right parenthesis space equals space 0
left parenthesis straight x space plus space 2 right parenthesis space left parenthesis straight x space minus space 3 right parenthesis space equals space 0
straight x space equals space minus 2 space or space straight x space equals space 3
but space straight x space can apostrophe straight t space be space negative comma space hence
box enclose straight x space equals space 3 end enclose
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=ebe9f443c58a3fb7a638341cdccaf737.png)
Question 8
If 2 is a root of the equation x2 + bx + 12 = 0 and the equation x2 + bx + q = 0 has equal roots, then q =
(a) 8
(b) -8
(c) 16
(d) -16Solution 8
It is given that 2 is a root of equation x2 + bx + 12 = 0
Hence
(2)2 + b(2) + 12 = 0
4 + 2b + 12 = 0
2b + 16 = 0
b = -8 ………(1)
It is also given that x2 + bx + q = 0 has equal root
so, b2 – 4(q) = 0 ……..(2)
from (1) & (2)
(-8)2 – 4q =0
q = 16
So, the correct option is (c).Question 9
![begin mathsize 12px style If space the space equation space open parentheses straight a squared space plus space straight b squared close parentheses space straight x squared space minus space 2 open parentheses ac space plus space bd close parentheses space straight x space plus space straight c squared space plus space straight d squared space equals space 0 space has space equal space roots comma space then space
open parentheses straight a close parentheses space ab space equals space cd
open parentheses straight b close parentheses space ad space equals space bc
open parentheses straight c close parentheses space ad space equals space square root of bc
open parentheses straight d close parentheses space ab space equals space square root of cd end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=198ffabbabb4a364e397a3cd540e3e82.png)
Solution 9
For any quadratic equation
ax2 + bx + c = 0
Having equal roots, the condition is
b2 – 4ac = 0
For the equation
(a2 + b2) x2 – 2 (ac + bd)x + c2 + d2 = 0
to have equal roots, we have
(-2(ac + bd))2 – 4 (c2 + d2) (a2 + b2) = 0
4 (ac + bd)2 – 4 (a2c2 + b2c2 + d2a2 + b2d2) = 0
(a2c2 + b2d2 + 2abcd) – (a2c2 + b2c2 + a2d2+ b2d2) = 0
2abcd – b2c2 – a2d2 = 0
b2c2 – a2d2 – 2abcd = 0
(bc – ad)2 = 0
bc = ad
So, the correct option is (b).Question 10
![begin mathsize 12px style If space the space roots space of space the space equation space left parenthesis straight a squared plus space straight b squared right parenthesis straight x squared space minus space 2 straight b left parenthesis straight a space plus space straight c right parenthesis straight x space plus space left parenthesis straight b squared plus space straight c squared right parenthesis space equals space 0 space are space equal comma space then
left parenthesis straight a right parenthesis space 2 straight b space equals space straight a space plus space straight c
open parentheses straight b close parentheses space straight b squared space equals space ac
open parentheses straight c close parentheses space straight b space equals fraction numerator 2 ac over denominator straight a space plus space straight c end fraction
open parentheses straight d close parentheses space straight b space equals space ac end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=1e20046aa644565ec3427ca3569a2fb1.png)
Solution 10
For any quadratic equation
ax2 + bx + c = 0
having equal roots, condition is
b2 – 4ac = 0
According to question, quadratic equation is
(a2 + b2)x2 – 2b(a + c)x + b2 + c2 = 0
having equal roots, so
(2b(a + c))2 – 4(a2 + b2) (b2 + c2) = 0
4b2 (a + c)2 – 4(a2b2 + a2c2 + b4 + b2c2) = 0
b2(a2 + c2 + 2ac) – (a2b2 + a2c2 + b2c2 + b4) = 0
a2b2 + b2c2 + 2acb2 – a2b2 – a2c2 – b2c2 – b4 = 0
2acb2 – a2c2 – b4 = 0
a2c2 + b4 – 2acb2 = 0
(ac – b2)2 = 0
ac = b2
So, the correct option is (b).Question 11
If the equation x2 – bx + 1 = 0 does not possess real roots, then
(a) -3 < b < 3
(b) -2 < b < 2
(c) b < 2
(d) b < -2Solution 11
For any quadratic equation ax2 + bx + c = 0 having no real roots, condition is
b2 – 4ac < 0
For the equation, x2 – bx + 1 = 0 having no real roots
b2 – 4 < 0
b2 < 4
-2 < b < 2
So, the correct option is (b).Question 12
If x = 1 is a common root of the equations ax2 + ax + 3 = 0 and x2 + x + b = 0, then ab =
(a) 3
(b) 3.5
(c) 6
(d) -3Solution 12
![begin mathsize 12px style If space straight x space equals space 1 space is space root space of space ax squared space plus space ax space plus space 3 space equals space 0
then
straight a open parentheses 1 close parentheses squared space plus space straight a open parentheses 1 close parentheses space plus space 3 space equals space 0
2 straight a space plus space 3 space equals space 0
straight a space equals space fraction numerator negative 3 over denominator 2 end fraction space space space space space space space space space space space space...... left parenthesis 1 right parenthesis
and space straight x space equals space 1 space is space also space root space of space straight x squared space plus space straight x space plus space straight b space equals 0
then
open parentheses 1 close parentheses squared space plus space open parentheses 1 close parentheses space plus space straight b space equals space 0
straight b space equals space minus 2 space space space space space space space space space.... left parenthesis 2 right parenthesis
from space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses
ab space equals space open parentheses fraction numerator negative 3 over denominator 2 end fraction close parentheses open parentheses negative 2 close parentheses
ab space equals 3
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=dc459734525da20cdc61c95fb294d1f6.png)
Question 13
If p and q are the roots of the equation x2 -px + q = 0, then
(a) p = 1, q = -2
(b) b = 0, q = 1
(c) p = -2, q = 0
(d) p = -2, q = 1Solution 13
If p, q are the roots of equation x2 – px + q = 0, then p and q satisfies the equation
Hence
(p)2 – p(p) + q = 0
p2 – p2 + q = 0
q = 0
and (q)2 – p(q) + q = 0
q2 – p(q) + q = 0
0 = 0
p can take any value
p = -2 and q = 0
So, the correct option is (c).Question 14
If a and b can take values 1, 2, 3, 4. Then the number of the equations of the form ax2 + bx + 1 = 0 having real roots is
(a) 10
(b) 7
(c) 6
(d) 12Solution 14
For the ax2 + bx + 1 = 0 having real roots condition is
b2 – 4(a) (1) ≥ 0
b2 ≥ 4a
For a = 1
b2 ≥ 4
b ≥ 2
b can take value 2, 3, 4
Here, 3 possible solutions are possible …….(1)
For a = 2
b2 ≥ 8
Here, b can take value 3, 4 ……(2)
Here, 2 solutions are possible
For a = 3
b2 ≥ 12
possible value of b is 4
Hence, only 1 possible solution ……(3)
For a = 4
b2 ≥ 16
possible value of b is 4
Hence, only 1 possible solution …….(4)
from (1), (2), (3), (4)
Total possible solutions are 7
So, the correct option is (b).Question 15
The number of quadratic equations having real roots and which do not change by squaring their roots is
(a) 4
(b) 3
(c) 2
(d) 1Solution 15
Any quadratic equation having roots 0 or 1 are only possible quadratic equation because on squaring 0 or 1, it remains same.
Hence, 2 solutions are possible, one having roots 1 and 1, while the other having roots 0 and 1.
So, the correct option is (c).
Chapter 4 – Quadratic Equations Exercise 4.84
Question 16
If (a2 + b2) x2 + 2(ab + bd) x + c2 + d2 = 0 has no real roots, then
(a) ad = bc
(b) ab = cd
(c) ac = bd
(d) ad ≠ bcSolution 16
If any quadratic equation ax2 + bx + c has no real roots then b2 – 4ac < 0 ……(1)
According to the question, the equation is
(a2 + b2) x2 + 2(ac + bd) x + c2 + d2 = 0
from (1)
4(ac + bd)2 – 4(a2 + b2) (c2 + d2) < 0
a2c2 + b2d2 + 2abcd – (a2c2 + a2d2 + b2c2 + b2d2) < 0
a2c2 + b2d2 + 2abcd – a2c2 – a2d2 – b2c2 – b2d2 < 0
2abcd – a2d2 – b2c2 < 0
-(ad – bc)2 < 0
(ad – bc)2 > 0
For this condition to be true ad ≠ bc
So, the correct option is (d).Question 17
![begin mathsize 12px style If space the space sum space of space the space roots space of space the space equation space straight x squared space minus space straight x space equals space straight lambda space open parentheses 2 straight x space minus space 1 close parentheses space is space zero comma space then space straight lambda space equals
open parentheses straight a close parentheses space minus 2
open parentheses straight b close parentheses space 2
open parentheses straight c close parentheses space fraction numerator negative 1 over denominator 2 end fraction
open parentheses straight d close parentheses space 1 half end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=2462ce741e0e492ae10b920c02a113be.png)
Solution 17
![begin mathsize 12px style We space know space for space any space quadratic space equation
ax squared space plus space bx space plus space straight c space equals space 0
sum space of space roots space equals fraction numerator negative straight b over denominator straight a end fraction
Now comma space according space to space the space question
straight x squared space minus space straight x space equals space straight lambda space open parentheses 2 straight x space minus space 1 close parentheses
straight x squared space minus space straight x space minus space 2 λx space plus space straight lambda space equals space 0
straight x squared space minus space straight x space open parentheses 1 space plus space 2 straight lambda close parentheses space plus space straight lambda space equals space 0
given space sum space of space roots space of space this space equation space is space zero
Hence comma space fraction numerator negative open parentheses negative open parentheses 1 space plus space 2 straight lambda close parentheses close parentheses over denominator 1 end fraction space equals space 0
space space space space space space space space space space space space space space space space space space 1 space plus space straight lambda 2 space equals space 0
space space space space space space space space space space space space space space space space space space space straight lambda space equals space fraction numerator negative 1 over denominator 2 end fraction
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=3c3aae16c7cf9c1a074cba0e6266d3a7.png)
Question 18
If x = 1 is a common root of ax2 + ax + 2 = 0 and x2 + x + b = 0 then, ab =
(a) 1
(b) 2
(c) 4
(d) 3Solution 18
It is given that x = 1 is root of equation ax2 + ax + 2 = 0
Hence, a(1)2 + a(1) + 2 = 0
2a + 2 = 0
a = -1 ……(1)
It is given that x = 1 is also root of x2 + x + b = 0
Hence, (1)2 + (1) + b = 0
b = -2 ……(2)
from (1) & (2)
ab = (-1) (-2)
ab = 2
So, the correct option is (b).Question 19
![begin mathsize 12px style The space value space of space straight c space for space which space the space equation space ax squared plus space 2 bx space plus space straight c space equals space 0 space has space equal space roots space is
left parenthesis straight a right parenthesis space straight b squared over straight a
left parenthesis straight b right parenthesis space fraction numerator straight b squared over denominator 4 straight a end fraction
open parentheses straight c close parentheses space straight a squared over straight b
open parentheses straight d close parentheses space fraction numerator straight a squared over denominator 4 straight b end fraction end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=6f9b209b664a39fe2c7c27b264a651e1.png)
Solution 19
![begin mathsize 12px style For space any space quadratic space equation space ax squared space plus space bx space plus space straight c space equals space 0 space having space equal space roots space condition space is
straight b squared space minus space 4 ac space equals space 0 space space space space space space space space space..... open parentheses 1 close parentheses
According space to space the space question comma space the space quadratic space equation space is
ax squared space plus space 2 bx space plus space straight c space equals space 0
from space left parenthesis 1 right parenthesis
open parentheses 2 straight b close parentheses squared space minus space 4 ac space equals space 0
4 straight b squared space minus space 4 ac space equals space 0
straight b squared space equals space ac
straight c space equals space straight b squared over straight a
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=d10766d71087d62cbf1081504297a0cb.png)
Question 20
![begin mathsize 12px style If space straight x squared space plus space straight k left parenthesis 4 straight x space plus space straight k space minus space 1 right parenthesis space plus space 2 space equals space 0 space has space equal space roots comma space then space straight k space equals
left parenthesis straight a right parenthesis space fraction numerator negative 2 over denominator 3 end fraction comma space 1
open parentheses straight b close parentheses space 2 over 3 comma space minus 1
open parentheses straight c close parentheses space 3 over 2 comma space 1 third
open parentheses straight d close parentheses space fraction numerator negative 3 over denominator 2 end fraction comma space fraction numerator negative 1 over denominator 3 end fraction end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=ebc3d2cd06efc24daebfc204c93b1f45.png)
Solution 20
![begin mathsize 12px style For space equation comma space straight x squared space plus space straight k open parentheses 4 straight x space plus space straight k space minus space 1 close parentheses space plus space 2 space equals space 0
space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight x squared space plus space 4 kx space plus space straight k open parentheses straight k space minus space 1 close parentheses space plus space 2 space equals space 0
Condition space for space equal space roots space is comma
open parentheses 4 straight k close parentheses squared space minus space 4 space open parentheses straight k open parentheses straight k space minus space 1 close parentheses space plus space 2 close parentheses space equals space 0
16 straight k squared space minus space 4 open parentheses straight k squared space minus space straight k space plus space 2 close parentheses space equals space 0
4 straight k squared space minus space open parentheses straight k squared space minus space straight k space plus space 2 close parentheses space equals space 0
4 straight k squared space minus space straight k squared space plus space straight k space minus space 2 space equals space 0
3 straight k squared space plus space straight k space minus 2 space equals space 0
3 straight k squared space plus space 3 straight k space minus space 2 straight k space minus space 2 space equals space 0
3 straight k space open parentheses straight k space plus space 1 close parentheses space minus space 2 open parentheses straight k space plus space 1 close parentheses space equals space 0
open parentheses 3 straight k space minus space 2 close parentheses space open parentheses straight k space plus space 1 close parentheses space equals space 0
straight k space equals space minus 1 space or space straight k space equals space 2 over 3
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=2c2363c25a46f4f0d4c0250c7de80fe0.png)
Question 21
![begin mathsize 12px style If space the space sum space and space product space of space the space roots space of space the space equation space kx squared space plus space 6 straight x space plus 4 straight k space equals space 0 space are space equal comma space then space straight k space equals
open parentheses straight a close parentheses space fraction numerator negative 3 over denominator 2 end fraction
open parentheses straight b close parentheses space 3 over 2
open parentheses straight c close parentheses space 2 over 3
open parentheses straight d close parentheses space fraction numerator negative 2 over denominator 3 end fraction end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=85d5451a39ccb4ddcf02d0021bd24948.png)
Solution 21
![begin mathsize 12px style For space any space quadratic space equation space ax squared space plus space bx space plus space straight c space equals space 0
sum space of space roots space equals space fraction numerator negative straight b over denominator straight a end fraction space and space product space of space roots space equals space straight c over straight a
For space equation comma space kx squared space plus space 6 straight x space plus space 4 straight k space equals space 0 comma space given space sum space and space product space of space roots space of space this space equation space are space equal comma
Hence comma
fraction numerator negative open parentheses 6 close parentheses over denominator straight k end fraction space equals space fraction numerator 4 straight k over denominator straight k end fraction
4 straight k space equals space minus 6
straight k space equals space fraction numerator negative 3 over denominator 2 end fraction
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=cb87f9f7e724c2944af777cc5a68042d.png)
Question 22
If sin α and cos α are the roots of the equation ax2 + bx + c = 0, then b2 =
(a) a2 – 2ac
(b) a2 + 2ac
(c) a2 – ac
(d) a2 + acSolution 22
![begin mathsize 12px style ax squared space plus space bx space plus space straight c space equals space 0
sin space straight alpha space plus space cos space straight alpha space equals space fraction numerator negative straight b over denominator straight a end fraction space space..... open parentheses 1 close parentheses space space space space and
sin space straight alpha space cosα space equals space straight c over straight a space space space space space space space...... open parentheses 2 close parentheses
On space squaring space we space get comma
open parentheses sin space straight alpha space plus space cos space straight alpha close parentheses squared space equals space straight b squared over straight a squared
sin squared straight alpha space plus space cos squared straight alpha space plus space 2 space sinα space cosα space equals space straight b squared over straight a squared space space space space space space space space space space space space.... open parentheses 3 close parentheses
We space know space sin squared space straight alpha space plus space cos squared space straight alpha space equals space 1 space space space space space space space..... open parentheses 4 close parentheses
from space open parentheses 2 close parentheses comma space open parentheses 3 close parentheses comma space open parentheses 4 close parentheses
1 space plus space 2 open parentheses straight c over straight a close parentheses space equals space straight b squared over straight a squared
straight b squared over straight a squared space equals space fraction numerator straight a space plus space 2 straight c over denominator straight a end fraction
straight b squared space equals straight a to the power of 2 space end exponent plus space 2 ac
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=69a2acd03b16ec86160a0cbf10f743f8.png)
Question 23
If 2 is a root of the equation x2 + ax + 12 = 0 and the quadratic equation x2 + ax + q = 0 has equal roots, then q =
(a) 12
(b) 8
(c) 20
(d) 16Solution 23
Given, 2 is a root of equation x2 + ax + 12 = 0
so (2)2 + a(2) + 12 = 0
4 + 2a + 12 = 0
a = -8 …..(1)
Given x2 + ax + q = 0 has equal roots so
a2 – 4q = 0 ……(2)
from (1) & (2)
(-8)2 – 4q = 0
4q = 64
q = 16
So, the correct option is (d).Question 24
If the sum of the roots of the equation x2 – (k + 6)x + 2(2k – 1) = 0 is equal to half of their product, then k =
(a) 6
(b) 7
(c) 1
(d) 5Solution 24
![begin mathsize 12px style For space any space quadratic space equation space straight x squared space minus space left parenthesis straight k space plus space 6 right parenthesis straight x space plus space 2 left parenthesis 2 straight k space minus space 1 right parenthesis space equals space 0
open parentheses sum space of space roots close parentheses space equals space fraction numerator negative open curly brackets negative open parentheses straight k space plus space 6 close parentheses close curly brackets over denominator 1 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals straight k space plus space 6 space space space space space space space......... open parentheses 1 close parentheses
open parentheses product space of space roots close parentheses space equals space fraction numerator 2 open parentheses 2 straight k space minus space 1 close parentheses over denominator 1 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 open parentheses 2 straight k space minus space 1 close parentheses space space space space space space space space space space space space space space...... open parentheses 2 close parentheses
Given space that space the space sum space of space roots space of space equation space equals space 1 half open parentheses product space of space roots close parentheses
from space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses
straight k space plus space 6 space equals space 1 half open parentheses 2 open parentheses 2 straight k space minus space 1 close parentheses close parentheses
straight k space plus space 6 space equals space 2 straight k space minus space 1
box enclose 7 space equals space straight k end enclose
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=bc343afa2d8470da784abd74bc468b85.png)
Question 25
If a and b are roots of the equation x2 + ax + b = 0, then a + b =
(a) 1
(b) 2
(c) -2
(d) -1Solution 25
If a and b are roots of the equation x2 + ax + b = 0
Then, sum of roots = -a
a + b = -a
2a + b = 0 …….(1)
product of roots = b
ab = b
ab – b = 0
b(a – 1) = 0 ……..(2)
from (1) and (2)
-2a(a – 1) = 0…(From (1), we have b = -2a)
a(a – 1) = 0
a = 0 or a = 1
if a = 0 b = 0
if a = 1 b = -2
Now a and b can’t be zero at same time, so correct solution is
a = 1 and b = -2
a + b = -1
So, the correct option is (d).Question 26
A quadratic equation whose one root is 2 and the sum of whose roots is zero, is
(a) x2 + 4 = 0
(b) x2 – 4 = 0
(c) 4x2 – 1 = 0
(d) x2 – 2 = 0Solution 26
Given sum of roots is zero and one root is 2.
So the other root must be -2
so any quadratic equation having root 2 and -2 is
(x – 2) (x – (-2)) = 0
(x – 2) (x + 2) = 0
x2 – 4 = 0
So, the correct option is (b).Question 27
If one root of the equation ax2 + bx + c = 0 is three times the other, then b2 : ac =
(a) 3 : 1
(b) 3 : 16
(c) 16 : 3
(d) 16 : 1Solution 27
![begin mathsize 12px style If space straight alpha comma space straight beta space are space roots space of space equation space ax squared space plus space bx space plus space straight c space equals space 0
Then comma space straight alpha space plus space straight beta space equals space fraction numerator negative straight b over denominator straight a end fraction space space space space space space space space space...... open parentheses 1 close parentheses
and space αβ space equals space straight c over straight a space space space space space space...... open parentheses 2 close parentheses
It space is space given space that space one space root space is space three space times space the space other space one
Let space straight alpha space equals space 3 straight beta space space space space space space space space space....... open parentheses 3 close parentheses
from space open parentheses 1 close parentheses space & space open parentheses 3 close parentheses
4 straight beta space equals space fraction numerator negative straight b over denominator straight a end fraction
straight beta space equals space fraction numerator negative straight b over denominator 4 straight a end fraction space space space space space space space space space space space...... open parentheses 4 close parentheses
from space open parentheses 2 close parentheses space & space open parentheses 3 close parentheses
3 straight beta squared space equals space straight c over straight a space space space space space space space space space.......... open parentheses 5 close parentheses
from space open parentheses 4 close parentheses space & space open parentheses 5 close parentheses
3 open parentheses fraction numerator negative straight b over denominator 4 straight a end fraction close parentheses squared space equals space straight c over straight a
fraction numerator 3 straight b squared over denominator 16 straight a squared end fraction equals straight c over straight a
straight b squared over ac equals space 16 over 3
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=8ed990553d0597b1948327989812ee58.png)
Question 28
If one root of the equation 2x2 + kx + 4 = 0 is 2, then the other root is
(a) 6
(b) -6
(c) -1
(d) 1Solution 28
![begin mathsize 12px style We space know space that comma space for space straight a space quadratic space equation space ax squared space plus space bx space plus space straight c space equals space 0
product space of space roots space equals space straight c over straight a space space space space space space space space space space space space space........ open parentheses 1 close parentheses
According space to space the space question comma space equation space is space 2 straight x squared space plus space kx space plus space 4 space equals space 0
from space open parentheses 1 close parentheses comma
product space of space roots space equals 4 over 2
αβ space equals space 2 space space space space space space space space space space....... open parentheses 2 close parentheses
given space one space root space is space 2. space Hence space from space open parentheses 2 close parentheses space other space root space is space 1.
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=8e8b51480e55480a3ae2022388df71a5.png)
Question 29
If one root of the equation x2 + ax + 3 = 0 is 1, then its other root is
(a) 3
(b) -3
(c) 2
(d) -2Solution 29
x2 + ax + 3 = 0
product of roots = 3
One root is 1. Hence other root is 3.
So, the correct option is (a).
Chapter 4 – Quadratic Equations Exercise 4.85
Question 30
If one root of the equation 4x2 – 2x + (λ – 4) = 0 is a reciprocal of the other, then λ =
(a) 8
(b) -8
(c) 4
(d) -4Solution 30
![begin mathsize 12px style Let space apostrophe straight a apostrophe space be space straight a space space root space of space given space equation.
Then space according space to space the space question comma space other space root space is space 1 over straight a.
4 straight x squared space minus space 2 straight x space plus space open parentheses straight lambda space minus space 4 close parentheses space equals space 0
product space of space roots space equals space fraction numerator straight lambda space minus space 4 over denominator 4 end fraction
straight a space cross times space 1 over straight a space equals space fraction numerator straight lambda space minus space 4 over denominator 4 end fraction
1 space equals fraction numerator space straight lambda space minus space 4 over denominator 4 end fraction
straight lambda space minus space 4 space equals space 4
straight lambda space equals space 8
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=8b7bc95784caada1253909b1d1299d9e.png)
Question 31
![begin mathsize 12px style If space straight y space equals space 1 space is space straight a space common space root space of space the space equations space ay squared plus space ay space plus space 3 space equals space 0 space and space straight y squared space plus space straight y space plus straight b space equals space 0 comma space then space ab space equals
open parentheses straight a close parentheses space 3
open parentheses straight b close parentheses space fraction numerator negative 7 over denominator 2 end fraction
open parentheses straight c close parentheses space 6
open parentheses straight d close parentheses space minus 3 end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=e7559f9901d0763c25d09cfc44d0e986.png)
Solution 31
![begin mathsize 12px style If space straight y space equals space 1 space is space straight a space root space of space equation space ay squared space plus space ay space plus space 3 space equals space 0
Then comma
straight a open parentheses 1 close parentheses squared space plus space straight a open parentheses 1 close parentheses space plus space 3 space equals space 0
2 straight a space plus space 3 space equals space 0
straight a space equals space fraction numerator negative 3 over denominator 2 end fraction space space space space space space space space space space space space space........ open parentheses 1 close parentheses
Also space straight y space equals space 1 space is space straight a space root space of space equation space straight y squared space plus space straight y space plus space straight b space equals space 0
Then comma
open parentheses 1 close parentheses squared space plus space 1 space plus space straight b space equals space 0
box enclose straight b space equals space minus 2 end enclose space space space space space space space space space space space space space space space space...... open parentheses 2 close parentheses
from space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses
ab space equals space open parentheses fraction numerator negative 3 over denominator 2 end fraction close parentheses open parentheses negative 2 close parentheses
space space space space space space equals space 3
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=a2b43b604db10eb7a2b498468e5da037.png)
Question 32
![begin mathsize 12px style The space values space of space straight k space for space which space the space quadratic space equation space 16 straight x squared space plus space 4 kx space plus space 9 space equals space 0 space has space real space and space equal space root space are
open parentheses straight a close parentheses space 6 comma space fraction numerator negative 1 over denominator 6 end fraction
open parentheses straight b close parentheses space 36 comma space minus 36
left parenthesis straight c right parenthesis space 6 comma space minus 6
open parentheses straight d close parentheses space 3 over 4 comma space fraction numerator negative 3 over denominator 4 end fraction end style](https://images.topperlearning.com/topper/tinymce/integration/showimage.php?formula=65929c552ff8a952a1a186e8661893ac.png)
Solution 32
Any quadratic equation, ax2 + bx + c = 0 has real and equal roots if b2 – 4ac = 0
For the question, equation is 16x2 + 4kx + 9 = 0,
(4k)2 – 4 × 16 × 9 = 0
16k2 – 36 × 16 = 0
k2 – 36 = 0
k2 = 36
k = ± 6
So, the correct option is (c).
Discover more from EduGrown School
Subscribe to get the latest posts sent to your email.