RD SHARMA SOLUTION CHAPTER- 4 Triangles| CLASS 10TH MATHEMATICS-EDUGROWN

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)

Solution 1(vi)

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

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

Solution 9

Question 10

Solution 10

Question 11

Solve the following quadratic equation by factorisation:

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solve the following quadratic equation by factorisation:

Solution 15

Question 16

Solution 16

Question 17

9x² – 6b²x – (a⁴ – b⁴) = 0Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solve the following quadratic equation by factorisation:

2x² + ax – a² = 0Solution 20

Question 21

Solve the following quadratic equation by factorisation:

Solution 21

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:

Solution 29

Question 30

Solve the following quadratic equation by factorisation:

Solution 30

Question 31

Solve the following quadratic equation by factorisation:

Solution 31

Question 32

Solve the following quadratic equation by factorisation:

Solution 32

Question 33

Solve the following quadratic equation by factorisation:

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solve the following quadratic equation by factorisation:

Solution 38

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:

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

Solution 46

Question 47

Solve the following quadratic equations by factorization:

Solution 47

Question 48

Solve the following quadratic equations by factorization:

Solution 48

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:

Solution 58

Question 59

Solve the following quadratic equation by factorisation:

Solution 59

Question 60

Solve the following quadratic equation by factorisation:

Solution 60

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 

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)

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)

3x² – 5x + 2 = 0Solution 2(xii)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solve for x:

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solve for x:

Solution 3(iv)

Question 3(v)

Solve for x:

Solution 3(v)

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:

Solution 1 (vi)

Given quadratic equation is 

Here, 

Therefore, we have

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)

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:

Solution 5(iv)

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)

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)

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

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

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

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

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

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

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

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

Question 35

The difference of two natural numbers is 3 and the difference of their reciprocal is . Find the numbers.Solution 35

Question 36

The sum of the squares of two consecutive odd numbers is 394. Find the numbers.Solution 36

Question 37

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.Solution 37

Question 38

The sum of the squares of two consecutive even numbers is 340. Find the numbers.Solution 38

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

Question 40

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.Solution 40

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

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

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

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

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

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

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

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

Chapter 4 – Quadratic Equations Exercise Ex. 4.10

Question 1

Solution 1

Question 2

Solution 2

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

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

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

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

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.

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

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

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

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

Solution 3

Question 4

Solution 4

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

Solution 7

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

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

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

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

Solution 17

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

Solution 19

Question 20

Solution 20

Question 21

Solution 21

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

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

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

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

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

Question 31

Solution 31

Question 32

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).