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Chapter 4 Quadratic Equations Important Questions for CBSE Class 10 Maths Board Exams

This page contains Important Questions for Class 10 Maths Chapter 4 Quadratic Equations, which will help the students to prepare for the CBSE Class 10 maths Board exam 2022-23. It help the doing better in their maths paper. Extra questions of Chapter 4 Quadratic Equations in Two Variables given here which are based on the pattern of CBSE NCERT book. Students will learn about the entire syllabus and learn how to solve problems in preparation for the exams.
Chapter 4 Quadratic Equations Important Questions for CBSE Class 10 Maths Board Exams

Important Questions for Chapter 4 Quadratic Equations Class 10 Maths


Important Questions Class 10 Maths contains all types of questions which could be asked in the examination. We have included short answer questions as well as long answer questions for Chapter 4 Quadratic Equations. For more exercise, students can also refer to the Quadratic Equations NCERT Solutions. Students can solve these questions and check their answers on the website. All the questions are solved in detail.

Quadratic Equations Class 10 Maths Important Questions Very Short Answer (1 Mark)

1. Find the positive root of √3x2 + 6 = 9.
Solution
Hence 5 is positive root.

2. If x = -1/2 is a solution of the quadratic equation 3x2 + 2kx -3 = 0, find the value of k.
Solution

3. Find the value of k, for which one root of the quadratic equation kx2 -14x + 8 = 0 is six times the other.

Solution

Let one root be α and other root be 6α.

Since k = 0 is not possible, therefore k = 3.


4. If 1 is a root of the equations ay2 + ay + 3 = 0 and y2 + y + b = 0, then find the value of ab.

Solution

ay2 + ay + 3 = 0
⇒ a(1)2 + a(1) + 3 = 0
⇒ 2a = -3
⇒ a = −32

y2 + y + b = 0
⇒ 12 + 1 + b = 0
⇒ b = -2
∴ ab = (−32)(−2) = 3


5. Find the roots of the equation x2 – 3x – m (m + 3) = 0, where m is a constant.

Solution

x2 – 3x – m(m + 3) = 0
⇒ D = b2 – 4ac
⇒ D = (- 3)2 – 4(1) [-m(m + 3)]
= 9 + 4m (m + 3)
= 4m2 + 12m + 9 = (2m + 3)2

∴ x = m + 3 or -m


Quadratic Equations Class 10 Maths Important Questions Short Answer (2 Marks)

6. For what values of k, the roots of the equation x2 + 4x + k = 0 are real?

Solution

Comparing the given equation with ax2 + bx + c = 0, we get a = 1, b = 4, c = k.

Since, given the equation has real roots,

D ≥ 0

⇒ b2 - 4ac ≥ 0

⇒ 42 - 4 × 1 × k ≥ 0

⇒ 4k ≤ 16

⇒ k ≤ 4


7. Find the value of m so that the quadratic equation mx (x – 7) + 49 = 0 has two equal roots.

Solution

We have, mx (x – 7) + 49 = 0
mx2 – 7mx + 49 = 0
Here, a = m, b = – 7m, c = 49
D = b2 – 4ac = 0 …[For equal roots]
⇒ (-7m)2 – 4(m) (49) = 0
⇒ 49m2 – 4m (49) = 0
⇒ 49m (m – 4) = 0
⇒ 49m = 0 or m – 4 = 0
m = 0 (rejected) or m = 4
∴ m = 4


8. Find the value of k for which the roots of the equations 3x2 - 10x + k = 0 are reciprocal of each other.

Solution

Comparing the given equation with ax2 + bx + c = 0 we get a = 3, b = -10, c = k

Let one root be α so other root is 1/α

Now,

product of roots, α × 1/α = c/a

Hence, the value of k is 3.


9. Find the roots of the following quadratic equation :

15x2 - 10√6x + 10 = 0

Solution

We have 15x2 - 10√6x + 10 = 0


10. Find the value of p so that the quadratic equation px(x – 3) + 9 = 0 has two equal roots.

Solution

We have, px (x – 3) + 9 = 0
px2 – 3px + 9 = 0 Here a = p, b = -3p,
D = 0
b2 – 4ac = 0
⇒ (-3p)2 – 4(p)(9) = 0
⇒ 9p2 – 36p = 0
⇒ 9p (p – 4) = 0
⇒ 9p = 0 or p – 4= 0
p = 0 (rejected) or p = 4
∴ p = 4 …(∵ Coefficient of x2 cannot be zero)


11. Solve for x:

36x2 – 12ax + (a2 – b2) = 0

Solution

We have, 36x2 – 12ax + (a2 – b2) = 0
⇒ (36x2 – 12ax + a2) – b2 = 0
⇒ [(6x)2 – 2(6x)(a) + (a)2] – b2 = 0
⇒ (6x – a)2 – (b)2 = 0 …[∵ x2 – 2xy + y2 = (x – y)2]
⇒ (6x – a + b) (6x – a – b) = 0 ...[∵ x2 – y2 = (x + y)(x – y)]
⇒ 6x – a + b = 0 or 6x – a – b = 0
⇒ 6x = a – b or 6x = a + b
⇒ x = a− b/6 or a+ b/6


12. Solve for x:

4√3x2 + 5x - 2√3 = 0

Solution

We have, 4√3x2 + 5x - 2√3 = 0

13. Find the roots of 4x2 + 3x + 5 = 0 by the method of completing the squares.

Solution

Here 4x2 + 3x + 5 = 0

But (2x+34)2 cannot be negative for any real value of x.


14. Solve for x : x2 - (√3+1)x + √3 = 0

Solution

x2 - (√3+1)x + √3 = 0

⇒ x2 -√3 x - 1x + √3 = 0

⇒ x(x-√3) - 1(x-√3) = 0

⇒ (x-√3) (x-1) = 0

Thus, x = √3, x = 1


15. Find the roots of the following quadratic equation :

(x+3)(x-1) = 3(x - 1/3)

Solution


Thus x = 2, -1


16. Solve the following quadratic equation for x :

9x2 - 6b2x - (a4 - b4) = 0

Solution

We have,

9x2 - 6b2x - (a4 - b4) = 0

Comparing with Ax2 + Bx + C

A = 9, B = -6b2, C = -(a4 - b4)


17. Solve for x (in terms of a and b) :
a/x-b + b/x-a = 2, x ≠ a,b
Solution

18. If x = 23 and x = -3 are roots of the quadratic equation ax2 + 7x + b = 0, find the values of a and b.

Solution

We have, ax2 + 7x + b = 0
Here ‘a’ = a, ‘b’ = 7, ‘c’ = b
Now, α = 23 and β = -3 …[Given]


19. Solve the following quadratic equation for x: 9x2 – 6b2x – (a4 – b4) = 0

Solution

The given quadratic equation can be written as
(9x2 – 6b2x + b4) – a4 = 0
⇒ (3x – b2)2 – (a2)2 = 0
⇒ (3x – b2 + a2) (3x – b2 – a2) = 0 …[x2 – y2 = (x + y) (x – y)]
⇒ 3x – b2 + a2 = 0 or 3x – b2 – a2 = 0
⇒ 3x = b2 – a2 or 3x = b2 + a2


20. 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

We have, 2x2 + px – 15 =0
Since (-5) is a root of the given equation
∴ 2(-5)2 + p(-5) – 15 = 0
⇒ 2(25) – 5p – 15 = 0
⇒ 50 – 15 = 5p
⇒ 35 = 5p
⇒ p = 7 …(i)

Now, p(x2 + x) + k
⇒ px2 + px + k = 0
7x2 + 7x + k = 0 …[From (i)]
Here, a = 7, b = 7, c = k

D = 0 …[Roots are equal]
⇒ b2 – 4ac = 0
⇒ (7)2 – 4(7)k = 0 ⇒ 49 – 28k = 0
⇒ 49 = 28k
∴ k = 49/28 = 7/4


21. Find the nature of the roots of the quadratic equation:

13√3 x2 + 10x + √3 = 0

Solution

Given,

13√3 x2 + 10x + √3 = 0

Comparing with ax2 + bx + c = 0, we get

a = 13√3, b = 10, c = √3

b2 - 4ac = 102 - 4(13√3)(√3)

=100 - 156

= -56

As D < 0, the equation has no real roots.


22. Find the positive value of k for which x2 - 8x + k = 0, will have real roots.

Solution

We have,

x2 - 8x + k = 0

Comparing with Ax2 + Bx + C = 0, we get

A = 1, B = -8, C= k

Since the given equation has real roots,

B2 - 4AC > 0

⇒ (-8)2 - 4(1)(k) ≥ 0

⇒ 64 - 4k ≥ 0

⇒ 16 - k ≥ 0

⇒ 16 ≥ k

Thus, k ≤ 16


23. If 2 is a root of the equation x2 + kx + 12 = 0 and the equation x2 + kx + q = 0 has equal roots, find the value of q.

Solution

We have,

x2 + kx + 12 = 0

If 2 is the root of above equation, it must satisfy it.

22 + 2k + 12 = 0

⇒ 2k + 16 = 0

⇒ k = -8

Substituting k = -8 in x2 + kx + q = 0 we have

x2 - 8x + q = 0

For equal roots,

(-8)2 - 4(1)q = 0

⇒ 64 - 4q = 0

⇒ 4q = 64

⇒ q = 16


24. Solve for x : √3 x2 + 10x + 7√3 = 0

Solution


25. Find the value of k for which the roots of the quadratic equation 2x2 + kx + 8 = 0 will have the equal roots?

Solution

We have,

2x2 + kx + 8 = 0

Comparing with ax2 + bx + c = 0, we get

a = 2, b = k and c = 8

For equal roots, D = 0,

b2 - 4ac = 0

k2 - 4×2×8 = 0

⇒ k2 = 64

⇒ k = ±√64

Thus k = ±8


Quadratic Equations Class 10 Maths Important Questions Short Answer-II (3 Marks)

26. Find the values of k for which the quadratic equation x2 + 2√2 kx + 18 = 0 has equal roots.

Solution

We have,

x2 + 2√2 kx + 18 = 0

Comparing it by ax2 + bx + c, we get a = 1, b = 2√2 k and c = 18.

Given that,

Equation x2 + 2√2 kx + 18 = 0 has equal roots.

b2 - 4ac = 0

⇒ (2√2 k)2 -4118 = 0

⇒ 8k2 - 72 = 0

⇒ 8k2 = 72

⇒ k2 = 72/8 = 9

⇒ k = ±3


27. If α and β are the zeroes of the polynomial f(x) = x2 - 4x - 5 then find the value of α2 + β2.

Solution

We have,

p(x) = x2 - 4x - 5

Comparing it by ax2 + bx + c, we get a = 1, b = -4 and c = -5.

Since, given α and β are the zeroes of the polynomial,


28. Find the quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. Hence find the zeroes.

Solution

Sum of zeroes,

α + β = -3 ...(1)

and product of zeroes, αβ = 2

Thus quadratic equation is

x2 - (α+β)x + αβ = 0

⇒ x2 - (-3)x + 2 = 0

⇒ x2 + 3x + 2 = 0

Thus quadratic equation is x2 + 3x + 2 = 0

Now, above equation can be written as

x2 + 2x + x + 2 = 0

⇒ x(x+2) + (x+2) = 0

⇒ (x+2) (x+1) = 0

Hence, zeroes are -2 and -1.


29. Find the roots of the following quadratic equation: 2√3 x2 – 5x + √3 = 0

Solution

We have, 2√3 x2 – 5x + √3 = 0

Here, a = 2√3, b = -5, c = √3

D = b2 – 4ac
∴ D = (-5)2 – 4 (2√3)(√3)
= 25 – 24 = 1


30. Solve for x: 4x2 – 4ax + (a2 – b2) = 0

Solution

4x2 – 4ax + (a2 – b2) = 0
⇒ [4x2 – 4ax + a2] – b2 = 0
⇒ [(2x)2 – 2(2x)(a) + (a)2] – b2 = 0
⇒ (2x – a)2 – (b)2 = 0
⇒ (2x – a + b) (2x – a – b) = 0
⇒ 2x – a + b = 1 or 2x – a – b = 0
2x = a – b or 2x = a + b


31. Solve for x: 4x2 – 4ax + (a2 – b2) = 0

Solution
4x2 – 4ax + (a2 – b2) = 0
⇒ [4x2 – 4ax + a2] – b2 = 0
⇒ [(2x)2 – 2(2x)(a) + (a)2] – b2 = 0
⇒ (2x – a)2 – (b)2 = 0
⇒ (2x – a + b) (2x – a – b) = 0
⇒ 2x – a + b = 1 or 2x – a – b = 0
2x = a – b or 2x = a + b

32. Find the value(s) of k so that the quadratic equation 2x2 + kx + 3 = 0 has equal roots.

Solution

Given: 2x2 + kx + 3 = 0
Here a = 2, b = k, c= 3
D = 0 …[Since roots are equal]
As b2 – 4ac = 0
∴ k2 – 4(2)(3) = 0
⇒ k2 – 24 = 0
⇒ k2 = 24


33. Find the value(s) of k so that the quadratic equation 3x2 – 2kx + 12 = 0 has equal roots.

Solution

Given: 3x2 – 2kx + 12 = 0
Here a = 3, b = -2k, c = 12
D = 0 …[Since roots are equal as b2 – 4ac = 0]
∴ (-2k)2 – 4(3) (12) = 0
⇒ 4k2 – 144 = 0
⇒ k2 = 144/4 = 36
∴ k = ±√36
∴ k = ±6


34. Find the quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. Hence find the zeroes.

Solution

Sum of zeroes α + β = -3

and product of zeroes αβ = 2

Thus quadratic equation is

x2 - (α+β)x + αβ = 0

⇒ x2 - (-3)x + 2 = 0

⇒ x2 + 3x + 2 = 0

Thus, quadratic equation is x2 + 3x + 2 = 0

Now, above equation can be written as

x2 + 2x + x + 2 = 0

⇒ x(x+2) + (x+2) = 0

⇒ (x+2) (x+1) = 0

Hence, zeroes are -2 and -1.


35. Find the zeroes of the quadratic polynomial 6x2 - 3 - 7x and verify the relationship between the zeroes and the coefficients.

Solution

We have,
p(x) = 6x2 - 3 - 7x
For zeroes of polynomial,
p(x) = 0
6x2 - 7x - 3 = 0
⇒ 6x2 - 9x + 2x - 3 = 0
⇒ 3x(2x - 3) + 1(2x-3) = 0
⇒ (2x-3) (3x+1) = 0
Thus, 2x - 3 = 0 and 3x + 1 = 0
Hence, x = 3/2 and x = -1/3
Therefore,
α = 3/2 and β = -1/3 are the zeroes of the given polynomial.
Verification:


36. Find the value of m for which the roots of the equation. mx (6x + 10) + 25 = 0, are equal.
Solution
We have, mx(6x + 10) + 25 = 0
6mx2 + 10mx + 25 = 0
Here a = 6m, b = 10m, c = 25
D = 0 …(Since roots are equal)
b2 – 4ac = 0
∴ (10m)2 – 4(6m) (25) = 0
⇒ 100m2 – 600m = 0
⇒ 100m (m – 6) = 0
⇒ 100m = 0 or m – 6 = 0
m = 0 or m = 6 …[Rejecting m = 0, as coefficient of x2 cannot be zero]
∴ m = 6

37. For what values of k, the roots of the quadratic equation (k + 4)x2 + (k + 1)x + 1 = 0 are equal?

Solution

We have,
(k + 4) x2 + (k + 1) x + 1 = 0
Here,
a = k + 4, b = k + 1, c = 1
D =0 …[∵ Roots are equal]
b2 – 4ac = 0
∴ (k + 1)2 – 4(k + 4)(1) = 0
k2 + 2k + 1 – 4k – 16 = 0
⇒ k2 – 2k – 15 = 0
⇒ k2 – 5k + 3k – 15 = 0
⇒ k(k – 5) + 3(k – 5) = 0
⇒ (k – 5)(k + 3) = 0
⇒ k – 5 = 0 or k + 3= 0
⇒ k = 5 or k = -3
∴ k = 5 and -3


38. Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.

Solution

Let

p(x) = x2 + 7x + 10

For zeroes of polynomial p(x) = 0,

x2 + 7x + 10 = 0

⇒ x2 + 5x + 2x + 10 = 0

⇒ x(x+5) + 2(x+5) = 0

⇒ (x+5) (X+2) = 0

So,

x = -2 and x = -5

Therefore, α = -2 and β = -5 are the zeroes of the given polynomial.

Verification:

Sum of zeroes,

α + β = -2 + (-5)


39. Solve for x: 1/x+4 - 1/x+7 = 11/30, x ≠ -4, -7
Solution

40. For what value of k, are the roots of the quadratic equation y2 + k2 = 2 (k + 1)y equal?

Solution

y2 + k2 = 2(k + 1)y
y2 – 2(k + 1)y + k2 = 0
Here a = 1, b = -2(k + 1), c = k2
D = 0 …[Roots are equal]
b2 – 4ac = 0
∴ [-2(k + 1)]2 – 4 × (1) × (k2) = 0
⇒ 4(k2 + 2k + 1) – 4k2 = 0
⇒ 4k2 + 8k + 4 – 4k2 = 0
⇒ 8k + 4 = 0
⇒ 8k = -4
∴ k = -4/8 = -1/2


41. Solve the equation 3/x+1 - 1/2 = 2/3x-1, x≠ -1, x≠ 1/3 for x.

Solution

⇒ 2(2x + 2) = (5 – x)(3x – 1)
⇒ 4x + 4 = 15x – 5 – 3x2 + x
⇒ 4x + 4 – 15x + 5 + 3x2 – x = 0
⇒ 3x2 – 12x + 9 = 0
⇒ x2 – 4x + 3 = 0 …[Dividing by 3]
⇒ x2 – 3x – x + 3 = 0
⇒ x(x – 3) – 1(x – 3) = 0
⇒ (x – 1) (x – 3) = 0
⇒ x – 1 = 0 or x – 3 = 0
∴ x = 1 or x = 3


42. Solve for x:

2x/x-3 + 1/2x+3 + 3x+9/(x+9)(2x+3) = 0, x ≠ 3, -3/2

Solution

2x(2x+3) + (x-3) + (3x+9) = 0
⇒ 4x2 + 6x + x - 3 + 3x + 9 = 0
⇒ 4x2 + 10x + 6 = 0
⇒ 2x2 + 5x + 3 =0
⇒ (x+1) (2x+3) = 0
Thus, x = -1, x = -3/2


43. Solve the following quadratic equation for x :

x22 + (a/a+b + a+b/a)x + 1 = 0

Solution


44. Solve for x: x2 + 5x -(a2 + a - 6) = 0
Solution
We have,
x2 + 5x -(a2 + a - 6) = 0
Thus x = a-2, x = -(a+3)

45. Solve the following quadratic equation for x: x2 + (2b-1)x + (b2-b-20) = 0

Solution

We have,

x2 - (2b-1)x + (b2-b-20) = 0

Comparing with Ax2 + Bx + C = 0 we have

A = 1, B = -(2b-1), C = (b2-b-20)

Thus x = b+4 and x = b-5


46. Find the roots of the equation 2x2 + x - 4 = 0, by the method of completing the squares.

Solution

We have,

2x2 + x - 4 = 0


47. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Find the numbers.

Solution

Let three consecutive natural numbers are x, x + 1, x + 2.
According to the question,
(x + 1)2 – [(x + 2)2 – x2] = 60
⇒ x2 + 2x + 1 – (x2 + 4x + 4 – x2) = 60
⇒ x2 + 2x + 1 – 4x – 4 – 60 = 0
⇒ x2 – 2x – 63 = 0
⇒ x2 – 9x + 7x – 63 = 0
⇒ x(x – 9) + 7(x – 9) = 0
⇒ (x – 9) (x + 7) = 0
⇒ x – 9 = 0 or x + 7 = 0
⇒ x = 9 or x = -7
Natural nos. can not be negative, ∴ x = 9
∴ Numbers are 9, 10, 11.


48. If the sum of two natural numbers is 8 and their product is 15, find the numbers.

Solution

Let the numbers be x and (8 – x).
According to the Question,
x(8 – x) = 15
⇒ 8x – x2 = 15
⇒ 0 = x2 – 8x + 15
⇒ x2 – 5x – 3x + 15 = 0
⇒ x(x – 5) – 3(x – 5) = 0
⇒ (x – 3)(x – 5) = 0
x – 3 = 0 or x – 5 = 0
x = 3 or x = 5
When x = 3, numbers are 3 and 5.
When x = 5, numbers are 5 and 3.


49. 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 k.

Solution

We have,

3x2 + px - 8 = 0

Since 2 is a root of above equation, it must satisfy it.

Substituting, x= 2 in we have

12 + 2p - 8 = 0

⇒ p = -2

Since,

4x2 - 2px + k = 0 has equal roots.

or, 4x2 + 4x + k = 0 has equal roots.

D = b2 - 4ac = 0

42 - 4(4)(k) = 0

⇒ 16 - 16k = 0

⇒ 16k = 16

⇒ k = 1


50. If -3 is a root of quadratic equation 2x2 + px - 15 = 0, while the quadratic equation x2 - 4px + k = 0 has equal roots. Find the value of k.

Solution

Given -3 is a root of quadratic equation.

We have,

2x2 + px - 15

Since 3 is a root of above equation, it must satisfy it.

Substituting x 3 = in above equation we have,

2(-3)2 + p(-3) - 15 = 0

⇒ 2×9 - 3p - 15 = 0

⇒ p = 1

Since,

x2 - 4px + k = 0 has equal roots

or, x2 - 4x + k  = 0 has equal roots.

b2 - 4ac = 0

⇒ (-4)2 - k = 0

⇒ 16 - 4k = 0

⇒ 4k = 16

⇒ k = 4


51. Solve 1/(a+b+x) = 1/a + 1/b + 1/x, a+b ≠ 0

Solution

We have,


Quadratic Equations Class 10 Maths Important Questions Long Answer (4 Marks)

52. Solve for x: (2x/x-5)2 + (2x/x-5) - 24 = 0, x≠5

Solution


⇒ 2x = -8x + 40

⇒ 10x = 40

⇒ x = 4

Hence, x = 15, 4


53. Solve for x: 1/x+1 + 2/x+2 = 4/x+4, x≠ -1,-2,-4

Solution


54. Find x in terms of , a b and c :
a/x-a + b/x-b = 2c/x-c, x≠ a,b,c
Solution

55. Find the zeroes of the quadratic polynomial 7y2 - 11/3 y - 2/3 and verify the relationship between the zeroes and the coefficients.
Solution
⇒ 21y2 - 14y + 3y  - 2 = 0
⇒ 7y(3y-2) + (3y-2) = 0
⇒ (3y-2) (7y+1) = 0
⇒ y = 2/3, -1/7
Hence, zeros of given polynomial are:
y = 2/3 and y = -1/7
Comparing the given equation with ax2 + bx + c = 0, we get a=21, b=-11 and c=-2

56. 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

(3k + 1)x2 + 2(k + 1) + 1 = 0
Here, a = 3k + 1, b = 2(k + 1), c = 1
D = 0 …[∵ Roots are equal]
As b2 – 4ac = 0
∴ [2(k + 1)]2 – 4(3k + 1)(1) = 0
⇒ 4(k + 1)2 – 4(3k + 1) = 0
⇒ 4(k2 + 2k + 1 – 3k – 1) = 0
⇒ (k2 – k) = 0/4
⇒ k(k – 1) = 0
⇒ k = 0 or k – 1 = 0
∴ k = 0 or k = 1
Roots are x = −b/2a ...[ Given, equal roots]


57. Find the value of p for which the quadratic equation (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.

Solution

(2p + 1)x2 – (7p + 2)x + (7p – 3) = 0
Here, a = 2p + 1, b = -(7p + 2), c = 7p – 3
D = 0 …[∵ Equal roots As h2 – 4ac = 0]
∴ [-(7p + 2)]2 – 4(2p + 1)(7p – 3) = 0
⇒ (7p + 2)2 – 4(14p2 – 6p + 7p – 3) = 0
⇒ 49p2 + 28p + 4 – 56p2 + 24p – 28p + 12 = 0
⇒ -7p2 + 24p + 16 = 0
⇒ 7p2 – 24p – 16 = 0 …[Dividing both sides by -1]
⇒ 7p2 – 28p + 4p – 16 = 0
⇒ 7p(p – 4) + 4(p – 4) = 0
⇒ (p – 4) (7p + 4) = 0
⇒ p – 4 = 0 or 7p + 4 = 0
⇒ p = 4 or p = −4/7


58. Solve for x : 1/a+b+x = 1/a + 1/b + 1/x
where, a+b+x≠0 and a,b,x 0
Solution
⇒ x2 + (a+b)x + ab = 0
⇒ (x+a) (x+b) = 0
⇒ x = -a, x = -b
Hence, x = -a or x = -b

59. Check whether the equation 5x2 - 6x - 2 = 0 has real roots if it has, find them by the method of completing the square. Also verify that roots obtained satisfy the given equation.

Solution

We have,

5x2 - 6x - 2 = 0

Comparing with ax2 + bx + c = 0 we get,

a = 5, b = -6 and c = -2

b2 - 4ac = (-6)2 - 4×5×(-2)

= 36 + 40 = 76 > 0

So the equation has real and two distinct roots.

5x2 - 6x = 2

Dividing both the sides by 5 we get


60. If -5 is a root of the quadratic equation 2x2 + px + 15 and the quadratic equation p(x2+x)+k = 0 has equal roots, then find the values of p and k.
Solution
Given,
2x2 + px -15 = 0
Since x = -15 is the root of above equation. It must satisfy it.
2(-5)2 + p(-5) - 15 = 0
⇒ 50 - 5p - 15 = 0
⇒ 5p = 35
⇒ p = 7
Now,
p(x2+x)+k = 0 has equal roots or,
7x2 + 7x + k = 0
Taking b2 - 4ac = 0, we have
72 - 4×7×k = 0
⇒ 7-4k = 0
⇒ k  = 7/4
Hence, p = 7 and k = 7/4

61. A shopkeeper buys some books for 80. If he had bought 4 more books for the same amount, each book would have cost ₹1 less. Find the number of books he bought.

Solution

Let the number of books he bought = x
Increased number of books he had bought = x +4
Total amount = ₹80
According to the problem,

⇒ x(x + 4) = 320
⇒ x2 + 4x – 320 = 0
⇒ x2 + 20x – 16x – 320 = 0
⇒ x(x + 20) – 16(x + 20) = 0
⇒ (x + 20) (x – 16) = 0
⇒ x + 20 = 0 or x – 16 = 0
⇒ x = -20 …(neglected) or x = 16
∴ Number of books he bought = 16


62.  Find the positive values of k for which quadratic equations x2 + kx + 64 = 0 and x2 - 8x + k = 0 both will have the real roots.

Solution

(i) For x2 + kx + 64 = 0 to have real roots

k2 - 256 ≥ 0

⇒ k2 ≥ 256

⇒ k ≥ 16 or k < -16

(ii) For x2 - 8x + k = 0 to have real roots

64 - 4k ≥ 0

⇒ 16 - k ≥ 0

⇒ 16 ≥ k

Therefore, For (i) and (ii) to hold simultaneously

k = 16


63. Sum of the areas of two squares is 400 cm2. If the difference of their perimeters is 16 cm, find the sides of the two squares.

Solution

Let the side of Large square = x cm
Let the side of small square = y cm
According to the Question,
x2 + y2 = 400 …(i) …[∵ area of square = (side)2]
4x – 4y = 16 …[∵ Perimeter of square = 4 sides]
⇒ x – y = 4 …[Dividing both sides by 4]
⇒ x = 4 + y …(ii)
Putting the value of x in equation (i),
(4 + y)2 + y22 = 400
⇒ y2 + 8y + 16 + y2 – 400 = 0
⇒ 2y2 + 8y – 384 = 0
⇒ y2 + 4y – 192 = 0 …[Dividing both sides by 2]
⇒ y2 + 16y – 12y – 192 = 0
⇒ y(y + 16) – 12(y + 16) = 0
⇒ (y – 12)(y + 16) = 0
⇒ y – 12 = 0 or y + 16 = 0
⇒ y = 12 or y = -16 …[Neglecting negative value]
∴ Side of small square = y = 12 cm and Side of large square = x = 4 + 12 = 16 cm


64. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

Solution

Let the length of shorter side be x m.
∴ length of diagonal = (x + 16) m
and length of longer side = (x + 14) m
Using pythagoras theorem,
(l)2 + (b)2 = (h)2
∴ x2 + (x + 14)22 = (x + 16)2
⇒ x2 + x2 + 196 + 28x = x2 + 256 + 32x
⇒ x2 – 4x – 60 = 0
⇒ x2 – 10x + 6x – 60 = 0
⇒ x(x – 10) + 6(x – 10) = 0
⇒ (x – 10) (x + 6) = 0
⇒ x – 10 = 0 or x + 6 = 0
⇒ x = 10 or x = -6 (Neglect as length cannot be negative])
⇒ x = 10 m 

Length of shorter side = x = 10 m
Length of diagonal = (x + 16) m = 26 m
Length of longer side = (x + 14)m = 24m
∴ Length of sides are 10 m and 24 m.


65. The sum of two numbers is 9 and the sum of their reciprocals is 1/2. Find the numbers.

Solution

Let the numbers be x and 9 – x.
According to the Question,


⇒ 18 = 9x – x2
⇒ x2 – 9x + 18 = 0
⇒ x2 – 3x – 6x + 18 = 0
⇒ x(x – 3) – 6(x – 3) = 0
⇒ (x – 3) (x – 6) = 0
⇒ x – 3 = 0 or x – 6 = 0
⇒ x = 3 or x = 6

When x = 3, numbers are 3 and 6.
When x = 6, numbers. are 6 and 3.


66. Find the nature of the roots of the quadratic equation 4x2 + 4√3x + 3 = 0.

Solution

Given,

4x2 + 4√3x + 3 = 0

Comparing the given equation with ax2 + bx + c = 0, we get

a = 4, b = 4√3 and c = 3.

Since, b2 - 4ac = 0, then roots of the given equation are real and equal.


67. The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.

Solution

Let three numbers in A.P. are a – d, a, a + d.
a – d + a + a + d = 12
⇒ 3a = 12
⇒ a = 4
(a – d)3 + (a)3 + (a + d)3 = 288
⇒ a3 – 3a2d + 3ad2 – d3 + a3 + a3 + 3a2d + 3ad2 + d3 = 288
⇒ 3a3 + 6ad2 = 288
⇒ 3a(a2 + 2d2) = 288
⇒ 3 × 4(42 + 2d2) = 288
⇒ (16 + 2d2) = 288/12
⇒ 2d2 = 24 – 16 = 8
⇒ d2 = 4
⇒ d = ± 2
When, a = 4, d = 2, numbers are:
a – d, a, a + d, i.e., 2, 4, 6
When, a = 4, d = -2, numbers are:
a – d, a, a + d, i.e., 6, 4, 2


68. Find the values of k for which the quadratic equations (k+4)x2 + (k+1)x + 1 = 0 has equal roots. Also, find the roots.

Solution

Given,

(k+4)x2 + (k+1)x + 1 = 0

Comparing with Ax2 + Bx + C = 0, we get

A = (k+4), B = (k+1), C = 1

If roots are equal, B2 - 4AC = 0

(k+1)2 - 4(k+4)(1) = 0

⇒ k2 + 1 + 2k - 4k - 16 = 0

⇒ k2 - 2k -15 = 0

⇒ (k-5) (k+3) = 0

⇒ k = 5, -3

For k = 5, equation becomes

9x2 + 6x + 1 = 0

⇒ (3x+1)2 = 0

⇒ x = -1/3

For k = -3, equation becomes

x2 - 2x + 1 = 0

⇒ (x-1)2 = 0

⇒ x = 1

Hence, roots are 1 and -1/3


69. The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.

Solution

Perimeter of right ∆ = 60 cm …[Given]
a + b + c = 60
⇒ a + b + 25 = 60
⇒ a + b = 60 – 25 = 35 …(i)
In rt. ∆ACB, AC2 + BC2 = AB2
b2 + a2 = (25)2 …[Pythagoras’ theorem]
⇒ a2 + b2 = 625 …(ii)
From (i), a + b = 35
(a + b)2 = (35) …[Squaring both sides]
⇒ a2 + b2 + 2ab = 1225
⇒ 625 + 2ab = 1225 …[From (ii)]
⇒ 2ab = 1225 – 625 = 600
⇒ ab = 300 …(iii)
Area of ∆ = 1/2 × base × corresponding altitude
= 1/2 × b × a = 1/2 (300) ...[From (iii)]
= 150 cm2


70. The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by 1/15. Find the fraction.

Solution

Let the denominator be x and the numerator be x – 3.
∴ Fraction =x−3/x
New denominator = x + 1
According to the Question,

⇒ 15x2 – 45x = 14x2 – 45x + 14x – 45
⇒ 15x2 – 14x2 – 14x + 45 = 0
⇒ x2 – 14x + 45 = 0
⇒ x2 – 5x – 9x + 45 = 0
⇒ x(x – 5) – 9(x – 5) = 0
⇒ (x – 5) (x – 9) = 0
⇒ x – 5 = 0 or x – 9 = 0
⇒ x = 5 or x = 9

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