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

**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.

## Important Questions for Chapter 4 Quadratic Equations Class 10 Maths

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

**1. Find the positive root of √3x**

^{2}+ 6 = 9.**Solution**

**2. If x = -1/2 is a solution of the quadratic equation 3x**

^{2}+ 2kx -3 = 0, find the value of k.**Solution**

**3. Find the value of k, for which one root of the quadratic equation kx**

^{2}-14x + 8 = 0 is six times the other.**Solution**

Let one root be Î± and other root be 6Î±.

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

**4. If 1 is a root of the equations ay ^{2} + ay + 3 = 0 and y^{2} + y + b = 0, then find the value of ab.**

**Solution**

ay^{2} + ay + 3 = 0

⇒ a(1)^{2} + a(1) + 3 = 0

⇒ 2a = -3

⇒ a = −32

y^{2} + y + b = 0

⇒ 1^{2} + 1 + b = 0

⇒ b = -2

∴ ab = (−32)(−2) = 3

**5. Find the roots of the equation x ^{2} – 3x – m (m + 3) = 0, where m is a constant.**

**Solution**

x^{2} – 3x – m(m + 3) = 0

⇒ D = b^{2} – 4ac

⇒ D = (- 3)^{2} – 4(1) [-m(m + 3)]

= 9 + 4m (m + 3)

= 4m^{2} + 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 x ^{2} + 4x + k = 0 are real?**

**Solution**

Comparing the given equation with ax^{2} + bx + c = 0, we get a = 1, b = 4, c = k.

Since, given the equation has real roots,

D ≥ 0

⇒ b^{2} - 4ac ≥ 0

⇒ 4^{2} - 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

mx^{2} – 7mx + 49 = 0

Here, a = m, b = – 7m, c = 49

D = b^{2} – 4ac = 0** …[For equal roots]**

⇒ (-7m)^{2} – 4(m) (49) = 0

⇒ 49m^{2} – 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 3x ^{2} - 10x + k = 0 are reciprocal of each other.**

**Solution**

Comparing the given equation with ax^{2} + bx + c = 0 we get a = 3, b = -10, c = k

Let one root be Î± so other root is 1/Î±

Now,

product of roots, Î± × 1/Î± = c/a

Hence, the value of k is 3.

**9. Find the roots of the following quadratic equation :**

**15x ^{2} - 10√6x + 10 = 0**

**Solution**

We have 15x^{2} - 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

px^{2} – 3px + 9 = 0 Here a = p, b = -3p,

D = 0

b^{2} – 4ac = 0

⇒ (-3p)^{2} – 4(p)(9) = 0

⇒ 9p^{2} – 36p = 0

⇒ 9p (p – 4) = 0

⇒ 9p = 0 or p – 4= 0

p = 0 (rejected) or p = 4

∴ p = 4 **…(∵ Coefficient of x ^{2} cannot be zero)**

**11. Solve for x:**

**36x ^{2} – 12ax + (a^{2} – b^{2}) = 0**

**Solution**

We have, 36x^{2} – 12ax + (a^{2} – b^{2}) = 0

⇒ (36x^{2} – 12ax + a^{2}) – b^{2} = 0

⇒ [(6x)^{2} – 2(6x)(a) + (a)^{2}] – b^{2} = 0

⇒ (6x – a)^{2} – (b)^{2} = 0 **…[∵ x ^{2} – 2xy + y^{2} = (x – y)^{2}]**

⇒ (6x – a + b) (6x – a – b) = 0

**...[∵ x**

^{2}– y^{2}= (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√3x ^{2} + 5x - 2√3 = 0**

**Solution**

^{2}+ 5x - 2√3 = 0

**13. Find the roots of 4x**

^{2}+ 3x + 5 = 0 by the method of completing the squares.**Solution**

Here 4x^{2} + 3x + 5 = 0

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

**14. Solve for x : x ^{2} - (√3+1)x + √3 = 0**

**Solution**

x^{2} - (√3+1)x + √3 = 0

⇒ x^{2} -√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 :**

**9x ^{2} - 6b^{2}x - (a^{4} - b^{4}) = 0**

**Solution**

We have,

9x^{2} - 6b^{2}x - (a^{4} - b^{4}) = 0

Comparing with Ax^{2} + Bx + C

A = 9, B = -6b^{2}, C = -(a^{4} - b^{4})

**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 ax**

^{2}+ 7x + b = 0, find the values of a and b.**Solution**

We have, ax^{2} + 7x + b = 0

Here ‘a’ = a, ‘b’ = 7, ‘c’ = b

Now, Î± = 23 and Î² = -3 **…[Given]**

**19. Solve the following quadratic equation for x: 9x**

^{2}– 6b^{2}x – (a^{4}– b^{4}) = 0**Solution**

The given quadratic equation can be written as

(9x^{2} – 6b^{2}x + b^{4}) – a^{4} = 0

⇒ (3x – b^{2})^{2} – (a^{2})^{2} = 0

⇒ (3x – b^{2} + a^{2}) (3x – b^{2} – a^{2}) = 0 **…[x ^{2} – y^{2} = (x + y) (x – y)]**

⇒ 3x – b

^{2}+ a

^{2}= 0 or 3x – b

^{2}– a

^{2}= 0

⇒ 3x = b

^{2}– a

^{2}or 3x = b

^{2}+ a

^{2}

**20. If -5 is a root of the quadratic equation 2x**

^{2}+ px – 15 = 0 and the quadratic equation p(x^{2}+ x) + k = 0 has equal roots, find the value of k.**Solution**

We have, 2x^{2} + 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(x^{2} + x) + k

⇒ px^{2} + px + k = 0

7x^{2} + 7x + k = 0 **…[From (i)]**Here, a = 7, b = 7, c = k

D = 0 **…[Roots are equal]**

⇒ b^{2} – 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 x ^{2} + 10x + √3 = 0**

**Solution**

Given,

13√3 x^{2} + 10x + √3 = 0

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

a = 13√3, b = 10, c = √3

b^{2} - 4ac = 10^{2} - 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 x ^{2} - 8x + k = 0, will have real roots.**

**Solution**

We have,

x^{2} - 8x + k = 0

Comparing with Ax^{2} + Bx + C = 0, we get

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

Since the given equation has real roots,

B^{2} - 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 x ^{2} + kx + 12 = 0 and the equation x^{2} + kx + q = 0 has equal roots, find the value of q.**

**Solution**

We have,

x^{2} + 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 x^{2} + kx + q = 0 we have

x^{2} - 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 2x**

^{2}+ kx + 8 = 0 will have the equal roots?**Solution**

We have,

2x^{2} + kx + 8 = 0

Comparing with ax^{2} + bx + c = 0, we get

a = 2, b = k and c = 8

For equal roots, D = 0,

b^{2} - 4ac = 0

k^{2} - 4×2×8 = 0

⇒ k^{2} = 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 x ^{2} + 2√2 kx + 18 = 0 has equal roots.**

**Solution**

We have,

x^{2} + 2√2 kx + 18 = 0

Comparing it by ax^{2} + bx + c, we get a = 1, b = 2√2 k and c = 18.

Given that,

Equation x^{2} + 2√2 kx + 18 = 0 has equal roots.

b^{2} - 4ac = 0

⇒ (2√2 k)^{2} -4118 = 0

⇒ 8k^{2} - 72 = 0

⇒ 8k^{2} = 72

⇒ k^{2} = 72/8 = 9

⇒ k = ±3

**27. If Î± and Î² are the zeroes of the polynomial f(x) = x ^{2} - 4x - 5 then find the value of Î±^{2} + Î²^{2}.**

**Solution**

We have,

p(x) = x^{2} - 4x - 5

Comparing it by ax^{2} + bx + c, we get a = 1, b = -4 and c = -5.

Since, given Î± and Î² 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,

Î± + Î² = -3 **...(1)**

and product of zeroes, Î±Î² = 2

Thus quadratic equation is

x^{2} - (Î±+Î²)x + Î±Î² = 0

⇒ x^{2} - (-3)x + 2 = 0

⇒ x^{2} + 3x + 2 = 0

Thus quadratic equation is x^{2} + 3x + 2 = 0

Now, above equation can be written as

x^{2} + 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 x ^{2} – 5x + √3 = 0**

**Solution**

We have, 2√3 x^{2} – 5x + √3 = 0

Here, a = 2√3, b = -5, c = √3

D = b^{2} – 4ac

∴ D = (-5)^{2} – 4 (2√3)(√3)

= 25 – 24 = 1

**30. Solve for x: 4x**

^{2}– 4ax + (a^{2}– b^{2}) = 0**Solution**

4x^{2} – 4ax + (a^{2} – b^{2}) = 0

⇒ [4x^{2} – 4ax + a^{2}] – b^{2} = 0

⇒ [(2x)^{2} – 2(2x)(a) + (a)^{2}] – b^{2} = 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: 4x ^{2} – 4ax + (a^{2} – b^{2}) = 0**

**Solution**

^{2}– 4ax + (a

^{2}– b

^{2}) = 0

^{2}– 4ax + a

^{2}] – b

^{2}= 0

⇒ [(2x)

^{2}– 2(2x)(a) + (a)

^{2}] – b

^{2}= 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 2x**

^{2}+ kx + 3 = 0 has equal roots.**Solution**

Given: 2x^{2} + kx + 3 = 0

Here a = 2, b = k, c= 3

D = 0 **…[Since roots are equal]**

As b^{2} – 4ac = 0

∴ k^{2} – 4(2)(3) = 0

⇒ k^{2} – 24 = 0

⇒ k^{2} = 24

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

**Solution**

Given: 3x^{2} – 2kx + 12 = 0

Here a = 3, b = -2k, c = 12

D = 0 **…[Since roots are equal as b ^{2} – 4ac = 0]**

∴ (-2k)

^{2}– 4(3) (12) = 0

⇒ 4k

^{2}– 144 = 0

⇒ k

^{2}= 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 Î± + Î² = -3

and product of zeroes Î±Î² = 2

Thus quadratic equation is

x^{2} - (Î±+Î²)x + Î±Î² = 0

⇒ x^{2} - (-3)x + 2 = 0

⇒ x^{2} + 3x + 2 = 0

Thus, quadratic equation is x^{2} + 3x + 2 = 0

Now, above equation can be written as

x^{2} + 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 6x ^{2} - 3 - 7x and verify the relationship between the zeroes and the coefficients.**

**Solution**

We have,

p(x) = 6x^{2} - 3 - 7x

For zeroes of polynomial,

p(x) = 0

6x^{2} - 7x - 3 = 0

⇒ 6x^{2} - 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,

Î± = 3/2 and Î² = -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**

6mx

^{2}+ 10mx + 25 = 0

Here a = 6m, b = 10m, c = 25

D = 0

**…(Since roots are equal)**

b

^{2}– 4ac = 0

∴ (10m)

^{2}– 4(6m) (25) = 0

⇒ 100m

^{2}– 600m = 0

⇒ 100m (m – 6) = 0

⇒ 100m = 0 or m – 6 = 0

m = 0 or m = 6

**…[Rejecting m = 0, as coefficient of x**

^{2}cannot be zero]∴ m = 6

**37. For what values of k, the roots of the quadratic equation (k + 4)x**

^{2}+ (k + 1)x + 1 = 0 are equal?**Solution**

We have,

(k + 4) x^{2} + (k + 1) x + 1 = 0

Here,

a = k + 4, b = k + 1, c = 1

D =0 **…[∵ Roots are equal]**

b^{2} – 4ac = 0

∴ (k + 1)^{2} – 4(k + 4)(1) = 0

k^{2} + 2k + 1 – 4k – 16 = 0

⇒ k^{2} – 2k – 15 = 0

⇒ k^{2} – 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 x ^{2} + 7x + 10, and verify the relationship between the zeroes and the coefficients.**

**Solution**

Let

p(x) = x^{2} + 7x + 10

For zeroes of polynomial p(x) = 0,

x^{2} + 7x + 10 = 0

⇒ x^{2} + 5x + 2x + 10 = 0

⇒ x(x+5) + 2(x+5) = 0

⇒ (x+5) (X+2) = 0

So,

x = -2 and x = -5

Therefore, Î± = -2 and Î² = -5 are the zeroes of the given polynomial.

**Verification:**

Sum of zeroes,

Î± + Î² = -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 y**

^{2}+ k^{2}= 2 (k + 1)y equal?**Solution**

y^{2} + k^{2} = 2(k + 1)y

y^{2} – 2(k + 1)y + k^{2} = 0

Here a = 1, b = -2(k + 1), c = k^{2}

D = 0 **…[Roots are equal]**

b^{2} – 4ac = 0

∴ [-2(k + 1)]^{2} – 4 × (1) × (k^{2}) = 0

⇒ 4(k^{2} + 2k + 1) – 4k^{2} = 0

⇒ 4k^{2} + 8k + 4 – 4k^{2} = 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 – 3x^{2} + x

⇒ 4x + 4 – 15x + 5 + 3x^{2} – x = 0

⇒ 3x^{2} – 12x + 9 = 0

⇒ x^{2} – 4x + 3 = 0 **…[Dividing by 3]**

⇒ x^{2} – 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

⇒ 4x^{2} + 6x + x - 3 + 3x + 9 = 0

⇒ 4x^{2} + 10x + 6 = 0

⇒ 2x^{2} + 5x + 3 =0

⇒ (x+1) (2x+3) = 0

Thus, x = -1, x = -3/2

**43. Solve the following quadratic equation for x :**

**x ^{2}2 + (a/a+b + a+b/a)x + 1 = 0**

**Solution**

**44. Solve for x: x**

^{2}+ 5x -(a^{2}+ a - 6) = 0**Solution**

^{2}+ 5x -(a

^{2}+ a - 6) = 0

**45. Solve the following quadratic equation for x: x**

^{2}+ (2b-1)x + (b^{2}-b-20) = 0**Solution**

We have,

x^{2} - (2b-1)x + (b^{2}-b-20) = 0

Comparing with Ax^{2} + Bx + C = 0 we have

A = 1, B = -(2b-1), C = (b^{2}-b-20)

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

**46. Find the roots of the equation 2x ^{2} + x - 4 = 0, by the method of completing the squares.**

**Solution**

We have,

2x^{2} + 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} – x^{2}] = 60

⇒ x^{2} + 2x + 1 – (x^{2} + 4x + 4 – x^{2}) = 60

⇒ x^{2} + 2x + 1 – 4x – 4 – 60 = 0

⇒ x^{2} – 2x – 63 = 0

⇒ x^{2} – 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 – x^{2} = 15

⇒ 0 = x^{2} – 8x + 15

⇒ x^{2} – 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 3x ^{2} + px - 8 = 0 and the quadratic equation 4x^{2} - 2px + k = 0 has equal roots, find k.**

**Solution**

We have,

3x^{2} + 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,

4x^{2} - 2px + k = 0 has equal roots.

or, 4x^{2} + 4x + k = 0 has equal roots.

D = b^{2} - 4ac = 0

4^{2} - 4(4)(k) = 0

⇒ 16 - 16k = 0

⇒ 16k = 16

⇒ k = 1

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

**Solution**

Given -3 is a root of quadratic equation.

We have,

2x^{2} + 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,

x^{2} - 4px + k = 0 has equal roots

or, x^{2} - 4x + k = 0 has equal roots.

b^{2} - 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 7y**

^{2}- 11/3 y - 2/3 and verify the relationship between the zeroes and the coefficients.**Solution**

^{2}- 14y + 3y - 2 = 0

^{2}+ 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)x**

^{2}+ 2(k + 1)x + 1 = 0 has equal roots. Also find the roots.**Solution**

(3k + 1)x^{2} + 2(k + 1) + 1 = 0

Here, a = 3k + 1, b = 2(k + 1), c = 1

D = 0 **…[∵ Roots are equal]**

As b^{2} – 4ac = 0

∴ [2(k + 1)]^{2} – 4(3k + 1)(1) = 0

⇒ 4(k + 1)^{2} – 4(3k + 1) = 0

⇒ 4(k^{2} + 2k + 1 – 3k – 1) = 0

⇒ (k^{2} – 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)x**

^{2}– (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.**Solution**

(2p + 1)x^{2} – (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(14p^{2} – 6p + 7p – 3) = 0

⇒ 49p^{2} + 28p + 4 – 56p^{2} + 24p – 28p + 12 = 0

⇒ -7p^{2} + 24p + 16 = 0

⇒ 7p^{2} – 24p – 16 = 0 **…[Dividing both sides by -1]**

⇒ 7p^{2} – 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**

^{2}+ (a+b)x + ab = 0

**59. Check whether the equation 5x ^{2} - 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,

5x^{2} - 6x - 2 = 0

Comparing with ax^{2} + bx + c = 0 we get,

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

b^{2} - 4ac = (-6)^{2} - 4×5×(-2)

= 36 + 40 = 76 > 0

So the equation has real and two distinct roots.

5x^{2} - 6x = 2

Dividing both the sides by 5 we get

**60. If -5 is a root of the quadratic equation 2x**

^{2}+ px + 15 and the quadratic equation p(x^{2}+x)+k = 0 has equal roots, then find the values of p and k.**Solution**

^{2}+ px -15 = 0

^{2}+ p(-5) - 15 = 0

^{2}+x)+k = 0 has equal roots or,

^{2}+ 7x + k = 0

^{2}- 4ac = 0, we have

**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

⇒ x^{2} + 4x – 320 = 0

⇒ x^{2} + 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 x ^{2} + kx + 64 = 0 and x^{2} - 8x + k = 0 both will have the real roots.**

**Solution**

(i) For x^{2} + kx + 64 = 0 to have real roots

k^{2} - 256 ≥ 0

⇒ k^{2} ≥ 256

⇒ k ≥ 16 or k < -16

(ii) For x^{2} - 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 cm ^{2}. 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,

x^{2} + y^{2} = 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}+ y2

^{2}= 400

⇒ y

^{2}+ 8y + 16 + y

^{2}– 400 = 0

⇒ 2y

^{2}+ 8y – 384 = 0

⇒ y

^{2}+ 4y – 192 = 0

**…[Dividing both sides by 2]**

⇒ y

^{2}+ 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}

∴ x^{2} + (x + 14)2^{2} = (x + 16)^{2}

⇒ x^{2} + x^{2} + 196 + 28x = x^{2} + 256 + 32x

⇒ x^{2} – 4x – 60 = 0

⇒ x^{2} – 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 – x

^{2}⇒ x

^{2}– 9x + 18 = 0

⇒ x

^{2}– 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 4x ^{2} + 4√3x + 3 = 0.**

**Solution**

Given,

4x^{2} + 4√3x + 3 = 0

Comparing the given equation with ax^{2} + bx + c = 0, we get

a = 4, b = 4√3 and c = 3.

Since, b^{2} - 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

⇒ a^{3} – 3a^{2}d + 3ad^{2} – d^{3} + a^{3} + a^{3} + 3a^{2}d + 3ad^{2} + d^{3} = 288

⇒ 3a^{3} + 6ad^{2} = 288

⇒ 3a(a^{2} + 2d^{2}) = 288

⇒ 3 × 4(4^{2} + 2d^{2}) = 288

⇒ (16 + 2d^{2}) = 288/12

⇒ 2d^{2} = 24 – 16 = 8

⇒ d^{2} = 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)x ^{2} + (k+1)x + 1 = 0 has equal roots. Also, find the roots.**

**Solution**

Given,

(k+4)x^{2} + (k+1)x + 1 = 0

Comparing with Ax^{2} + Bx + C = 0, we get

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

If roots are equal, B^{2} - 4AC = 0

(k+1)^{2} - 4(k+4)(1) = 0

⇒ k^{2} + 1 + 2k - 4k - 16 = 0

⇒ k^{2} - 2k -15 = 0

⇒ (k-5) (k+3) = 0

⇒ k = 5, -3

For k = 5, equation becomes

9x^{2} + 6x + 1 = 0

⇒ (3x+1)^{2} = 0

⇒ x = -1/3

For k = -3, equation becomes

x^{2} - 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, AC^{2} + BC^{2} = AB^{2}

b^{2} + a^{2} = (25)^{2} **…[Pythagoras’ theorem]**

⇒ a^{2} + b^{2} = 625** …(ii)**

From (i), a + b = 35

(a + b)^{2} = (35) **…[Squaring both sides]**

⇒ a^{2} + b^{2} + 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 cm^{2}

**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,

⇒ 15x^{2} – 45x = 14x^{2} – 45x + 14x – 45

⇒ 15x^{2} – 14x^{2} – 14x + 45 = 0

⇒ x^{2} – 14x + 45 = 0

⇒ x^{2} – 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