Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

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Q1. If the roots of the quadratic equation  ax² + bx + c = 0 are equal then show that b² = 4ac.

Sol. ∵ For equal roots, we have
b² − 4ac =0
∴ b² = 4ac

Q2. Find the value of ‘k’ for which the quadratic equation kx² − 5x + k = 0 have real roots.

Sol. Comparing kx² − 5x + k = 0 with ax² + bx + c = 0, we have:

a = k
b = − 5
c = k

∴ b² − 4ac =(− 5)² − 4 (k) (k)
= 25 − 4k² 

For equal roots, b² − 4ac = 0
∴ 25 − 4k² = 0
⇒ 4k² = 25

⇒ k² = 25/4
⇒ k = ± √ 254  = ± 52 

Q3. 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.

Sol. Since, 2 is a root of x² + kx + 12 = 0
∴ (2)² + k(2) + 12 = 0
or 4 + 2k + 12 = 0
⇒ 2k = −16 or  k = − 8
Roots of x² + kx + q = 0 are equal
∴ k² − 4(1) (q) = 0  or  k² − 4q = 0
But k = −8, so (−8)² = 4q or q = 16

Q4. If − 4 is a root of the quadratic equation x² + px − 4 = 0 and x² + px + k = 0 has equal roots, find the value of k.

Sol. ∵ (–4) is a root of x² + px − 4 = 0
∴ (− 4)² + p (− 4) - 4 = 0
⇒ 16 − 4p − 4 = 0
⇒ 4p = 12 or p = 3
Now, x² + px + k = 0

⇒ Putting the value of p = 3
⇒ x² + 3x + k = 0 [∵ p = 3]
Now, a = 1, b = 3 and c = + k
∴ b² − 4ac = (3)² − 4 (1) (k)
= 9 − 4k
For equal roots, b² − 4ac = 0

⇒ 9 − 4k =0 ⇒ 4k = 9

⇒ k = 9/4

Q5. If one root of the quadratic equation 2x² − 3x + p = 0 is 3, find the other root of the quadratic equation. Also, find the value of p.

Sol. We have: 2x² − 3x + p = 0 ...(1)
∴ a = 2, b = − 3 and c = p

Since, the sum of the roots = -b/a
 = -(-3)2 = 32
∵ One of the roots = 3
∴ The other root = 32 - 3  = 3 - 6 2 = -3 2 

Now, substituting x = 3 in (1), we get

2 (3)² − 3 (3) + p =0
⇒ 18 − 9 + p = 0
⇒ 9 + p = 0
⇒ p  = − 9

Q6. If one of the roots of x² + px − 4 = 0 is − 4 then find the product of its roots and the value of p.

Sol. If − 4 is a root of the quadratic equation,

x² + px − 4=0
∴ (− 4)² + (− 4) (p) − 4 = 0
⇒ 16 − 4p − 4 = 0
⇒ 12 − 4p = 0
⇒ p  = 3

Now, in ax² + bx + c = 0, the product of the roots = c/a

∴ Product of the roots in x² +  px − 4= 0 

= -4/1 = -4

Q7. For what value of k, does the given equation have real and equal roots? (k + 1) x² − 2 (k − 1) x + 1 = 0.

Sol. Comparing the given equation with ax² + bx + c = 0, we have:

a = k + 1
b = − 2 (k − 1)
c = 1

For equal roots, b² − 4ac = 0

∴ [− 2 (k − 1)]² − 4 (k + 1) (1) = 0
⇒ 4 (k − 1)²  − 4 (k + 1) = 0
⇒ 4 (k²  + 1 − 2k) − 4k − 4 = 0
⇒ 4k²  + 4 − 8k − 4k − 4 = 0
⇒ 4k²  − 12k = 0
⇒ 4k (k − 3) = 0
⇒ k = 0 or k = 3

Q8. Using quadratic formula, solve the following quadratic equation for x: 

x² − 2ax + (a² − b²) = 0

Sol. Comparing x² − 2ax + (a² − b²) = 0, with ax² + bx + c = 0, we have:

 a = 1, b = − 2a, c = a² − b²

x =  - b ± (√b² - 4ac)2a

x =  - (-2a)  ± (√(2a)² - 4(1)(a²- b²))2(1)

x =  2a  ± (√4a² - 4a² + 4b²)2

x =  2a  ± √ 4b²2

x =  2a  ±  2b2

x = a ± b 

x = a + b , a - b

∴ x =( a + b) or x = (a − b)

Q9. If one of the roots of the quadratic equation 2x² + kx − 6 = 0 is 2, find the value of k. Also, find the other root.

Sol. Given equation:

2x² + kx − 6= 0
one root = 2
Substituting x = 2 in 2x² + kx − 6 = 0
We have:

2 (2)² + k (2) − 6= 0
⇒ 8 + 2k − 6= 0
⇒ 2k + 2 = 0  ⇒ k = − 1
∴ 2x² + kx − 6 = 0  ⇒ 2x² − x − 6 = 0

Sum of the roots  = -b/a = 1/2
∴ other root =   1 2 - 2 
=  - 3 2

Q10. Determine the value of k for which the quadratic equation 4x² − 4kx + 1 = 0 has equal roots.

Sol. We have:

4x² − 4kx + 1 = 0
Comparing with ax² + bx + c = 0,
we have
a = 4, b = − 4k and c = 1
∴ b² − 4ac =(− 4k)² − 4 (4k) (1)
= 16k² − 16

For equal roots

b² − 4ac = 0

∴ 16k² − 16 = 0

⇒ 16k² = 16 ⇒ k² = 1
⇒ k = ± 1

Q11. For what value of k, does the quadratic equation x² − kx + 4 = 0 have equal roots?

Sol. Comparing  x² − kx + 4 = 0 with ax² + bx + c = 0, we get
a = 1
b = − k
c = 4

∴ b² − 4ac =(− k)² − 4 (1) (4) = k² − 16

For equal roots,

b² − 4ac =0
⇒ k² − 16 = 0
⇒ k² = 16
⇒ k = ±  4

Q12. What is the nature of roots of the quadratic equation 4x² − 12x + 9 = 0?

Sol. Comparing 4x² − 12x + 9 = 0 with ax² + bx + c = 0 we get

a = 4
b = − 12
c = 9

∴ b² − 4ac =(− 12)² − 4 (4) (9)
= 144 − 144 = 0

Since b² − 4ac = 0

∴ The roots are real and equal.

Q13. Write the value of k for which the quadratic equation x² − kx + 9 = 0 has equal roots.

Sol. Comparing x² − kx + 9 = 0 with ax² + bx + c = 0, we get

a = 1
b = − k
c = 9

∴ b² − 4ac = (− k)² − 4 (1) (9)
= k² − 36

For equal roots, b² − 4ac = 0
⇒ k² − 36 = 0  ⇒ k² = 36
⇒ k = ±  6

Q14. For what value of k are the roots of the quadratic equation 3x² + 2kx + 27 = 0 real and equal?

Sol. Comparing 3x² + 2 kx + 27 = 0 with ax² + bx + c = 0, we have:

a = 3
b = 2k
c = 27

∴ b² − 4ac = (2k)² − 4 (3) (27)
= 4k² − (12 × 27)

For the roots to be real and equal

b² − 4ac = 0
⇒ 4k² − (12 × 27) = 0
⇒ 4k² = 12 × 27

⇒ k²  =  12  x  274

 ⇒k² =  81

⇒ k = ± 9

Q15. For what value of k are the roots of the quadratic equation kx² + 4x + 1 = 0 equal and real?

Sol. Comparing kx² + 4x + 1 = 0, with ax² + bx + c = 0, we get

a = k
b = 4
c = 1

∴ b² − 4ac = (4)² − 4 (k) (1)
= 16 − 4k

For equal and real roots, we have

b² − 4ac =0
⇒ 16 − 4k = 0
⇒ 4k = 16

⇒ k = 16/4 = 4

Q16. For what value of k does (k − 12) x² + 2 (k − 12) x + 2 = 0 have equal roots?

Sol. Comparing (k − 12) x² + 2 (k − 12) x + 2 = 0 with ax² + bx + c = 0, we have:

a = (k − 12)
b = 2 (k − 12)
c = 2

∴ b² − 4ac = [2 (k − 12)]² − 4 (k − 12) (2)
= 4 (k − 12)² − 8 (k − 12)
= 4 (k − 12) [k − 12 − 2]
= 4 (k − 12) (k − 14)

For equal roots,

b² − 4ac =0
⇒ 4 (k − 12) [k − 14] = 0
⇒ Either 4 (k − 12) = 0 ⇒ k = 12

or k − 14 = 0 ⇒ k = 14

But k = 12 makes k − 12 = 0 which is not required

∴ k ≠ 12
However, k = 12 leads to a scenario where a = 0, which is not acceptable for a quadratic equation. Thus, we conclude: k = 14 is the only valid solution.

Q17. For what value of k does the equation 9x² + 3kx + 4 = 0 has equal roots?

Sol. Comparing 9x² + 3kx + 4 = 0 with ax² + bx + c = 0, we get

a = 9
b = 3k
c = 4

∴ b² − 4ac =(3k)² − 4 (9) (4)
= 9k² − 144
For equal roots,
b² − 4ac = 0
⇒ 9k² − 144 = 0
⇒9 k² = 144

⇒ k²  =  1449

⇒  k²  = 16
⇒ k = ±  4