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

⇒ k= 25/4

∴    b2 = 4ac

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 x+ px − 4 = 0
∴ (− 4)2 + p (− 4) = 0
⇒ 16 − 4p − 4= 0
⇒ 4p = 12 or p = 3
Now, x2 + px + k = 0
⇒ 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

∵ One of the roots = 3
∴ The other root

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

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 =( 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 2x+ 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
= 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 x− 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 × 17) = 0
⇒ 4k2 = 12 × 27

⇒ 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)− 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
⇒ k = 14

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

⇒ k = ±  4

The document Class 10 Maths Chapter 4 Question Answers - Quadratic Equations is a part of the Class 10 Course Mathematics (Maths) Class 10.
All you need of Class 10 at this link: Class 10

## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 4 Question Answers - Quadratic Equations

 1. What is a quadratic equation?
Ans. A quadratic equation is a polynomial equation of the second degree, which means it contains a variable raised to the power of 2. The general form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants.
 2. How do you solve a quadratic equation?
Ans. There are several methods to solve a quadratic equation, including factoring, completing the square, and using the quadratic formula. Factoring involves finding two numbers that multiply to give the constant term and add up to give the coefficient of the x-term. Completing the square involves manipulating the equation to create a perfect square trinomial. The quadratic formula is a formula that gives the solutions of any quadratic equation.
 3. What are the roots of a quadratic equation?
Ans. The roots of a quadratic equation are the values of x that satisfy the equation and make it equal to zero. A quadratic equation can have two real roots, two complex roots, or one real root depending on the discriminant (b^2 - 4ac). If the discriminant is positive, the equation has two distinct real roots. If it is zero, the equation has one real root. If it is negative, the equation has two complex roots.
 4. Can a quadratic equation have only one root?
Ans. Yes, a quadratic equation can have only one root. This happens when the discriminant (b^2 - 4ac) of the equation is equal to zero. In this case, the equation has a repeated root, also known as a double root. The graph of the equation will touch the x-axis at a single point.
 5. What is the significance of quadratic equations in real life?
Ans. Quadratic equations have various applications in real-life situations. They are used in physics to model the motion of objects under the influence of gravity. They are also used in engineering to calculate the trajectory of projectiles. In finance, quadratic equations are used to calculate the maximum profit or minimum cost. Additionally, quadratic equations are used in architecture and design to determine the shape of structures and objects.

## Mathematics (Maths) Class 10

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