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**Q1. If the roots of the quadratic equation − ax ^{2} + bx + c = 0 are equal then show that b^{2} = 4ac.**

**Sol. **∵ For equal roots, we have

b^{2} − 4ac =0

∴ b^{2} = 4ac**Q2. Find the value of ‘k’ for which the quadratic equation kx ^{2} − 5x + k = 0 have real roots.**

**Sol. **Comparing kx^{2} − 5x + k = 0 with ax^{2} + bx + c = 0, we have:

a = k

b = − 5

c = k

∴ b^{2} − 4ac =(− 5)^{2} − 4 (k) (k)

= 25 − 4k^{2}

For equal roots, b^{2} − 4ac = 0

∴ 25 − 4k^{2} = 0

⇒ 4k^{2} = 25

⇒ k^{2 }= 25/4

⇒

∴ b^{2} = 4ac**Q3. If 2 is a root of the equatio**n x^{2} + kx + 12 = 0 and the equation x^{2} + kx + q = 0 has equal roots, find the value of q.

**Sol. **Since, 2 is a root of x^{2} + kx + 12 = 0

∴ (2)^{2} + k(2) + 12 = 0

or 4 + 2k + 12 = 0

⇒ 2k = −16 or k = − 8

Roots of x^{2} + kx + q = 0 are equal

∴ k^{2} − 4(1) (q) = 0 ^{or }k^{2} − 4q = 0

But k = −8, so (−8)^{2} = 4q or q = 16**Q4. If − 4 is a root of the quadratic equation** x^{2} + px − 4 = 0 and x^{2} + px + k = 0 has equal roots, find the value of k.

**Sol. **∵ (–4) is a root of x^{2 }+ px − 4 = 0

∴ (− 4)^{2} + p (− 4) = 0

⇒ 16 − 4p − 4= 0

⇒ 4p = 12 or p = 3

Now, x^{2} + px + k = 0

⇒ x^{2} + 3x + k = 0 [∵ p = 3]

Now, a = 1, b = 3 and c = + k

∴ b^{2} − 4ac = (3)^{2} − 4 (1) (k)

= 9 − 4k

For equal roots, b^{2} − 4ac = 0

⇒ 9 − 4k =0 ⇒ 4k = 9

⇒ k = 9/4**Q5. If one root of the quadratic equation 2x ^{2} − 3x + p = 0 is 3, find the other root of the quadratic equation. Also, find the value of p.**

**Sol. **We have: 2x^{2} − 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 x ^{2} + 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,

x^{2} + px − 4=0

∴ (− 4)^{2} + (− 4) (p) − 4 = 0

⇒ 16 − 4p − 4 = 0

⇒ 12 − 4p = 0

⇒ p = 3

Now, in ax^{2} + bx + c = 0, the product of the roots = c/a

∴ Product of the roots in x^{2} − px − 4= 0

= -4/1 = -4**Q7. For what value of k, does the given equation have real and equal roots? (k + 1) x ^{2} − 2 (k − 1) x + 1 = 0.**

**Sol. **Comparing the given equation with ax^{2} + bx + c = 0, we have:

a = k + 1

b = − 2 (k − 1)

c = 1

For equal roots, b^{2} − 4ac = 0

∴ [− 2 (k − 1)]^{2 }− 4 (k + 1) (1) = 0

⇒ 4 (k − 1)^{2 } − 4 (k + 1) = 0

⇒ 4 (k^{2 } + 1 − 2k) − 4k − 4 = 0

⇒ 4k^{2 } + 4 − 8k − 4k − 4 = 0

⇒ 4k^{2 } − 12k = 0

⇒ 4k (k − 3) = 0

⇒ k = 0 or k = 3**Q8. Using quadratic formula, solve the following quadratic equation for x: **

**x ^{2} − 2ax + (a^{2} − b^{2}) = 0**

**Sol. **Comparing x^{2} − 2ax + (a^{2} − b^{2}) = 0, with ax^{2} + bx + c = 0, we have:

a = 1, b = − 2a, c = a^{2} − b^{2}

∴ x =( a + b) or x = (a − b)**Q9. If one of the roots of the quadratic equation 2x ^{2} + kx − 6 = 0 is 2, find the value of k. **

**Sol. **Given equation:

2x^{2} + kx − 6= 0

one root = 2

Substituting x = 2 in 2x^{2 }+ kx − 6 = 0

We have:

2 (2)^{2} + k (2) − 6= 0

⇒ 8 + 2k − 6= 0

⇒ 2k + 2 = 0 ⇒ k = − 1

∴ 2x^{2} + kx − 6 = 0 ⇒ 2x^{2} − 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 4x ^{2} − 4kx + 1 = 0 has equal roots.**

**Sol.** We have:

4x^{2} − 4kx + 1 = 0

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

we have

a = 4, b = − 4k and c = 1

∴ b^{2} − 4ac =(− 4k)^{2} − 4 (4k) (1)

= 16k^{2} − 16

For equal roots

b^{2} − 4ac = 0

∴ 16k^{2} − 16 = 0

⇒ 16k^{2} = 16 ⇒ k^{2} = 1

⇒ k = ± 1**Q11. For what value of k, does the quadratic equation x ^{2} − kx + 4 = 0 have equal roots?**

**Sol. **Comparing x^{2} − kx + 4 = 0 with ax^{2} + bx + c = 0, we get

a = 1

b = − k

c = 4

∴ b^{2} − 4ac =(− k)^{2} − 4 (1) (4) = k^{2} − 16

For equal roots,

b^{2} − 4ac =0

⇒ k^{2} − 16 = 0

⇒ k^{2} = 16

⇒ k = ± 4**Q12. What is the nature of roots of the quadratic equation 4x ^{2} − 12x + 9 = 0?**

**Sol.** Comparing 4x^{2} − 12x + 9 = 0 with ax^{2} + bx + c = 0 we get

a = 4

b = − 12

c = 9

∴ b^{2} − 4ac =(− 12)^{2} − 4 (4) (9)

= 144 − 144 = 0

Since b^{2} − 4ac = 0

∴ The roots are real and equal.

**Q13. Write the value of k for which the quadratic equation x ^{2} − kx + 9 = 0 has equal roots.**

**Sol.** Comparing x^{2 }− kx + 9 = 0 with ax^{2} + bx + c = 0, we get

a = 1

b = − k

c = 9

∴ b^{2} − 4ac = (− k)^{2} − 4 (1) (9)

= k^{2} − 36

For equal roots, b^{2} − 4ac = 0

⇒ k^{2} − 36 = 0 ⇒ k^{2} = 36

⇒ k = ± 6**Q14. For what value of k are the roots of the quadratic equation 3x ^{2} + 2kx + 27 = 0 real and equal?**

**Sol.** Comparing 3x^{2} + 2 kx + 27 = 0 with ax^{2} + bx + c = 0, we have:

a = 3

b = 2k

c = 27

∴ b^{2} − 4ac = (2k)^{2} − 4 (3) (27)

= 4k^{2} − (12 × 27)

For the roots to be real and equal

b^{2} − 4ac = 0

⇒ 4k^{2} − (12 × 17) = 0

⇒ 4k^{2} = 12 × 27

⇒

⇒ k = ± 9**Q15. For what value of k are the roots of the quadratic equation kx ^{2} + 4x + 1 = 0 equal and real?**

**Sol.** Comparing kx^{2} + 4x + 1 = 0, with ax^{2} + bx + c = 0, we get

a = k

b = 4

c = 1

∴ b^{2} − 4ac = (4)^{2 }− 4 (k) (1)

= 16 − 4k

For equal and real roots, we have

b^{2} − 4ac =0

⇒ 16 − 4k = 0

⇒ 4k = 16

⇒ k = 16/4 = 4**Q16. For what value of k does (k − 12) x ^{2} + 2 (k − 12) x + 2 = 0 have equal roots?**

**Sol.** Comparing (k − 12) x^{2} + 2 (k − 12) x + 2 = 0 with ax^{2} + bx + c = 0, we have:

a = (k − 12)

b = 2 (k − 12)

c = 2

∴ b^{2} − 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,

b^{2} − 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 9x ^{2} + 3kx + 4 = 0 has equal roots?**

**Sol.** Comparing 9x^{2} + 3kx + 4 = 0 with ax^{2} + bx + c = 0, we get

a = 9

b = 3k

c = 4

∴ b^{2} − 4ac =(3k)^{2} − 4 (9) (4)

= 9k^{2} − 144

For equal roots,

b^{2} − 4ac = 0

⇒ 9k^{2} − 144 = 0

⇒9 k^{2} = 144

⇒

⇒ k = ± 4

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