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**VERY SHORT ANSWER TYPE QUESTIONS**

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

**Sol. **âˆµ For equal roots, we have

b^{2} âˆ’ 4ac =0

âˆ´ b^{2} = 4ac**Q2. Find the value of â€˜kâ€™ for which the quadratic equation kx ^{2} âˆ’ 5x + k = 0 have real roots.**

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

a = k

b = âˆ’ 5

c = k

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 5)^{2} âˆ’ 4 (k) (k)

= 25 âˆ’ 4k^{2}

For equal roots, b^{2} âˆ’ 4ac = 0

âˆ´ 25 âˆ’ 4k^{2} = 0

â‡’ 4k^{2} = 25

â‡’ k^{2 }= 25/4

â‡’

âˆ´ 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

âˆ´ (2)^{2} + k(2) + 12 = 0

or 4 + 2k + 12 = 0

â‡’ 2k = âˆ’16 or k = âˆ’ 8

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

âˆ´ k^{2} âˆ’ 4(1) (q) = 0 ^{or }k^{2} âˆ’ 4q = 0

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

**Sol. **âˆµ (â€“4) is a root of x^{2 }+ px âˆ’ 4 = 0

âˆ´ (âˆ’ 4)^{2} + p (âˆ’ 4) = 0

â‡’ 16 âˆ’ 4p âˆ’ 4= 0

â‡’ 4p = 12 or p = 3

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

â‡’ x^{2} + 3x + k = 0 [âˆµ p = 3]

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

âˆ´ b^{2} âˆ’ 4ac = (3)^{2} âˆ’ 4 (1) (k)

= 9 âˆ’ 4k

For equal roots, b^{2} âˆ’ 4ac = 0

â‡’ 9 âˆ’ 4k =0 â‡’ 4k = 9

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

**Sol. **We have: 2x^{2} âˆ’ 3x + p = 0 ...(1)

âˆ´ a = 2, b = âˆ’ 3 and c = p

Since, the sum of the roots = -b/a

âˆµ One of the roots = 3

âˆ´ The other root

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

2 (3)^{2} âˆ’ 3 (3) + p =0

â‡’ 18 âˆ’ 9 + p = 0

â‡’ 9 + p = 0

â‡’ p = âˆ’ 9**Q6. If one of the roots of x ^{2} + px âˆ’ 4 = 0 is âˆ’ 4 then find the product of its roots and the value of p.**

**Sol.** If âˆ’ 4 is a root of the quadratic equation,

x^{2} + px âˆ’ 4=0

âˆ´ (âˆ’ 4)^{2} + (âˆ’ 4) (p) âˆ’ 4 = 0

â‡’ 16 âˆ’ 4p âˆ’ 4 = 0

â‡’ 12 âˆ’ 4p = 0

â‡’ p = 3

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

âˆ´ Product of the roots in x^{2} âˆ’ px âˆ’ 4= 0

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

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

a = k + 1

b = âˆ’ 2 (k âˆ’ 1)

c = 1

For equal roots, b^{2} âˆ’ 4ac = 0

âˆ´ [âˆ’ 2 (k âˆ’ 1)]^{2 }âˆ’ 4 (k + 1) (1) = 0

â‡’ 4 (k âˆ’ 1)^{2 } âˆ’ 4 (k + 1) = 0

â‡’ 4 (k^{2 } + 1 âˆ’ 2k) âˆ’ 4k âˆ’ 4 = 0

â‡’ 4k^{2 } + 4 âˆ’ 8k âˆ’ 4k âˆ’ 4 = 0

â‡’ 4k^{2 } âˆ’ 12k = 0

â‡’ 4k (k âˆ’ 3) = 0

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

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

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

a = 1, b = âˆ’ 2a, c = a^{2} âˆ’ b^{2}

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

**Sol. **Given equation:

2x^{2} + kx âˆ’ 6= 0

one root = 2

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

We have:

2 (2)^{2} + k (2) âˆ’ 6= 0

â‡’ 8 + 2k âˆ’ 6= 0

â‡’ 2k + 2 = 0 â‡’ k = âˆ’ 1

âˆ´ 2x^{2} + kx âˆ’ 6 = 0 â‡’ 2x^{2} âˆ’ x âˆ’ 6 = 0

Sum of the roots = -b/a = 1/2

âˆ´ other root

= 3/2**Q10. Determine the value of k for which the quadratic equation 4x ^{2} âˆ’ 4kx + 1 = 0 has equal roots.**

**Sol.** We have:

4x^{2} âˆ’ 4kx + 1 = 0

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

we have

a = 4, b = âˆ’ 4k and c = 1

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 4k)^{2} âˆ’ 4 (4k) (1)

= 16k^{2} âˆ’ 16

For equal roots

b^{2} âˆ’ 4ac = 0

âˆ´ 16k^{2} âˆ’ 16 = 0

â‡’ 16k^{2} = 16 â‡’ k^{2} = 1

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

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

a = 1

b = âˆ’ k

c = 4

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ k)^{2} âˆ’ 4 (1) (4) = k^{2} âˆ’ 16

For equal roots,

b^{2} âˆ’ 4ac =0

â‡’ k^{2} âˆ’ 16 = 0

â‡’ k^{2} = 16

â‡’ k = Â± 4**Q12. What is the nature of roots of the quadratic equation 4x ^{2} âˆ’ 12x + 9 = 0?**

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

a = 4

b = âˆ’ 12

c = 9

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 12)^{2} âˆ’ 4 (4) (9)

= 144 âˆ’ 144 = 0

Since b^{2} âˆ’ 4ac = 0

âˆ´ The roots are real and equal.

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

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

a = 1

b = âˆ’ k

c = 9

âˆ´ b^{2} âˆ’ 4ac = (âˆ’ k)^{2} âˆ’ 4 (1) (9)

= k^{2} âˆ’ 36

For equal roots, b^{2} âˆ’ 4ac = 0

â‡’ k^{2} âˆ’ 36 = 0 â‡’ k^{2} = 36

â‡’ k = Â± 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

âˆ´ b^{2} âˆ’ 4ac = (2k)^{2} âˆ’ 4 (3) (27)

= 4k^{2} âˆ’ (12 Ã— 27)

For the roots to be real and equal

b^{2} âˆ’ 4ac = 0

â‡’ 4k^{2} âˆ’ (12 Ã— 17) = 0

â‡’ 4k^{2} = 12 Ã— 27

â‡’

â‡’ k = Â± 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

âˆ´ b^{2} âˆ’ 4ac = (4)^{2 }âˆ’ 4 (k) (1)

= 16 âˆ’ 4k

For equal and real roots, we have

b^{2} âˆ’ 4ac =0

â‡’ 16 âˆ’ 4k = 0

â‡’ 4k = 16

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

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

a = (k âˆ’ 12)

b = 2 (k âˆ’ 12)

c = 2

âˆ´ b^{2} âˆ’ 4ac = [2 (k âˆ’ 12)]^{2} âˆ’ 4 (k âˆ’ 12) (2)

= 4 (k âˆ’ 12)^{2} âˆ’ 8 (k âˆ’ 12)

= 4 (k âˆ’ 12) [k âˆ’ 12 âˆ’ 2]

= 4 (k âˆ’ 12) (k âˆ’ 14)

For equal roots,

b^{2} âˆ’ 4ac =0

â‡’ 4 (k âˆ’ 12) [k âˆ’ 14] = 0

â‡’ Either 4 (k âˆ’ 12) = 0 â‡’ k = 12

or k âˆ’ 14 = 0 â‡’ k = 14

But k = 12 makes k âˆ’ 12 = 0 which is not required

âˆ´ k â‰ 12

â‡’ 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

âˆ´ b^{2} âˆ’ 4ac =(3k)^{2} âˆ’ 4 (9) (4)

= 9k^{2} âˆ’ 144

For equal roots,

b^{2} âˆ’ 4ac = 0

â‡’ 9k^{2} âˆ’ 144 = 0

â‡’9 k^{2} = 144

â‡’

â‡’ k = Â± 4

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