Question

Q. No 8
Find the value of k for which the quadratic equations

(k-12)x²+2(k-12)x+2=0

k²x²-2(k-1)x+4=0

have real and equal roots. ​

Answer 1

The given equation is

(k - 12)x2 + 2(k - 12)x + 2 = 0

Here, a = k - 12, b = 2(k - 12) and c = 2

Since, the given equation has two equal real roots

then we must have b2 - 4ac = 0

⇒ [2(k - 12)]2 - 4(k - 12) x 2 = 0

⇒ 4(k - 12)2 - 8(k - 12) = 0

⇒ 4(k - 12) {k - 12 - 2} = 0

⇒ (k - 12) (k - 14) = 0

⇒ k - 12 = 0 or k - 14 = 0

⇒ k = 12 or k = 14.

Note: But at k = 12, terms of x2 and x in the equation vanish hence only k = 14 is acceptable.

Answer 2

Answer:

see the above attachment for the answer.