Question

Find the value of k for wnich the equation (k- 12)x2 + 2(k - 12)x + 2 = 0 has real and equal roots.​

Answer 1

Answer:

The given equation is

(k - 12)x^2 + 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 b^2 - 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 x^2 and x in the equation vanish hence only k = 14 is acceptable.