Question

The integral value of k for which the equation (k-12)x2+2(k-12)x+2=0 possess no real solution is

Answer 1

Answer:

Given :-

➪ (k - 12)x² + 2(k - 12)x + 2 = 0

Solution :-

➪ (k - 12)x² + 2(k - 12)x + 2 = 0

Comparing it with quadratic equation

➭ ax² + bx + c = 0

a = (k - 12)

b = 2(k - 12)

c = 2

Discriminant (D)

➭ b² - 4ac

➭ 4(k - 12)² - 4 × (k - 12) × 2

➭ 4(k - 12) [k - 12 - 2]

➭ 4(k - 12) (k - 24)

Given equation will have real and equal roots, if discriminant = 0

D = 0

➭ 4(k - 12) (k - 14) = 0

➭ (k - 12) = 0

➭ k - 12 = 0

➭ k = 12

Either,

➭ (k - 14) = 0

➭ k - 14 = 0

➭ k = 14

∴ The value of k is 12, 14