Question

If x² - bx/ax - c = k - 1/k + 1 has roots which are numerically equal but of opposite signs, then k is equal to

(a)  a -b/a +b

(b)  a + b/a - b

(c)  c

(d)  1/c

Answer 1

Correct option (a)  a - b/a + b

The equation  x² - bc/ax - c = k - 1/k + 1 can be written as

(k + 1)² - [a(k - 1)+ b(k + 1)]x + c(k - 1) = 0

The roots are equal in magnitude but opposite in sign means sum of the roots = 0. Therefore,

a(k - 1)+ b(k + 1)  = 0

⇒ k = a - b/a + b