Question

Find the value of ‘K’ for which x = 3 is a solution of the quadratic equation

`(K + 2)x^2 - kx + 6 = 0`

Answer 1

`(K + 2)x^2 - Kx + 6 = 0`     ...(1)

Substituting x = 3 in equation (1), we get

`(K + 2) (3)^2 - K(3) + 6 = 0`

`:. 9(K +2) - 3K + 6 = 0`

`:. 9K + 18 - 3k + 6 = 0`

`:. 6k + 24 = 0`

`:. K= -4`

Now, substituting K = –4 in equation (1), we get

`(-4 + 2)x^2 - (-4)x + 6 = 0`

`:. -2x^2 + 4x + 6 = 0`

`:. x^2 - 2x - 3 = 0`

`:. x^2 -  3x + x - 3 = 0`

`:. x(x - 3) + 1(x- 3) = 0`

`:. (x + 1) (x + 3) = 0`

So, the roots are x = –1 and x = 3.

Thus the other root of the equation is x = -1.