Question

In each of the following find the values of k of which the given value is a solution of the given equation:
x² + 3ax + k = 0; x = a.

Answer 1

x² + 3ax + k = 0; x = a.
Putting x = -a in L.H.S. of equation
L.H.S = (-a)² + 3a x (-a) + k = 0
⇒ a² - 3a² + k = 0
⇒ -2a² + k = 0
Hence, k = 2a².