Question

Show that one root of the quadratic equation x² + (3 – 2a)x – 6a = 0 is –3. Hence, find its other root.

Answer 1

Given equation be x² + (3 – 2a)x – 6a = 0   ...(1)

Since –3 be the root of given equation (1)

∴ Putting x = –3 in equation (1); we have

9 + (3 – 2a)(–3) – 6a = 0

`\implies` 9 – 9 + 6a – 6a = 0

`\implies` 0 = 0, which is true

∴ From (1); x² + 3x – 2ax – 6a = 0

`\implies` x(x + 3) – 2a(x + 3) = 0

Either x + 3 = 0 or x – 2a = 0

i.e. x = –3 or x = 2a

∴ x = –3, 2a