Question

Find the roots: ax² + (4a² − 3b) x − 12ab = 0

Answer 1

Given:

ax² + (4a² − 3b) x − 12ab = 0

We look for two terms whose product is:

a × (−12ab) = −12a²b

whose sum is: 4a² − 3b

ax² + 4a²x − 3bx − 12ab = 0

(ax² + 4a²x) − (3bx + 12ab) = 0

ax(x + 4a) − 3b(x + 4a) = 0

(x + 4a) (ax − 3b) = 0

x + 4a = 0

⇒ x = −4a

ax − 3b = 0

⇒ x = `(3b)/a`

x = −4a and x = `(3b)/a`