Question

Solve by completing the square:

x² − 2ax + 3x − 6a = 0

Answer 1

Given:

x² − 2ax + 3x − 6a = 0

Complete the square:

`x^2 + (3 − 2a)x = 6a `

Add `[(3 − 2a)/2]^2` to both sides:

`(x + (3 − 2a)/2)^2 = 6a + (3 − 2a)^2/4`

Compute the right-hand side:

6`a + (9 − 12a + 4a^2)/4 = (4a^2 + 12a + 9)/4 = ((2a + 3)/2)^2`

So `(x + (3 − 2a)/2)^2 = ((2a + 3)/2)^2`

Take square roots:

`x + (3 − 2a)/2 = ±(2a + 3)/2`

x = 2a or x = −3