Question

How do you solve `2x ^ { 2} + 12x + 9= 0`?

Answer 1

`x = -3+-(3sqrt(2))/2`

Explanation:

Complete the square then use the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=2x+6` and `b=3sqrt(2)` as follows:

`0 = 2(2x^2+12x+9)`

`color(white)(0) = 4x^2+24x+18`

`color(white)(0) = (2x)^2+2(2x)(6)+(6)^2-18`

`color(white)(0) = (2x+6)^2-(3sqrt(2))^2`

`color(white)(0) = ((2x+6)-3sqrt(2))((2x+6)+3sqrt(2))`

`color(white)(0) = (2x+6-3sqrt(2))(2x+6+3sqrt(2))`

Hence:

`x = 1/2(-6+-3sqrt(2)) = -3+-(3sqrt(2))/2`