Question

How do you solve `x^2-3x-20=0` by completing the square?

Answer 1

`x = 3/2+-sqrt(89)/2`

Explanation:

The difference of squares identity can be written:

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

Use this with `a=2x-3` and `b=sqrt(89)` as follows:

`0 = 4(x^2-3x-20)`

`color(white)(0) = 4x^2-12x-80`

`color(white)(0) = (2x)^2-2(2x)(3)+9-89`

`color(white)(0) = (2x-3)^2-(sqrt(89))^2`

`color(white)(0) = ((2x-3)-sqrt(89))((2x-3)+sqrt(89))`

`color(white)(0) = (2x-3-sqrt(89))(2x-3+sqrt(89))`

Hence:

`x = 1/2(3+-sqrt(89)) = 3/2+-sqrt(89)/2`