Question

How do you solve `2x^2-3x-14=0` by completing the square?

Answer 1

`x=7/2" "` or `" "x=-2`

Explanation:

The difference of squares identity can be written:

`A^2-B^2 = (A-B)(A+B)`

We will use this with `A=(4x-3)` and `B=11` later.

`color(white)()`
Given:

`2x^2-3x-14 = 0`

To avoid fractions as much as possible, let us premultiply this quadratic by `8 = 2*2^2`. One factor of `2` makes the leading term a perfect square, then the other `2^2` deals with the `2` denominator in "`b/(2a)`" which would otherwise cause us to have to work with `1/2`'s.

So:

`0 = 8(2x^2-3x-14)`

`color(white)(0) = 16x^2-24x-112`

`color(white)(0) = (4x)^2-2(4x)(3)+9-121`

`color(white)(0) = (4x-3)^2-11^2`

`color(white)(0) = ((4x-3)-11)((4x-3)+11)`

`color(white)(0) = (4x-14)(4x+8)`

`color(white)(0) = (2(2x-7))(4(x+2))`

`color(white)(0) = 8(2x-7)(x+2)`

Hence:

`x=7/2" "` or `" "x=-2`