Question

How do you solve `2x^2+3x-5=0` by completing the square?

Answer 1

`x = 1" "` or `" "x = -5/2`

Explanation:

The difference of squares identity can be written:

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

We use this later with `a = (4x+3)` and `b = 7`

`color(white)()`
Given:

`2x^2+3x-5 = 0`

To delay having to work with fractions, first multiply this by `8`.

Why `8`?

We want to multiply by `2` to make the leading term into a perfect square, namely `4x^2 = (2x)^2`. Then the middle term becomes `6x=3(2x)`. We want the coefficient `3` of `(2x)` to be even too, so multiply by another factor of `2^2 = 4` to keep the leading term a perfect square.

Then we find:

`0 = 8(2x^2+3x-5)`

`color(white)(0) = 16x^2+24x-40`

`color(white)(0) = (4x)^2+2(3)(4x)+9-49`

`color(white)(0) = (4x+3)^2-7^2`

`color(white)(0) = ((4x+3)-7)((4x+3)+7)`

`color(white)(0) = (4x-4)(4x+10)`

`color(white)(0) = 4(x-1)2(2x+5)`

Hence:

`x = 1" "` or `" "x = -5/2`