Question

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

Answer 1

`x = 1/2` or `x=-3`

Explanation:

We will use the difference of squares identity, which can be written:

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

First, to reduce the amount of arithmetic involving fractions, first multiply through by `2^3 = 8` to get:

`0 = 16x^2+40x-24`

`= (4x)^2+2(5)(4x)-24`

`= (4x+5)^2-25-24`

`=(4x+5)^2-7^2`

`=((4x+5)-7)((4x+5)+7)`

`=(4x-2)(4x+12)`

`=(2(2x-1))(4(x+3))`

`=8(2x-1)(x+3)`

Hence:

`x = 1/2` or `x=-3`

`color(white)()`
Why did I premultiply by `8`?

One factor of `2` makes the leading term into a perfect square, then the additional factor `2^2 = 4` compensates for the middle term being an odd number.

Otherwise, we might proceed as follows:

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

`=2(x^2+5/2x-3/2)`

`=2((x+5/4)^2-25/16-3/2)`

`=2((x+5/4)^2-49/16)`

`=2((x+5/4)^2-7^2/4^2)`

`=2((x+5/4)-7/4)((x+5/4)+7/4)`

`=2(x-2/4)(x+12/4)`

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

Hence:

`x = 1/2` or `x=-3`

`color(white)()`
Which method do you prefer?