Question

How do you solve `2x^2+5x-1=0` by completing the square?

Answer 1

`2(x+1.25)^2-4.125=0`

Explanation:

We first take the first two terms and factor out the coefficient of `x^2`:
`(2x^2)/2+(5x)/2=2(x^2+2.5x)`

Then we divide by `x`, half the integer and square what remains:
`2(x^2/x+2.5x/x)2=2(x+2.5)`
`2(x+2,5/2)=2(x+1.25)`
`2(x+1..25)^2`

Expand out the bracket:
`2x^2+2.5x+2.5x+2(1.25^2)=2x^2+5x+3.125`

Make it equal the original equations:
`2x^2+5x+3.125+a=2x^2+5x-1`

Rearrange to find `a`:
`a=-1-3.125=-4.125`

Put in `a` to the factorised equation:
`2(x+1.25)^2-4.125=0`