Question

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

Answer 1

`x = 1/4+-sqrt(41)/4`

Explanation:

Premultiply by `8` (to avoid some fractions), complete the square and use the difference of squares identity:

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

with `A=(4x-1)` and `B=sqrt(41)` as follows:

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

`color(white)(0) = 16x^2-8x-40`

`color(white)(0) = (4x)^2-2(4x)+1-41`

`color(white)(0) = (4x-1)^2-(sqrt(41))^2`

`color(white)(0) = ((4x-1)-sqrt(41))((4x-1)+sqrt(41))`

`color(white)(0) = (4x-1-sqrt(41))(4x-1+sqrt(41))`

So:

`4x = 1+-sqrt(41)`

and:

`x = 1/4+-sqrt(41)/4`

Note that multiplying by `8` has a couple of effects:

• It makes the leading term into a perfect square ready for completing the square.

• It makes the coefficient of `x` into a multiple of twice the square root of the leading coefficient, thus avoiding having to use fractions until the end.