Question

How do you solve `y=5x^2+20x+23` using the completing square method?

Answer 1

`x = -2+-sqrt((y-3)/5)`

Explanation:

I will assume that we want to solve the equation in the sense of expressing possible values of `x` in terms of `y`.

If you actually just wanted to know the zeros, we can then put `y=0` and evaluate the result.

Given:

`y = 5x^2+20x+23`

We will perform a sequence of steps to isolate `x` on one side.

First divide both sides by `5` to make the coefficient of `x^2` into `1`:

`y/5 = x^2+4x+23/5`

If we take half of the coefficient of `x` then we get the value `2`, so note that:

`(x+2)^2 = x^2+4x+4`

which matches the right hand side of our equation in the first two terms.

So we can proceed as follows:

`y/5 = x^2+4x+23/5`

`color(white)(y/5) = x^2+4x+4-4+23/5`

`color(white)(y/5) = (x+2)^2+3/5`

Subtracting `3/5` from both ends we get:

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

Transposed:

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

Take the square root of both sides, allowing for the possibility of either sign of square root to get:

`x+2 = +-sqrt((y-3)/5)`

Finally subtract `2` from both sides to find:

`x = -2+-sqrt((y-3)/5)`

If we put `y=0` to find the zero of the quadratic then we get a pair of non-Real Complex values:

`x = -2+-sqrt((0-3)/5) = -2+-sqrt(-3/5) = -2+-sqrt(3/5)i = -2+-sqrt(15)/5i`