Question

How do you solve `q(x) = -(x+2)^2+3` using the quadratic formula?

Answer 1

`x = -2+-sqrt3`

Explanation:

First, expand the polynomial.

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

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

`-x^2 -4x -1 = 0`

You can multiply by -1 to make things easier.

`x^2 + 4x + 1 = 0`

So we see that

`a=1`
`b=4`
`c=1`

Finally, plug these values into the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`x = (-4+-sqrt(4^2-4(1)(1)))/(2(1))`

`x = -2 +- sqrt12/2`

`x = -2 +- (2sqrt3)/2`

`x = -2 +- sqrt3`

Final Answer

Answer 2

The quadratic formula is a fine way to solve this problem, but if we want to find the roots of `q`, when `q(x)=0`, this problem is already set up very well for a quick and easy algebraic manipulation to find the roots.

If you're familiar with completing the square to find the roots of a quadratic equation, this is essentially already in the "completed square" form.

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

Add `(x+2)^2` to both sides.

`(x+2)^2=3`

Take the square root of both sides, not forgetting the `pm` sign.

`x+2=pmsqrt3`

Subtract `2` from both sides.

`x=-2pmsqrt3`