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

Jun 11, 2017

$x = - 2 \pm \sqrt{3}$

Explanation:

First, expand the polynomial.

$- {\left(x + 2\right)}^{2} + 3 = 0$

$- \left({x}^{2} + 4 x + 4\right) + 3 = 0$

$- {x}^{2} - 4 x - 1 = 0$

You can multiply by -1 to make things easier.

${x}^{2} + 4 x + 1 = 0$

So we see that

$a = 1$
$b = 4$
$c = 1$

Finally, plug these values into the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(1\right) \left(1\right)}}{2 \left(1\right)}$

$x = - 2 \pm \frac{\sqrt{12}}{2}$

$x = - 2 \pm \frac{2 \sqrt{3}}{2}$

$x = - 2 \pm \sqrt{3}$

Jun 12, 2017

The quadratic formula is a fine way to solve this problem, but if we want to find the roots of $q$, when $q \left(x\right) = 0$, 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.

$0 = - {\left(x + 2\right)}^{2} + 3$

Add ${\left(x + 2\right)}^{2}$ to both sides.

${\left(x + 2\right)}^{2} = 3$

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

$x + 2 = \pm \sqrt{3}$

Subtract $2$ from both sides.

$x = - 2 \pm \sqrt{3}$