# What are the x-intercepts of x²-y+2x=6 ?

Dec 5, 2017

The x-intercepts are at $\left(- 1 + \sqrt{7} , 0\right)$ and $\left(- 1 - \sqrt{7} , 0\right)$

#### Explanation:

We first isolate y.

${x}^{2} - y + 2 x = 6$
$- y + 2 x = 6 - {x}^{2}$
$- y = 6 - {x}^{2} - 2 x$
$y = - 6 + {x}^{2} + 2 x$
$y = {x}^{2} + 2 x - 6$

We now try to find the zeros. Since we cannot factor this equation in any quick way, we find the two zeros using the quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 \left(a\right) \left(c\right)}}{2 \left(a\right)}$
$a = 1$
$b = 2$
$c = - 6$

The zeros are:
$\frac{- 2 \pm \sqrt{{2}^{2} - 4 \left(1\right) \left(- 6\right)}}{2 \left(1\right)}$
$\frac{- 2 \pm \sqrt{4 + 24}}{2}$
$\frac{- 2 \pm \sqrt{28}}{2}$
$\frac{- 2 \pm 2 \sqrt{7}}{2}$
$- 1 \pm \sqrt{7}$

$- 1 \pm \sqrt{7}$ are the x values for the x-intercepts.

Note that we know that(and it will, if you are correct) when we plug in the zeros, we get zero for the equation.

*Here is the idea of how the zeros are the answer. Since the y value is 0, we know that the point must lie on the x coordinate, which means those points are the x intercepts.