Roots of parabola

Question

I have a parabola with the equation $y = x^2 + 6x + 7$ and I am trying to calculate the $x$-intercept points.

Here is my working so far...

1. let $y = 0$,
2. $x^2 + 6x + 7 = 0$
3. $x(x + 6) = -7$

After this I have no idea where to go next. Any help is appreciated.

Answer 1

You should probably find the roots by using this formula. For a quadratic equation $ax^{2}+bx+c$ we have the root's as: $$x = \frac{-b \pm \sqrt{D}}{2a}$$

where $D$ is the discriminant given by $D=b^{2}-4ac$. So in your case note that $a=1$, $b=6$ and $c=7$. Hope you can find your way from here. So $D=6^{2}-4\times 7 \times 1=8$. So you have the value of the roots as: $$ x= \frac{-6 \pm \sqrt{8}}{2 \times 1} = \frac{-6 \pm 2\sqrt{2}}{2} = -3 \pm \sqrt{2}$$

For more information regarding Quadratic Equations Please see:

• http://en.wikipedia.org/wiki/Quadratic_equation

Here is the Image of how your parabola will look like: