# Need help with quadratic equation.

• Aug 31st 2011, 07:43 AM
bseymore
Can anyone show me how to solve this using the quadratic equation?

e^3=x^2 + 6x + 8

It says in the solution you must group e^3 with 8. Therefore making the problem...

x^2 + 6x -(e^3-8)=0

I'm just stuck on how you put this into the quadratic equation.
• Aug 31st 2011, 07:50 AM
Siron
Re: Need help with quadratic equation.
Notice $\displaystyle e^3$ is also a constant so rewrite the equation as:
$\displaystyle x^2+6x+(8-e^3)=0$

EDIT:
I notice you have modified your post.
Calculate $\displaystyle D=b^2-4ac=6^2-4.1.(8-e^3)=...$

Afterwards you just have to substitute the discriminant into:
$\displaystyle x_1,x_2=\frac{-b\pm \sqrt{D}}{2a}$

Where $\displaystyle x_1$ and $\displaystyle x_2$ are the two solutions of the equation.
• Aug 31st 2011, 08:53 AM
bseymore
Re: Need help with quadratic equation.
Can you possibly show the work to solve for x? The answer key says....

x= -3 + sqrt(1+e^3)

When I work out the problem, I get....

x= -3 + sqrt(4e^3)

What am I doing incorrectly?
• Aug 31st 2011, 09:07 AM
Siron
Re: Need help with quadratic equation.
If you calculate the discriminant you get:
$\displaystyle D=(6)^2-4\cdot (8-e^3)=36-32+4e^3=4+4e^3=4(1+e^3)$
So the solutions are:
$\displaystyle x_1=\frac{-6+\sqrt{4(1+e^3)}}{2}=\frac{-6}{2}+\frac{2\sqrt{1+e^3}}{2}=-3+\sqrt{1+e^3}$
$\displaystyle x_2=\frac{-6-\sqrt{4(1+e^3)}}{2}=-3-\sqrt{1+e^3}$
• Aug 31st 2011, 10:54 AM
Prove It
Re: Need help with quadratic equation.
Quote:

Originally Posted by bseymore
Can anyone show me how to solve this using the quadratic equation?

e^3=x^2 + 6x + 8

It says in the solution you must group e^3 with 8. Therefore making the problem...

x^2 + 6x -(e^3-8)=0

I'm just stuck on how you put this into the quadratic equation.

\displaystyle \displaystyle \begin{align*} x^2 + 6x + 8 &= e^3 \\ x^2 + 6x &= e^3 - 8 \\ x^2 + 6x + 3^2 &= e^3 - 8 + 3^2 \\ (x + 3)^2 &= e^3 + 1 \\ x + 3 &= \pm \sqrt{e^3 + 1} \\ x &= -3 \pm \sqrt{e^3 + 1} \end{align*}