# A question. Finding the intercepts in -5x^2+20x+2... Having some issues.

• Aug 21st 2012, 11:58 AM
OverlyFrustrated
A question. Finding the intercepts in -5x^2+20x+2... Having some issues.
So I've got -5x^2+20x+2, and have to determine the x intercepts.

I can use partial factoring to get -5x(x-4)+2

from which I can get

0=-5x
0/-5=x
0=x
AND

0=x-4
4=x

Which gives me 0 and 4.

My issues that when I enter -5x^2+20x+2 into a graphing calculator, the x-intercept coordinates I am given are -0.098 and 4.098. I'd like to know why. Have I done something wrong?

Thanks so much for the help!
• Aug 21st 2012, 12:22 PM
ebaines
Re: A question. Finding the intercepts in -5x^2+20x+2... Having some issues.
The x-interceopt are the values for x where -5x^2+20x+2 = 0. The mistake you made is going from -5x(x-4)+2=0 to -5x(x-4) = 0; what happened to the 2? You should always double check your answers to see if they work. If you plug x=0 or x = 4 into the original equation you get:

$-5(0^2)+20(0)+2 = 0$,
$2 = 0$

and:

$-5(4^2)+20(4)+2 =0$,
$2 = 0$

so clearly these answers are not right.

Easiest way to find the roots is to apply the quadratic equation:

$x = \frac {-20 \pm \sqrt {20^2-4(-5)(2)}} {2(-5)}$
• Aug 21st 2012, 12:27 PM
Plato
Re: A question. Finding the intercepts in -5x^2+20x+2... Having some issues.
Quote:

Originally Posted by OverlyFrustrated
So I've got -5x^2+20x+2, and have to determine the x intercepts.

Well the roots are $2\pm$ ${\sqrt{110}/{5}$
• Aug 21st 2012, 01:06 PM
OverlyFrustrated
Re: A question. Finding the intercepts in -5x^2+20x+2... Having some issues.
Hi there,

Thanks for taking the time to respond.

I've read the question I was basing this answer off of again and realize that there was no C value in the example I was provided.