# x intercept of parabola

• Sep 16th 2011, 12:22 PM
David Green
x intercept of parabola
Just a bit confused here and would appreciate a little guidance please.

Find the x intercept when y=0

0 = -x^2 + 4x +5

x^2 - 4x - 5 = 0

(x-1)(x-5) = 0 (Something here I think is not quite right)?

multiply out the brackets;

x^2 - 5x - x +5 = 0

x^2 - 6x + 5 = 0 (This can't be right it does'nt equal the original equation)?
• Sep 16th 2011, 12:28 PM
Siron
Re: x intercept of parabola
Note: Recalculate the discriminant. It has to be:
$-x^2+4x+5=(x+1)\cdot (x-5)$
Because:
$(x+1)\cdot (x-5)=x^2-5x+x-5=x^2-4x-5$
• Sep 16th 2011, 12:32 PM
David Green
Re: x intercept of parabola
Thanks for that Siron, I see where I was going wrong now, I said (x - 1) you pointed out (x + 1)

So now it works

Thanks
• Sep 16th 2011, 12:35 PM
Siron
Re: x intercept of parabola
You're welcome! In this case using the discriminant is not necessary, you can just factore it directly. But if you want, you'll see -1 is a zero in stead of 1.
• Sep 16th 2011, 12:51 PM
David Green
Re: x intercept of parabola
Quote:

Originally Posted by Siron
Note: Recalculate the discriminant. It has to be:
$-x^2+4x+5=(x+1)\cdot (x-5)$
Because:
$(x+1)\cdot (x-5)=x^2-5x+x-5=x^2-4x-5$

This I think I found a problem?

(x + 1)(x - 5) = x^2 -5x + x - 5

= x^2 -4x - 5 ?

should minus 5 not be +5?

(x - 1)(x + 5) = x^2 +5x - x + 5

= x^2 + 4x + 5

Sorry you were right before I change things, if I do this;

- x^2 - 4x - 5 = 0

It's wrong.
• Sep 16th 2011, 01:05 PM
Siron
Re: x intercept of parabola
But (+1)(-5)=-5 and (-1)(+5)=-5
• Sep 16th 2011, 01:05 PM
skeeter
Re: x intercept of parabola
$-x^2 + 4x + 5 = 0$

$-(x^2 - 4x - 5) = 0$

$x^2 - 4x - 5 = 0$

$(x-5)(x+1) = 0$

$x = 5$ , $x = -1$