# Thread: x intercept of parabola

1. ## 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)?

2. ## Re: x intercept of parabola

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

3. ## 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

4. ## 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.

5. ## Re: x intercept of parabola

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

6. ## Re: x intercept of parabola

But (+1)(-5)=-5 and (-1)(+5)=-5

7. ## Re: x intercept of parabola

$\displaystyle -x^2 + 4x + 5 = 0$

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

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

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

$\displaystyle x = 5$ , $\displaystyle x = -1$