Re: x intercept of parabola

Note: Recalculate the discriminant. It has to be:

$\displaystyle -x^2+4x+5=(x+1)\cdot (x-5)$

Because:

$\displaystyle (x+1)\cdot (x-5)=x^2-5x+x-5=x^2-4x-5$

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

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.

Re: x intercept of parabola

Quote:

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.

Re: x intercept of parabola

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

Re: x intercept of parabola

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

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

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

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

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