# Thread: Finding the x- intercept

1. ## Finding the x- intercept

I would just like a second opinion please to ensure I am doing this correctly.

When y = 0

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

x = - 1.25

2. ## Re: Finding the x- intercept

Originally Posted by David Green
I would just like a second opinion please to ensure I am doing this correctly.

When y = 0

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

x = - 1.25
The x^2 term does not disappear.

\displaystyle \begin{align*} 0 &= -x^2 + 4x + 5 \\ x^2 - 4x - 5 &= 0 \\ (x + 1)(x - 5) &= 0 \\ x + 1 = 0 \textrm{ or } x - 5 &= 0 \\ x = -1 \textrm{ or } x &= 5 \end{align*}

3. ## Re: Finding the x- intercept

What happened with the $-x^2$?

4. ## Re: Finding the x- intercept

Originally Posted by Siron
What happened with the $-x^2$?
Not sure if this works?

- b + or - square root b^2 - 4ac / 2a

b = 4, a = -1, and c = 5

4^2 - 4 x 1 x 5 = 16 - 20 = - 4

square root of - 4 = E?

5. ## Re: Finding the x- intercept

Originally Posted by David Green
Not sure if this works?

- b + or - square root b^2 - 4ac / 2a

b = 4, a = -1, and c = 5

4^2 - 4 x 1 x 5 = 16 - 20 = - 4

square root of - 4 = E?
The discriminant is $\displaystyle 4^2 - 4(-1)(5) = 16 + 20 = 36$, so there are two solutions.

Of course, the easier method is to look at my post above and realise that the quadratic easily factorises.

6. ## Re: Finding the x- intercept

Originally Posted by Prove It
The discriminant is $\displaystyle 4^2 - 4(-1)(5) = 16 + 20 = 36$, so there are two solutions.

Of course, the easier method is to look at my post above and realise that the quadratic easily factorises.
I do keep getting my minus and a minus confused?

Thanks

7. ## Re: Finding the x- intercept

In the second degree equation:
$-x^2+4x+5=0$
is $a=(-1), b=4, c=5$ that means if we substitute this values in the formula $D=b^2-4ac$ we get:
$D=(4)^2-4(-1)(5)=36$

8. ## Re: Finding the x- intercept

Originally Posted by Siron
In the second degree equation:
$-x^2+4x+5=0$
is $a=(-1), b=4, c=5$ that means if we substitute this values in the formula $D=b^2-4ac$ we get:
$D=(4)^2-4(-1)(5)=36$
Interestingly I still get confused with my rules?

36 squared = 6 / 2 = 3, 3 + (-4) = -1 first root

36 squared = 6 / 2 = 3, 3 - ( - 4) = 7 ?

How do I get the root 5?

9. ## Re: Finding the x- intercept

There're two solutions (if $D\geq0$) for a quadratic equation by using the quadratic formula:
$x_1,x_2=\frac{-b\pm\sqrt{D}}{2a}$
So in this case:
$x_1=\frac{-4+6}{-2}=-1$
$x_2=\frac{-4-6}{-2}=\frac{-10}{-2}=5$

10. ## Re: Finding the x- intercept

Originally Posted by Siron
There're two solutions (if $D\geq0$) for a quadratic equation by using the quadratic formula:
$x_1,x_2=\frac{-b\pm\sqrt{D}}{2a}$
So in this case:
$x_1=\frac{-4+6}{-2}=-1$
$x_2=\frac{-4-6}{-2}=\frac{-10}{-2}=5$
I can't say that I can follow the reasoning there yet, what happend to the rest of the formula?

11. ## Re: Finding the x- intercept

What do you mean with the 'rest of the formula'? Don't you understand how I got the solutions? ...

12. ## Re: Finding the x- intercept

I understand the two roots and how you got them, but the formula has been modified.

- b + or - square root D / 2a

See the changes I have not seen before?

Capital D not squared, and - 4ac missing?

Like I said before this subject is new to me and is a new learning curve.

13. ## Re: Finding the x- intercept

I don't think the formula has been modified.
In general if you have a quadratic equation:
$ax^2+bx+c=0$
Then this equation has:
two different solutions (or roots) if $D=b^2-4ac>0$
$x_1=\frac{-b+\sqrt{D}}{2a}$
$x_2=\frac{-b-\sqrt{D}}{2a}$

If $D=0$ then the two identical solutions are:
$x_1,x_2=\frac{-b}{2a}$

If $D<0$:
There're no real solutions.

Is this clear? If you need a prove of this then you can ask it.

14. ## Re: Finding the x- intercept

So the capital D, this I now understand to be the solution of;

b^2 - 4ac

then the answer = D which I take the square root value.

15. ## Re: Finding the x- intercept

D is the discriminant, and indeed $D=b^2-4ac$.

I think you get the point now. Do you?

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