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

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- Sep 3rd 2011, 12:46 PMDavid GreenFinding 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 - Sep 3rd 2011, 12:51 PMProve ItRe: Finding the x- intercept
- Sep 3rd 2011, 12:52 PMSironRe: Finding the x- intercept
What happened with the $\displaystyle -x^2$?

Use the quadratic formula. - Sep 3rd 2011, 01:03 PMDavid GreenRe: Finding the x- intercept
- Sep 3rd 2011, 01:05 PMProve ItRe: Finding the x- intercept
- Sep 3rd 2011, 01:10 PMDavid GreenRe: Finding the x- intercept
- Sep 3rd 2011, 01:14 PMSironRe: Finding the x- intercept
In the second degree equation:

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

is $\displaystyle a=(-1), b=4, c=5$ that means if we substitute this values in the formula $\displaystyle D=b^2-4ac$ we get:

$\displaystyle D=(4)^2-4(-1)(5)=36$ - Sep 3rd 2011, 02:25 PMDavid GreenRe: Finding the x- intercept
- Sep 3rd 2011, 02:28 PMSironRe: Finding the x- intercept
There're two solutions (if $\displaystyle D\geq0$) for a quadratic equation by using the quadratic formula:

$\displaystyle x_1,x_2=\frac{-b\pm\sqrt{D}}{2a}$

So in this case:

$\displaystyle x_1=\frac{-4+6}{-2}=-1$

$\displaystyle x_2=\frac{-4-6}{-2}=\frac{-10}{-2}=5$ - Sep 3rd 2011, 02:32 PMDavid GreenRe: Finding the x- intercept
- Sep 3rd 2011, 02:41 PMSironRe: Finding the x- intercept
What do you mean with the 'rest of the formula'? Don't you understand how I got the solutions? ...

- Sep 3rd 2011, 02:49 PMDavid GreenRe: 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. - Sep 3rd 2011, 02:56 PMSironRe: Finding the x- intercept
I don't think the formula has been modified.

In general if you have a quadratic equation:

$\displaystyle ax^2+bx+c=0$

Then this equation has:

two different solutions (or roots) if $\displaystyle D=b^2-4ac>0$

$\displaystyle x_1=\frac{-b+\sqrt{D}}{2a}$

$\displaystyle x_2=\frac{-b-\sqrt{D}}{2a}$

If $\displaystyle D=0$ then the two identical solutions are:

$\displaystyle x_1,x_2=\frac{-b}{2a}$

If $\displaystyle D<0$:

There're no real solutions.

Is this clear? If you need a prove of this then you can ask it. - Sep 3rd 2011, 02:59 PMDavid GreenRe: 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. - Sep 3rd 2011, 03:03 PMSironRe: Finding the x- intercept
D is the discriminant, and indeed $\displaystyle D=b^2-4ac$.

I think you get the point now. Do you?