# Math Help - help pls and hurry my test is tomorrow quadratic

1. ## help pls and hurry my test is tomorrow quadratic

help pls and hurry my test is tomorrow,

im having issues with solving quadratic equations (paralobas)

the problem is x^-2x-8=0 i have tried everything i can not solve this

if u can pls map this out for me showing me how to get x, and y and the corresponding y-values to obtain the points for the paraloba...

i just need this one problem totaly worked out and i will get it,

i just simply can solve it, i keep getting 4 but book says answer is 4 and -2, thanks in advance.....

p.s. pls give steps and the rule si mut follow when solving for a paraloba pls,

thanks again and in advance

2. Originally Posted by thepcnerd
help pls and hurry my test is tomorrow,

im having issues with solving quadratic equations (paralobas)

the problem is x^-2x-8=0 i have tried everything i can not solve this

if u can pls map this out for me showing me how to get x, and y and the corresponding y-values to obtain the points for the paraloba...

i just need this one problem totaly worked out and i will get it,

i just simply can solve it, i keep getting 4 but book says answer is 4 and -2, thanks in advance.....

p.s. pls give steps and the rule si mut follow when solving for a paraloba pls,

thanks again and in advance
$x^2 - 2x - 8 = 0$

When in doubt use the quadratic formula. (We don't really need to here, but if you run out of other ideas then this ALWAYS works.)
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}$

$x = \frac{2 \pm \sqrt{4 + 32}}{2}$

$x = \frac{2 \pm \sqrt{36}}{2}$

$x = \frac{2 \pm 6}{2}$

$x = 4, -2$

A quick note: Whenever you solve a quadratic you will get one of the three possibilities:
1) Two unequal real roots
2) One real root (actually two identical real roots)
3) Two complex roots (they will be complex conjugates of each other.)

I'm not sure why you are having a problem graphing. All you need to do is pick some x value and plug it into:
$y = x^2 - 2x - 8$
and the corresponding y value comes out.

-Dan

3. Originally Posted by thepcnerd
help pls and hurry my test is tomorrow,

im having issues with solving quadratic equations (paralobas)

the problem is x^-2x-8=0 i have tried everything i can not solve this

if u can pls map this out for me showing me how to get x, and y and the corresponding y-values to obtain the points for the paraloba...

i just need this one problem totaly worked out and i will get it,

i just simply can solve it, i keep getting 4 but book says answer is 4 and -2, thanks in advance.....

p.s. pls give steps and the rule si mut follow when solving for a paraloba pls,

thanks again and in advance
x^-2x-8=0 <== I am assuming you mean:

x^2 - 2x - 8 = 0

Factor.

(x - 4)*(x + 2) = 0

Thus, x = -2 and x = 4.

Plug in values to graph it.

That is:

(-2, 0) <-- solved for that alrdy
(-1, 5)
(0, -8)
(1, -9)
(2, -8)
(3, -5)
(4, 0) <-- solved for that too

The graph is concave up.

(1, -9) is the stationary point or where it's at its minimum.

4. ## im not h aving issues graphing im having issues

im not h aving issues graphing im having issues with

getting the y values the one where u do the chart, book says its 4 and -2 but when i plug in value 4 in place of x and do equaition i get 0 as answer and not -2, im trying to get the pts for y values to plot on the parabola

5. but u showed me that thanks!!!!!

6. ## ??

topsquark how did u get y(-2) where did we get this number, how do we decide that -2 is the num ber to use.???

thanks

7. Just for fun, here's a 'completing the square' method I always use to expedite things along.

You have: $x^{2}-2x-8=0$

a=1, b=-2, c=-8

Just like in the quadratic formula.

When completing the square, you can use the following formula:

$a(x+\frac{b}{2a})^{2}+\underbrace{c-\frac{b^{2}}{4a}}_{\text{constant}}$

Filling in a, b, and c we get

$(x-1)^{2}-9=0$

This can easily be solved for $x=-2 \;\ and \;\ x=4$

The above formula is particularly helpful when the leading coefficient is not 1.

8. Originally Posted by thepcnerd
im not h aving issues graphing im having issues with

getting the y values the one where u do the chart, book says its 4 and -2 but when i plug in value 4 in place of x and do equaition i get 0 as answer and not -2, im trying to get the pts for y values to plot on the parabola
You're not really making sense here. Are you surprised that you don't get -2 when you plug in 4? You're not supposed to. You're supposed to get 0 when you plug in 4 or -2. Did you mean that you don't get 0 when you plug in -2? You should: (-2)^2-2*(-2)-8 = 4+4-8 = 0.

I don't know what you mean by "getting the y values the one where you do the chart" or "the pts for y values to plot on the parabola". If you want to plot the graph, you just have to pick a bunch of numbers that you plug into that expression, like AfterShock did. Just make sure that some of the numbers you plug in are close to 1, because that's where x^2-2x-8 has it's minimum. You see this by taking the derivative. It's 2x-2, which is =0 when x=1.

9. Originally Posted by thepcnerd
topsquark how did u get y(-2) where did we get this number, how do we decide that -2 is the num ber to use.???
Originally Posted by topsquark
$x = \frac{2 \pm 6}{2}$

$x = 4, -2$
Use the "+" sign to get the first root:
$x = \frac{2 + 6}{2} = \frac{8}{2} = 4$

Use the "-" sign to get the second root:
$x = \frac{2 - 6}{2} = \frac{-4}{2} = -2$

-Dan