1. pre-calc stress

I've been staring at the following problems for at least half an hour. Would anyone help me?

2x^2+8x=0
-9x^2+12x-4=0
4x^2-4x+2=0
3x^2+4x-1=0
x^3+2x^2-4x-8=0

2. Do you know how to use the quad formula?

$\frac{-b_-^+ \sqrt{b^2-4ac}}{2a}$

3. I've tried, but I keep getting answers I'm sure are way off from being right.

4. What kind of answers do you keep getting?

5. for the first one i got .396 and 8.48

6. I'm going to do one for you for an example

$-9x^2+12x-4=0$

$Ax^2+Bx+C=0$

$\frac{-12_-^+ \sqrt{144-4(-9)(-4)}}{-18}$

$\frac{-12_+^- 0}{-18}$

$\frac{1}{3}$

7. Thank you so much. I think I have an idea of how to do the rest.

8. Ok, I got stuck on x^3+2x^2-4x-8=0

9. Originally Posted by 11rdc11
I'm going to do one for you for an example

$-9x^2+12x-4=0$

$Ax^2+Bx+C=0$

$\frac{-12_-^+ \sqrt{144-4(-9)(-4)}}{-18}$

$\frac{-12_+^- 0}{-18}$

$\frac{1}{3}$
Yo

Next time, you can use $\pm$ (\pm) and $\mp$ (\mp)

10. Originally Posted by nina9
Ok, I got stuck on x^3+2x^2-4x-8=0
Try x - 2 as a factor. $x^3 + 2x^2 - 4x - 8 = (x - 2)(x^2 + 4x + 4)$ using long or synthetic division.

11. For 2x^2+8x=0, both the terms have x in them so this can be factorised out:

x(2x+8)=0

I was going to do the rest of them by finding a factor, but they all appear to be decimals or surds apart from the last one. I'll show you the method so you can use it on other questions:

f(x)=x^3+2x^2-4x-8

f(2)=2^3+2^3-8-8
f(2)=8+8-8-8
f(2)=0

Since f(2)=0, f(x) has a factor x-2 ie. when x=2 f(x)=0.

This can then be factorised using this (see icemanfan's post).

A graphical method can help:

If in doubt, draw the graph and see if that helps.