pre-calc stress

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September 14th 2008, 06:54 PM
nina9

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
September 14th 2008, 06:56 PM
11rdc11

Do you know how to use the quad formula?

$\frac{-b_-^+ \sqrt{b^2-4ac}}{2a}$
September 14th 2008, 06:59 PM
nina9

I've tried, but I keep getting answers I'm sure are way off from being right.
September 14th 2008, 07:04 PM
JoshHJ

What kind of answers do you keep getting?
September 14th 2008, 07:12 PM
nina9

for the first one i got .396 and 8.48
September 14th 2008, 07:14 PM
11rdc11

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

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

$Ax^2+Bx+C=0$

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

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

$\frac{1}{3}$
September 14th 2008, 07:30 PM
nina9

Thank you so much. I think I have an idea of how to do the rest.
September 14th 2008, 07:46 PM
nina9

Ok, I got stuck on x^3+2x^2-4x-8=0
September 15th 2008, 12:58 PM
Moo

Quote:

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) :)
September 15th 2008, 03:58 PM
icemanfan

Quote:

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.
September 16th 2008, 04:37 AM
Showcase_22

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:

http://i116.photobucket.com/albums/o...sproblem56.jpg

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