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

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- September 14th 2008, 06:54 PMnina9pre-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 PM11rdc11
Do you know how to use the quad formula?

- September 14th 2008, 06:59 PMnina9
I've tried, but I keep getting answers I'm sure are way off from being right.

- September 14th 2008, 07:04 PMJoshHJ
What kind of answers do you keep getting?

- September 14th 2008, 07:12 PMnina9
for the first one i got .396 and 8.48

- September 14th 2008, 07:14 PM11rdc11
I'm going to do one for you for an example

- September 14th 2008, 07:30 PMnina9
Thank you so much. I think I have an idea of how to do the rest.

- September 14th 2008, 07:46 PMnina9
Ok, I got stuck on x^3+2x^2-4x-8=0

- September 15th 2008, 12:58 PMMoo
- September 15th 2008, 03:58 PMicemanfan
- September 16th 2008, 04:37 AMShowcase_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.