Finding Solutions of Equations Help

[cole92]

My Pre-Cal class is still working on some review, and I have encountered something else I need help with remembering please. There are about 10 problems similar to the few listed below, all of which I need help with. Any step-by-step help would be awesome! I'm not sure if I was taking the right approach, but any of my work is below:

DIRECTIONS: Find all solutions of the equation. Use a graphin utility to verify the solutions graphically
115. 4x^4 - 18x^2 = 0
116. 20x^3 - 125x = 0

I think once I can do one problem I will be able to do all the rest. I tried using u = x^2 in number 115 but it got me to 4u^2 - 18u and I didn't know what to do after that. A similar situation in number 116. Any help would be great!

Thanks

 

[Loren]

>4x^4 - 18x^2 = 0

The first thing you always want to do is look for common monomial factors. In this case you have 2x^2 to factor out. Remember that if you multiply two or more quantities and get a product of zero, at least one of them must be a zero. In your case you will get an equation of the form a(b+c)=0. So, we know that either a = 0 or b+c=0. We solve each of these for the variable to get all the possible solutions. In your case you will get four solutions. One of them is a double solution which is verified when you graph the original equation. If you need more help, show us how far you got.

 

[skeeter]

\(\displaystyle 4x^4 - 18x^2 = 0\)

factor ...

\(\displaystyle 2x^2(2x^2 - 9) = 0\)

factor again ...

\(\displaystyle 2x^2(\sqrt{2} \, x - 3)(\sqrt{2} \, x + 3) = 0\)

use the zero product property ...

\(\displaystyle 2x^2 = 0\) ... \(\displaystyle x = 0\)

\(\displaystyle \sqrt{2} \, x - 3 = 0\) ... \(\displaystyle x = \frac{3}{\sqrt{2}}\)

\(\displaystyle \sqrt{2} \, x + 3 = 0\) ... \(\displaystyle x = -\frac{3}{\sqrt{2}}\)



solve \(\displaystyle 20x^3 - 125x = 0\) the same way.

 

[Denis]

Easier by factoring once:

4x^2 - 18x^2 = 0
2x^2(2x^2 - 9) = 0
So:
2x^2 = 0 ; x = 0
or
2x^2 - 9 = 0
x^2 = 9/2
x = +- 3/sqrt(2)

 

[cole92]

Thank all of you very much, I think I got it down now. Just took a jump start to get my memory going.