Solving Equations

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Melba

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Sep 21, 2007
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So... I'm stuck!
9x^4=25x^2-16 Set problem to zero...
9x^4+25x^2-16=0 Right?
This is where I get stuck. I know I need to factor and will have 4 possible answers. What I can't get the hang of is factoring the the problem correctly. If I could learn how to do this part, I know how to solve for the rest of the problem.
 
\(\displaystyle \L\\9x^{4}=25x^{2}-16\)

That should be:

\(\displaystyle \L\\9x^{4}-25x^{2}+16=0\), you have the incorrect signs.

Rewrite as:

\(\displaystyle \L\\9x^{4}\underbrace{-9x^{2}-16x^{2}}_{\text{-25x^2}}+16=0\)

Group and factor out something common:

\(\displaystyle \L\\9x^{2}(x^{2}-1)-16(x^{2}-1)=0\)

\(\displaystyle \L\\(9x^{2}-16)(x^{2}-1)=0\)

It factors further. Finish?.
 
Okay, I saw my mistake with the signs, so does it continue on like this?

9x^2-16=0 or x^2-1=0

9x^2=16
divide 9 from both sides, so
x^2=16/9
sqrt x^2=+- sqrt 16/9
x=4/3, -4/3, 1, -1

Crude, but how's that?
 
Yes, good, those are the solutions all right.

But, I was getting at factoring \(\displaystyle \L\\(9x^{2}-16)(x^{2}-1)=0\) a little further by noticing the difference of two squares.

\(\displaystyle \L\\x^{2}-1=(x+1)(x-1)\)

\(\displaystyle \L\\9x^{2}-16=(3x+4)(3x-4)\)

Then you have \(\displaystyle \L\\(x+1)(x-1)(3x+4)(3x-4)=0\)

which gives the solutions you found.
 
Okay, I'm feeling really dense right now.

Say the problem was 9x^4-24x^2+16=0.

What is it that you're looking at specifically to come up with the grouping? When it's shown to me, I get it. Looking at a different example, I don't see it again.
 
Melba said:
Okay, I'm feeling really dense right now.

Say the problem was 9x^4-24x^2+16=0.

(3x<sup>2</sup> - 4)<sup>2</sup>

What is it that you're looking at specifically to come up with the grouping? When it's shown to me, I get it. Looking at a different example, I don't see it again.
Click to expand...
 
Oh, I think I finally see it; would the final answer finally be:
x=2sqrt3, x=-2sqrt3, repeated
 
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