Honors Algebra 2 Solving rational equations

Bloodprincess15 · **Joined:** Thu, 5 May 2011 00:39:23 UTC **Posts:** 6

Please help me solve this:

(1/x^2)-x^2
___________= 3/2

(1/x)+x

The teacher gave us a hint, but I still do not understand how to solve this problem.
(Hint: You may need to use factoring by division to solve this.)

Shadow · **Posted:** Fri, 6 May 2011 02:58:04 UTC

Bloodprincess15 wrote:

Please help me solve this:

(1/x^2)-x^2
___________= 3/2

(1/x)+x

The teacher gave us a hint, but I still do not understand how to solve this problem.
(Hint: You may need to use factoring by division to solve this.)

Clear denominators then solve the resulting equation which is a polynomial.

Bloodprincess15 · **Joined:** Thu, 5 May 2011 00:39:23 UTC **Posts:** 6

what do you multiply the equation by to clear the denominators?

Shadow · **Posted:** Fri, 6 May 2011 03:05:38 UTC

Bloodprincess15 wrote:

what do you multiply the equation by to clear the denominators?

x.

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

x^2 actually.

Shadow · **Posted:** Fri, 6 May 2011 03:09:55 UTC

Matt wrote:

x^2 actually.

Ah yes, silly numerator fractions.

Bloodprincess15 · **Joined:** Thu, 5 May 2011 00:39:23 UTC **Posts:** 6

So then would you get:

1-x^4
_____=3x^2/2x^2
x+x^3

Or did I multiply wrong?

Shadow · **Posted:** Fri, 6 May 2011 03:15:12 UTC

Bloodprincess15 wrote:

So then would you get:

1-x^4
_____=3x^2/2x^2
x+x^3

Or did I multiply wrong?

You didn't multiply the right side by x^2, you multiplied it by x^2/x^2, which is not the same thing.

Bloodprincess15 · **Joined:** Thu, 5 May 2011 00:39:23 UTC **Posts:** 6

so then it would be:

....=3x^2/2

Right?

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9632

Multiplying by x^2 on the left-hand side creates an equivalent fraction.
This has nothing to do with the right-hand side. The result is:

$$\frac{1-x^4}{x+x^3}=\frac32$

Denis · **Posted:** Fri, 6 May 2011 10:56:15 UTC

Princess, the "left side multiplication by x^2" means the numerator
AND denominator are multiplied; so that's really a multiplication by
x^2 / x^2 which equals 1, so no need to adjust right side...OK?

Soroban · **Posted:** Fri, 6 May 2011 14:02:31 UTC

Hello, Bloodprincess15!

Another approach . . .

Quote:

Solve: . $$\frac{\frac{1}{x^2} - x^2}{\frac{1}{x} + x} \:=\:\frac{3}{2}$

Note that the numerator is a difference-of-squares.

$$\text{Factor: }\;\frac{\left(\frac{1}{x} - x\right)\left(\frac{1}{x} + x\right)}{\frac{1}{x} + x} \;=\;\frac{3}{2}$

Reduce: . $$\frac{1}{x} - x \:=\:\frac{3}{2}$ . . (Note that: $\frac{1}{x}+x \,\ne\,0$ for real

)

Multiply by $2x\!:\;\; 2 - 2x^2 \:=\:3x \quad\Rightarrow\quad 2x^2 + 3x - 2 \:=\:0$

Therefore: . $(x + 2)(2x-1) \:=\:0 \quad\Rightarrow\quad x \:=\:-2,\,\frac{1}{2}$

Mr. Mathturo · **Joined:** Tue, 13 Mar 2007 09:00:13 UTC **Posts:** 534

I would have first cross multiplied, and then multiplied both sides by x^2.

Just another way to accomplish the same result.

Denis · **Posted:** Thu, 12 May 2011 07:37:29 UTC

Mr. Mathturo wrote:

I would have first cross multiplied, .........

YA!
Take that, TKHunny :shock:

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Honors Algebra 2 Solving rational equations

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