# 0.4 Rational expression concepts -- rational equations

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This module introduces rational expressions in equations.

## Rational equations

A rational equation means that you are setting two rational expressions equal to each other. The goal is to solve for x; that is, find the x value(s) that make the equation true.

Suppose I told you that:

$\frac{x}{8}=\frac{3}{8}$

If you think about it, the x in this equation has to be a 3. That is to say, if x=3 then this equation is true; for any other x value, this equation is false.

This leads us to a very general rule.

A very general rule about rational equations

If you have a rational equation where the denominators are the same, then the numerators must be the same.

This in turn suggests a strategy: find a common denominator, and then set the numerators equal.

Example: Rational Equation
$\frac{3}{{x}^{2}+\text{12}x+\text{36}}=\frac{4x}{{x}^{3}+{4x}^{2}-\text{12}x}$ Same problem we worked before, but now we are equating these two fractions, instead of subtracting them.
$\frac{3\left(x\right)\left(x-2\right)}{\left(x+6{\right)}^{2}\left(x\right)\left(x-2\right)}=\frac{4x\left(x+6\right)}{x\left(x+6{\right)}^{2}\left(x-2\right)}$ Rewrite both fractions with the common denominator.
$3x\left(x-2\right)=4x\left(x+6\right)$ Based on the rule above—since the denominators are equal, we can now assume the numerators are equal.
$3{x}^{2}–6x=4{x}^{2}+24x$ Multiply it out
${x}^{2}+30x=0$ What we’re dealing with, in this case, is a quadratic equation. As always, move everything to one side...
$x\left(x+30\right)=0$ ...and then factor. A common mistake in this kind of problem is to divide both sides by $x$ ; this loses one of the two solutions.
$\mathrm{x=}0$ or $\mathrm{x=}-30$ Two solutions to the quadratic equation. However, in this case, $x=0$ is not valid, since it was not in the domain of the original right-hand fraction. (Why?) So this problem actually has only one solution, $x=–30$ .

As always, it is vital to remember what we have found here. We started with the equation $\frac{3}{{x}^{2}+\text{12}x+\text{36}}=\frac{4x}{{x}^{3}+{4x}^{2}-\text{12}x}$ . We have concluded now that if you plug $x=–30$ into that equation, you will get a true equation (you can verify this on your calculator). For any other value, this equation will evaluate false.

To put it another way: if you graphed the functions $\frac{3}{{x}^{2}+\text{12}x+\text{36}}$ and $\frac{4x}{{x}^{3}+{4x}^{2}-\text{12}x}$ , the two graphs would intersect at one point only: the point when $x=–30$ .

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yes that's correct
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I think
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research.net
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