<< Chapter < Page | Chapter >> Page > |
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:
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 | |
---|---|
$\frac{3}{{x}^{2}+\text{12}x+\text{36}}=\frac{\mathrm{4x}}{{x}^{3}+{\mathrm{4x}}^{2}-\text{12}x}$ | Same problem we worked before, but now we are equating these two fractions, instead of subtracting them. |
$\frac{3(x)(x-2)}{(x+6{)}^{2}(x)(x-2)}=\frac{\mathrm{4x}(x+6)}{x(x+6{)}^{2}(x-2)}$ | Rewrite both fractions with the common denominator. |
$3x(x-2)=4x(x+6)$ | Based on the rule above—since the denominators are equal, we can now assume the numerators are equal. |
$3{x}^{2}\u20136x=4{x}^{2}+24x$ | 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... |
$x(x+30)=0$ | ...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, $x=\u201330$ . |
As always, it is vital to remember what we have found here. We started with the equation $\frac{3}{{x}^{2}+\text{12}x+\text{36}}=\frac{\mathrm{4x}}{{x}^{3}+{\mathrm{4x}}^{2}-\text{12}x}$ . We have concluded now that if you plug $x=\u201330$ 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 $\frac{\mathrm{4x}}{{x}^{3}+{\mathrm{4x}}^{2}-\text{12}x}$ , the two graphs would intersect at one point only: the point when $x=\u201330$ .
Notification Switch
Would you like to follow the 'Math 1508 (lecture) readings in precalculus' conversation and receive update notifications?