# Common Core Standards: Math

#### The Standards

# High School: Algebra

### Reasoning with Equations and Inequalities HSA-REI.A.2

**2. Solve simple rational and radical equations in one variable, and give examples showing how extraneous solutions may arise.**

Rational equations might sometimes be a bit more radical than students would like. Those square root signs can instill terror in the heart of any student unprepared for them. If students overcame their fear of monsters under the bed (and hopefully they did), they'll get over their fear of radicals too.

Rational equations mean that fractions are involved. Radical equations mean that square roots are involved. Students should know how to deal with both separately and together.

A **radical equation** is one in which the variable is under the radical sign. When solving radical equations, it's usually best to leave the radicals for last unless there's a quick and easy way to get rid of all of them. If only there were a quick and easy way to get rid of monsters.

For example, is a radical equation. To solve, students should add 4 to both sides and get , or . All that's left is squaring both sides to get *x* = 256. Not too scary, right?

Of course, solving radical equations means students have to understand how to combine radicals. For example, since both terms have the same thing under the radical. We cannot combine and this same way. We could, however, rewrite as in which case . Not as pretty, but nonetheless doable.

Students should know how to combine, manipulate, and rewrite radical expressions. This usually takes practice and repetition. When all else fails, tell students to treat radicals like they'd treat variables. This is the one time the Golden Rule doesn't apply.

Students should already know how to solve rational equations. They should be able to find *x* faster than a pirate with a treasure map. And if he has an eye patch then it shouldn't even be a contest.

A rational equation might look like

In this case, let's say it does. First, cross-multiply to get 3(*x* – 4) = (*x* – 4)(*x* + 2). Since we have (*x* – 4) on both sides, we can reduce the equation to 3 = *x* + 2. Our final answer is *x* = 1.

Students should know that sometimes, algebraic manipulation produces extraneous solutions. For example, multiplying the fairly simple equation *x* + 5 = 0 by *x* will give *x*^{2} + 5*x* = 0. Now, both *x* = 0 and *x* = -5 will satisfy that quadratic. However, looking at *x* + 5 = 0, we can see that *x* = 0 won't work for the original equation (because 0 + 5 ≠ 0). That means *x* = 0 is an extraneous solution.

To check for extraneous solutions, students should plug in their final answers back into the original equation. If the equation produces an incorrect statement (like 5 = 0), then they'll know that solution didn't really exist. Much like the monsters under the bed…we hope.

### Aligned Resources

- Solving Rational Equations - Math Shack
- Evaluating Expressions with Negative Exponents - Math Shack
- Solving Basic Rational Equations - Math Shack
- Solving Radical Equations - Math Shack
- Solving Radical Equations Involving Fractions - Math Shack
- Solving Complicated Equations - Math Shack
- Solving Rational Inequalities - Math Shack