# Heart of Algebra – Radical Equations and Extraneous Solutions

*By Jeffrey Dalton*

Try to solve this sample SAT math question above. If you are a “savvy” test taker, then you might have noticed that the easiest way to solve it is to take the answer choices and substitute them back into the original equation in place of x. For example, you can take answer choice A) and say to yourself, “Let’s see if x = 2.”

If we substitute x in as 2 and simplify the equation, we get -3 = 3. So, let’s now ask ourselves, “does -3 equal 3?” Hopefully, you can conclude that that is not true. Since the equation did not work when x = 2, then 2 does not satisfy the equation. Eliminate A and try the remaining answer choices. You can see from the work above, that answer choice C simplified out to “2 = 2,” which is a true statement, as you are undoubtedly aware. Thus the correct answer is C.

However, for the sake of the argument, let’s find another way to solve this question. Even though the above method is the easiest and fastest, I promise you that it’s important to know how to solve this without substituting the answer choices back in.

The equation above is called a “radical equation” because there is a variable underneath a square root sign. In order to “cancel out” or “remove” the radical, we have to square both sides of the equation, like this:

If we continue the process by expanding our equation and combining our “like terms,” we get the quadratic equation x squared minus 9x plus 14. Hopefully, you remember how to factor a quadratic equation: we need to find 2 numbers that multiply out to 14 and add up to 9.

Let’s take a look at a simple Logical Transition question in order to learn the rule and technique.

If we factor and solve, we get two different values of x: 7 and 2, which perfectly matches answer choice D. But wait! We solved this equation earlier. The answer was C, not D! What is happening here?! Stay calm. Take a deep breath. The correct answer is still C. You can remember that when we substituted 2 back into the original equation, 2 was not a valid solution. But 7 was.

So why did solving this equation produce 2 values of x? Well, the problem occurred in the very first step of solving this equation, when we squared both sides!

Let’s unpack what happened. Now, a regular, single-variable equation (in which the highest exponent is 1) usually has 1 solution. If I asked you to solve 2x + 1 = 9, you would get a single value of x. That makes sense. You might remember from Math class that the “degree” (or highest exponent) of an equation indicates the greatest number of possible solutions.

A quadratic equation, like x2+7x + 10 = 0 has a maximum of two possible solutions. So, when we squared both sides of our original radical equation, we changed the equation from a radical equation (in which the highest exponent was 1) into a quadratic equation (in which the highest exponent was 2).

We created a quadratic. We introduced an “extraneous” solution. This was OUR fault!

An extraneous solution is a solution to an equation that emerges from the process of solving but is not a valid solution to the problem.

So there you have it. You’re probably thinking, “Ok, I’ll just plug the answer choices back into the original equation every time. Why did I have to learn about extraneous solutions? My head hurts.”

What if you saw this question below on your SAT test.

Checkmate. You’re in trouble now. This question isn’t asking for values of x that satisfy the equation. It’s literally asking for a value of x that “arises from the problem as you solve it but is not a valid solution to the equation.” Give this a try before you scroll down to the explanation below.

If we square both sides, combine like terms, and factor our resulting quadratic equation, we get values of 5 and 10. One of these is the correct answer. One of these is the extraneous solution. Can you figure out which? Substitute 5 and 10 back into the original equation and see which one gives you a true statement. I’ll wait.

You can see from the working above, that an x value of 5 gives you a true statement: 2 = 2. An x value of 10 gives you a false statement: -3 = 3. Therefore, x = 5 is a valid solution! Hurray, let’s rush and pick answer choice B. Wait, wait, wait. The question is asking us for the extraneous solution, so the correct answer is actually D.

Here’s one final question to test your knowledge of what you’ve read about today. Go ahead and try this one. Send your answer to [email protected] or let me know if you have any questions!

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