Explainer: Radical Equations

In this explainer, we will learn how to solve equations that include radical expressions, where the variable is the radicand, by raising each side to a power.

Consider the following equation: 2π‘₯+√π‘₯=3.

Here, we have an equation that contains a radical or, more precisely, a square root. The natural way to solve this is to isolate the square root term, square both sides, and solve the resulting quadratic. This method, however, comes with a caveat: it will give rise to extraneous solutions.

Definition: Extraneous Solution

A number that comes from a solution method but that does not satisfy the original equation is called an β€œextraneous solution.”

Before looking at this in more detail, let us introduce an area which often causes confusion.

As an aside, we have to be careful with how we define a square root. We often think of a square root as the inverse of a square (and vice versa). If we ask for solutions for the equation π‘₯=9, it is quickly identified (by taking square roots) that π‘₯=3 is a solution. However, we also have that π‘₯=βˆ’3 is a solution, and because of this point it is sometimes (incorrectly, or perhaps informally) written that √9=Β±3. To be clear, √9=3 and βˆ’βˆš9=βˆ’3. To hammer home the point, we can look at another example. If we consider evaluating √64, we get √64=8, but we do not get √64=βˆ’8. If we want to consider the negative square root, we need to make this clear in the question; that is, evaluate βˆ’βˆš64. In this case, the answer would be βˆ’8.

With this definition of the square root in mind, we will investigate the claim from earlier further. Recall the equation 2π‘₯+√π‘₯=3. Subtracting 2π‘₯ from both sides, we get √π‘₯=3βˆ’2π‘₯.

Squaring both sides, we get π‘₯=(3βˆ’2π‘₯).

Expanding the right-hand side, we get π‘₯=ο€Ή9βˆ’12π‘₯+4π‘₯.

Subtracting π‘₯ from each side, we get 4π‘₯βˆ’13π‘₯+9=0.

And, finally, factoring we get (π‘₯βˆ’1)(4π‘₯βˆ’9)=0.

This has two solutions: π‘₯=1 and π‘₯=94.

Now, if we substitute these two values of π‘₯ into the left-hand side of our original equation, for π‘₯=1, we get 2(1)+√1=2+1=3 which is correct. For π‘₯=94 we get 2ο€Ό94+ο„ž94=92+32=6 which is incorrect. Note here that had we said that ο„ž94=βˆ’32, then our solution would have appeared correct. However, this is not the definition of the square root as outlined earlier, so this is in fact an extraneous solution.

Let us now look at some more formal examples.

Example 1: Solving a Simple Equation Involving a Square Root

Solve √49=π‘₯.


With this question, we need to recall that 7=49.

We can see that the square root of 49 is 7 and that our solution is π‘₯=7.

It is worth noting at this point that (βˆ’7) is also 49, and a common mistake would be to also state that π‘₯=βˆ’7 is a solution. This, however, is not the case as this would be the solution to the equation βˆ’βˆš49=π‘₯.

We define that the √π‘₯ is always positive.

Example 2: Solving a More Complicated Equation Involving a Square Root

The natural way to solve √π‘₯+9βˆ’2π‘₯=3 introduces an extraneous solution. What is that extraneous solution?


With this question, we have been explicitly asked to look for an extraneous solution. The easiest way to do this is to solve the equation by squaring and then identifying which of the solutions does not satisfy the original equation.

Our first step is to add 2π‘₯ to both sides to isolate the square root:


Then, we square both sides to get π‘₯+9=(3+2π‘₯).

Expanding and simplifying the right-hand side, we get π‘₯+9=9+12π‘₯+4π‘₯, and subtracting π‘₯ and 9 from each side, we get 4π‘₯+11π‘₯=0.

Factoring, we find that π‘₯(4π‘₯+11)=0.

This has solutions of π‘₯=0 and π‘₯=βˆ’114. We then need to substitute our solutions into the left- and right-hand sides of equation (1). For π‘₯=0, we get √9=3, which is correct; for π‘₯=βˆ’114, we get ο„ž254=βˆ’52, which is incorrect. So, our extraneous solution is βˆ’114.

Let us have a look at an example that explores the definition of the square root further and the discovery of extraneous solutions.

Example 3: Identifying an Error in a Calculation for Solving a Radical Equation

Consider the following argument for solving βˆ’3=√π‘₯:

  1. βˆ’3=√π‘₯
  2. (βˆ’3)=ο€Ίβˆšπ‘₯ο†οŠ¨οŠ¨
  3. 9=π‘₯.

What mistake was made?


With this question, if we follow the steps of working it out, the logic of squaring both sides is correct for step 2, and the simplification is done correctly in step 3. The issue that arises is how the square root is defined. If we substitute the answer π‘₯=9 into the right-hand side of the original equation we get √9. If we evaluate this, we get an answer of 3, not βˆ’3. Therefore, βˆ’3 is an extraneous solution and the mistake in working out the equation is the failure to check for extraneous solutions.

Let us finish by looking at three further examples.

Example 4: Solving an Equation Involving Square Roots given a Prescribed Method

In order to solve the equation π‘₯+√π‘₯=6, Noah started with π‘₯+√π‘₯=6π‘₯+√π‘₯βˆ’6=0π‘₯βˆ’6=βˆ’βˆšπ‘₯(π‘₯βˆ’6)=π‘₯.

What solutions did Noah find?


If we continue Noah’s calculations and expand the left-hand side, we get π‘₯βˆ’12π‘₯+36=π‘₯.

Subtracting π‘₯ from each side of the equation, this simplifies to π‘₯βˆ’13π‘₯+36=0.

If we factor the quadratic, we get (π‘₯βˆ’9)(π‘₯βˆ’4)=0, which has solutions of π‘₯=9 and π‘₯=4. This answers the question, but one additional step that Noah should do at this point is check if he has any extraneous solutions. If we substitute π‘₯=4 into the left-hand side of our original equation, we get 4+√4=4+2=6, which is correct. If we substitute π‘₯=9, we get 9+√9=9+3=12, which is incorrect, so π‘₯=9 is an extraneous solution and, therefore, the only solution to the equation is π‘₯=4.

Example 5: Solving an Applied Problem Involving Square Roots

For what value of π‘˜ is π‘₯=1 introduced as an extraneous solution to the equation π‘₯=√π‘₯+3+π‘˜ when it is solved in the natural way?


The easiest way to approach this problem is to start by rearranging the equation to isolate the square root. If we subtract π‘˜ from each side, we get


If we then square both sides of the equation, we get π‘₯+3=(π‘₯βˆ’π‘˜). As an aside, it is worth noting the following: if we take a positive number π‘₯ and square it, we get π‘₯; if we then take the number βˆ’π‘₯ and square it, we also get π‘₯. Therefore, if we look at the right-hand side of our expression, once we have squared it, we do not know if the expression was positive or negative. It is at this point that we introduce an extraneous solution.

Continuing with our solution, we are trying to find values of π‘˜ which produce a solution of π‘₯=1, so we can substitute this to get 4=(1βˆ’π‘˜).

Expanding the right-hand side we get 4=1βˆ’2π‘˜+π‘˜, and subtracting 4 from each side we get π‘˜βˆ’2π‘˜βˆ’3=0. Factoring this expression, we find that (π‘˜βˆ’3)(π‘˜+1)=0, which has solutions π‘˜=3 and π‘˜=βˆ’1.

To determine which value of π‘˜ gives an extraneous solution, we can use equation (2). If we substitute π‘₯=1 into equation (2), we get √4=1βˆ’π‘˜.

We then need to identify which value of π‘˜ makes the right-hand side of this expression not equal to the left-hand side. Substituting π‘˜=3, we get 1βˆ’3=βˆ’2, which is incorrect as √4=2. Therefore, the only solution is π‘˜=3.

Example 6: Solving Equations Involving Radicals

What is the solution set of the equation π‘₯+√π‘₯+20π‘₯+100=5?


We isolate the radical, √π‘₯+20π‘₯+100=5βˆ’π‘₯, and then square both sides: π‘₯+20π‘₯+100=(5βˆ’π‘₯)π‘₯+20π‘₯+100=π‘₯βˆ’10π‘₯+25π‘₯+20π‘₯+100βˆ’ο€Ήπ‘₯βˆ’10π‘₯+25=030π‘₯+75=0, which gives the single solution π‘₯=βˆ’7530=βˆ’52.

Since we squared, we check whether this is a true, not extraneous, solution: βˆ’52+ο„Ÿο€Όβˆ’52+20ο€Όβˆ’52+100=βˆ’52+ο„ž254βˆ’50+100=βˆ’52+ο„ž2254=βˆ’52+152=102=5.

Key Points

  1. Recall that, for a perfect square π‘₯, the √π‘₯ is always a positive integer.
  2. A number that comes from a solution method but that does not satisfy the original equation is called an β€œextraneous solution.”
  3. When solving a radical equation, we need to be careful when checking for any extraneous solutions that arise as part of the calculation. For example, given the equation √9=π‘₯, the only solution is π‘₯=3 as π‘₯=βˆ’3 is an extraneous solution.

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