Simplifying Square Roots of Variable Expressions

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Lesson Priority: VIP Knowledge

 
Objectives
  • Quickly review the root product property
  • Know how to simplify the square root of a variable raised to a power, and know the important difference between applying a square root to a variable with an even exponent versus an odd exponent
  • Extend our knowledge of square roots of integers to square roots of monomials
  • Practice taking square roots of monomials with and without coefficients inside and outside of the radicand
  • Learn why square roots of variables raised to powers behave differently whether they are standalone expressions or part of an equation
Lesson Description

Until now, learning about square roots has involved numbers only. For the first time, we will look at how to take the square root of variable monomials, and some minute differences between the setting of standalone expressions and the setting of equations. We will also understand how to handle coefficients properly in order to present fully simplified answers.

 
Practice Problems

Practice problems and worksheet coming soon!

 
You Should Know
Looking to learn about roots of variable expressions that involve fractions? We'll look at that in the next lesson» .

The Root of the Problem

At this point, you're now up to speed on simplifying integer roots»  as well as applying the all-important concept of rationalizing»  for numerical fractions that involve square root denominators. Now, as you might expect in Algebra, we are going to apply the same principles to variables instead of integers.Simplifying roots of variables is just an extension of what you already know, and many students actually find it easier. It starts with our core root rule.Recall the Root Product Property from the recent lesson on simplifying:
The Root Product PropertyThe square root of a product is equal to the product of each factor's square root, and vice versa.In math symbols:$$\sqrt{xy} = \sqrt{x}\sqrt{y}$$and$$\sqrt{x}\sqrt{y} = \sqrt{xy}$$
This rule applies whether we are working with integers or variables. We'll use it shortly.Let's also recall the idea of perfect square integers. These are integers that result from squaring smaller integers. For example, $25$ is a perfect square integer because$$5^2 = 25$$but $40$ is not a perfect square integer, because there is no integer that, when squared, gives you $40$ as a result.The same thing happens with variables expressions, but since variables are unknown values, the only thing we have to be concerned with is the exponent to which the variable is raised.

Square Roots of Variables

As we just said, variables are unknown values, so we will not be concerned with simplifying to equivalent forms like we did with integers. $\sqrt{x}$ is just $\sqrt{x}$, because we don't know what $x$ is. Instead, our job with square roots of variables revolves around simplifying the result if the variable is raised to a power.The simplest case is when a variable is raised to the second power. By the definition»  of square roots, we can make the following assertion:$$\sqrt{x^2} = x$$As we have recently seen, this is because squaring and square rooting are opposite operations.
Remember!
If you square root a perfect square, or if you square a square root, the radicand "goes away" because these operations are inverses of one another.$$\sqrt{x^2} = x$$$$\left( \sqrt{x}\right)^2 = x$$
If the variable is raised to an exponent higher than two, we can generalize a rule that tells us what the square root of that expression is, based on the definition of square roots and whether the exponent is even or odd.The rule for even exponents is much easier to digest than the rule for odd, because square roots are the opposite of squaring, and squaring always results in even exponent answers. e.g.$$\left(x^2\right)^2 = x^4$$$$\left(x^5\right)^2 = x^{10}$$$$\left(x^{11}\right)^2 = x^{22}$$
Square Root of a Single Variable TermFor any variable raised to an integer exponent, the square root of that object will be determined by whether the exponent is even or odd.Let $\displaystyle x^p$ represent any variable $x$ raised to an integer power $p$.$\displaystyle \sqrt{x^p} = x^{p/2}$ if $p$ is even.$\displaystyle \sqrt{x^p} = x^{(p-1)/2}\cdot \sqrt{x}$ if $p$ is odd.
While the case for odd exponents is less clean than the case for even exponents (for the reason we just noted regarding how squaring and square rooting are opposites) there is fairly clean way to get answers in practice, using the Root Product Property.Let's look at an example.
 
Example 1Simplify $\displaystyle \sqrt{x^6}$ and $\displaystyle \sqrt{x^{11}}$.$\blacktriangleright$ The first expression involves an even exponent, and so, by the rule:$$\sqrt{x^6} = x^3$$If we follow the rule for odd exponents verbatim for the second expression, we have$$\sqrt{x^{11}} = x^{(11-1)/2} \sqrt{x} = x^5 \sqrt{x}$$Let's look at each from a concept perspective.For the first, it makes sense that $\displaystyle \sqrt{x^6} = x^3$ because, by the rule of exponents,$$\left( x^3 \right)^2 = x^6$$For the second, let's look at an approach using the Root Product Property. If you re-write $\displaystyle x^{11}$ as $\displaystyle x^{10} \cdot x$, then$\displaystyle \sqrt{x^{11}} = \sqrt{x^{10}\cdot x}$$\displaystyle = \sqrt{x^{10}} \cdot \sqrt{x}$ (Root Product Property) $\displaystyle = x^5 \sqrt{x}$ (root rule for even exponent variables)For odd exponents, I think this approach is more intuitive and helpful when it comes to practice and getting it right on quizzes and tests. I think of it as "peeling off" an $x$ from the odd exponent variable, leaving me with an even exponent piece (which just uses the one-half even exponent rule) and a standalone $\displaystyle \sqrt{x}$ piece. The answer is the same.

Square Roots of Any Monomial

As soon as you've mastered the process for a single variable, handling square roots of entire monomials is just an extension. Thanks to the Root Product Property, we can simply handle each piece of the monomial on its own, and multiply the results.
 
Example 2Simplify $\displaystyle \sqrt{w^4 z^9}$.$\blacktriangleright$ By the Root Product Property, we know that$$\sqrt{w^4 z^9} = \sqrt{w^4} \cdot \sqrt{z^9}$$Now we just simplify each one.$$\sqrt{w^4} = w^2$$$$\sqrt{z^9} = z^4 \cdot \sqrt{z}$$Therefore, putting this all together we find that$$\sqrt{w^4 z^9} = w^2 z^4 \sqrt{z} \;\; \blacksquare$$
 
Often times problems have coefficients in front. We will handle the problem by examining one piece of the monomial at a time, but be aware that the square root of the coefficient must be simplified down just like any integer, which may involve the typical perfect square factor process.
 
Example 3Simplify $\displaystyle $\sqrt{32 x^3 y^4 z^5}$.$\blacktriangleright$ Tackle each piece of the monomial and put the results together.$$\sqrt{32} = \sqrt{16 \cdot 2} = 4 \sqrt{2}$$$$\sqrt{x^3} = \sqrt{x^2 \cdot x} = x \sqrt{x}$$$$\sqrt{y^4} = y^2$$$$\sqrt{z^5} = z^2 \sqrt{z}$$All in all,$$\sqrt{32 x^3 y^4 z^5} = 4x y^2 z^2 \sqrt{2xz} \;\; \blacksquare$$
 
Pro Tip
As you practice and become more confident with the variable square root process, you'll notice that essentially, everything that "comes out" of the square root gets thrown together in the final answer, and everything that "stays under" the square root does as well. This is a good perspective to use to check your work, because that is always going to happen.

Unusual Suspects

Usually when we work these types of problems, your teacher or your instructions will tell you to assume all variable values are positive quantities.Very rarely, I have seen teachers who do not state this, and who enforce the "bars" rule. That is, technically speaking:$$\sqrt{x^2} = |x|$$The reason the bars are needed here is, that if $x$ was a negative number, we couldn't square root it first. Inverse operations are supposed to work backward and forward. For example, with a positive number like $3$, we can say$\displaystyle 3^2 = 9$ and $\displaystyle \sqrt{9} = 3$However, if we use negative 3 instead for this statement:$\displaystyle (-3)^2 = 9$ and $\displaystyle \sqrt{9} = 3$This is because we always assume square roots give positive answers only, when working on expressions (if we worked on equations, as we'll see, then $\sqrt{9} = \pm 3$).Don't worry too much about this, but just be aware that some teachers expect bars around the variable for the square root of a squared variable. It's fairly rare for a teacher to care about this technicality.

Put It To The Test

Simplify each of the following expressions.
 
Example 4$\displaystyle \sqrt{r^7}$$\blacktriangleright$ Use the rule verbatim, or if you prefer, use my trick of "peeling off" one of the variables, leaving you with something digestible.$$\sqrt{r^7} = \sqrt{r^6 \cdot r}$$$$ = \sqrt{r^6} \cdot \sqrt{r}$$$$= r^3 \sqrt{r} \;\;\; \blacksquare$$
 
Example 5$\displaystyle \sqrt{a^4 b^6}$$\blacktriangleright$ Handle each of the terms of the monomial separately.$$\sqrt{a^4} = a^2$$$$\sqrt{b^6} = b^3$$Therefore,$$\sqrt{a^4 b^6} = a^2 b^3 \;\;\; \blacksquare$$
 
Example 6$\displaystyle \sqrt{2 x^3 y^{12}}$$\blacktriangleright$ Figure out the square root of each piece, including the coefficient. Here, the coefficient cannot be further simplified anyway.$$\sqrt{x^3} = x\sqrt{x}$$$$\sqrt{y^{12}} = y^6$$$$\therefore \sqrt{2 x^3 y^{12}} = x y^6 \sqrt{2x} \;\;\; \blacksquare$$
 
Example 7$\displaystyle \sqrt{18 t^5 u^6 v^3}$$$\blacktriangleright \;\;\; \sqrt{18} = 3\sqrt{2}$$$$\sqrt{t^5} = t^2 \sqrt{t}$$$$\sqrt{u^6} = u^3$$$$\sqrt{v^3} = v \sqrt{v}$$$$\therefore \sqrt{18 t^5 u^6 v^3} = 3t^2 u^3 v \sqrt{2tv} \;\;\; \blacksquare$$
 
Example 8$\displaystyle \sqrt{80 x y^2 z^{13}}$$$\blacktriangleright \;\;\; \sqrt{80} = 4\sqrt{5}$$$$\sqrt{x} = \sqrt{x}$$$$\sqrt{y^2} = y$$$$\sqrt{z^{13}} = z^6 \sqrt{z}$$$$\therefore \sqrt{80 x y^2 z^{13}} = 4y z^6 \sqrt{5xz} \;\;\; \blacksquare$$
 
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