Learning Outcomes
- Simplify square roots with variables
- Recognize that by definition [latex]\sqrt{x^{2}}[/latex] is always nonnegative
Key words
- Radical expression: an expression that contains radicals
Radical expressions are expressions that contain radicals. Radical expressions come in many forms, from simple and familiar, such as[latex] \sqrt{16}[/latex], to more complicated, as in [latex] \sqrt[3\;]{250{{x}^{4}}y}[/latex]. In this section we will discover how to simplify expressions containing square roots.
Simplifying Square Roots
We have already investigated simplifying a radical expression with integers by using factoring and the product property of square roots. For example, [latex]\sqrt{45}=\sqrt{9\cdot 5}=\sqrt{9}\cdot\sqrt{5}=3\sqrt{5}[/latex]. We can use this same method to simplify a radical term that contains variables. However, we have to be careful.
Consider the expression [latex] \sqrt{{{x}^{2}}}[/latex]. This looks like it should be equal to [latex]x[/latex], right? Let’s test some values for[latex]x[/latex] and see what happens.
Suppose, [latex]x=5[/latex]. Then [latex] \sqrt{{{x}^{2}}}=\sqrt{{{5}^{2}}}=\sqrt{25}=5=x[/latex]. In this case, [latex] \sqrt{{{x}^{2}}}=x[/latex].
Now suppose [latex]x=-3[/latex]. Then [latex] \sqrt{{{x}^{2}}}=\sqrt{{{(-3)}^{2}}}=\sqrt{9}=3\neq x[/latex]. In this case, [latex] \sqrt{{{x}^{2}}}\neq x[/latex].
The table shows more examples.
| [latex]x[/latex] | [latex]x^{2}[/latex] | [latex]\sqrt{x^{2}}[/latex] | [latex]\left|x\right|[/latex] |
|---|---|---|---|
| [latex]−5[/latex] | [latex]25[/latex] | [latex]5[/latex] | [latex]5[/latex] |
| [latex]−2[/latex] | [latex]4[/latex] | [latex]2[/latex] | [latex]2[/latex] |
| [latex]0[/latex] | [latex]0[/latex] | [latex]0[/latex] | [latex]0[/latex] |
| [latex]6[/latex] | [latex]36[/latex] | [latex]6[/latex] | [latex]6[/latex] |
| [latex]10[/latex] | [latex]100[/latex] | [latex]10[/latex] | [latex]10[/latex] |
Notice that in cases where [latex]x[/latex] is a negative number, [latex]\sqrt{x^{2}}\neq{x}[/latex]! (This happens because the process of squaring the number loses the negative sign, since a negative times a negative is a positive.) However, in all cases [latex]\sqrt{x^{2}}=\left|x\right|[/latex]. We need to consider this fact when simplifying radicals that contain variables, because by definition [latex]\sqrt{x^{2}}[/latex] is always nonnegative.
When we square any exponential term, we multiply the exponent by 2. For example, [latex]\left ( y^3\right )^2=y^{3\cdot 2}=y^6[/latex]. This means that in order to take the square root of an exponential term, the exponent must be even.
Let’s consider [latex]\sqrt{y^6}[/latex]. We can write [latex]y^6[/latex] as [latex]\left (y^3\right )^2[/latex]. Then [latex]\sqrt{y^6}=\sqrt{\left (y^3\right )^2}=\left | x^3 \right |[/latex]. We need the absolute value because the square root must be non-negative.
Notice that when we square an exponential term, we multiply the exponent by [latex]2[/latex]. Since taking the square root “undoes” squaring, we can divide the exponent by 2 t take the square root, provided the exponent is even.
the Square Root Of variables
[latex]\sqrt{x^{2}}=\left|x\right|[/latex]
[latex]\sqrt{x^{2n}}=\left | x^n \right |[/latex]
Now that we know this, we can simplify a radical expression by using factoring and the product property of square roots.
The goal is to find factors under the radical that are perfect squares so that we can simplify.
Example
Simplify. [latex] \sqrt{9{{x}^{6}}}[/latex]
Solution
Write the radicand as perfect square factors. Note how we use the power rule for exponents to write [latex]x^6[/latex] as a perfect square: [latex]\left ({x^3}\right )^2[/latex]
[latex] \sqrt{{{3}^{2}}\cdot {{\left( {{x}^{3}} \right)}^{2}}}[/latex]
Separate into individual radicals using the product property.
[latex] \sqrt{{{3}^{2}}}\cdot \sqrt{{{\left( {{x}^{3}} \right)}^{2}}}[/latex]
Take the square roots, remembering that [latex] \sqrt{{{x}^{2}}}=\left|x\right|[/latex].
[latex] 3\left|{{x}^{3}}\right|[/latex]
Answer
[latex] \sqrt{9{{x}^{6}}}=3\left|{{x}^{3}}\right|[/latex] We have to keep the absolute value since square roots are never negative.
Let’s try to simplify another radical expression.
Example
Simplify. [latex] \sqrt{100{{x}^{2}}{{y}^{4}}}[/latex]
Solution
Find perfect squares under the radical: exponents that are even.
[latex] \sqrt{10^2\cdot {x}^{2}\cdot {y}^{4}}[/latex]
Separate the perfect squares into individual radicals.
[latex] \sqrt{100}\cdot \sqrt{{x}^{2}}\cdot \sqrt{({y}^{2})^{2}}[/latex]
Simplify each radical by taking the square root. Remember to put absolute values on variable terms.
[latex]10\cdot\left|x\right|\cdot \left |{y}^{2}\right |[/latex]
Simplify. Since [latex]y^2\geq 0[/latex], [latex]\left |{y}^{2}\right | = y^2[/latex].
[latex]10\left|x\right|y^{2}[/latex]
Answer
[latex] \sqrt{100{{x}^{2}}{{y}^{4}}}=10\left| x \right|{{y}^{2}}[/latex]
Remember that we can check always check our answer by squaring it to be sure it equals [latex] 100{{x}^{2}}{{y}^{4}}[/latex].
Example
Simplify. [latex] \sqrt{49{{x}^{10}}{{y}^{8}}}[/latex]
Solution
Look for perfect squares: numbers and variables.
49 is a perfect square since [latex]7^2=49[/latex]; [latex]x^{10}[/latex] and [latex]y^8[/latex] are perfect squares since their exponents are even. [latex]x^{10}=\left (x^5\right )^2[/latex] and [latex]y^8=\left (y^4\right )^2[/latex].
Separate the squared factors into individual radicals.
[latex] \sqrt{7^2}\cdot\sqrt{({x^5})^2}\cdot\sqrt{({y^4})^2}[/latex]
Take the square root of each radical using the rule that [latex] \sqrt{{{x}^{2}}}=\left|x\right|[/latex].
[latex] 7\cdot\left|{{x}^{5}}\right|\cdot\left |{{y}^{4}}\right |[/latex]
Simplifly: [latex]\left |{{y}^{4}}\right |=y^4[/latex], since [latex]y^4\geq 0[/latex].
[latex] 7\left|{{x}^{5}}\right|{{y}^{4}}[/latex]
Answer
[latex] \sqrt{49{{x}^{10}}{{y}^{8}}}=7\left|{{x}^{5}}\right|{{y}^{4}}[/latex]
In order to check this calculation, we could square [latex]7\left|{{x}^{5}}\right|{{y}^{4}}[/latex], hoping to arrive at [latex] 49{{x}^{10}}{{y}^{8}}[/latex]. And, in fact, we would get this expression if we evaluated [latex] {\left({7\left|{{x}^{5}}\right|{{y}^{4}}}\right)^{2}}[/latex].
Try It
Simplify. [latex] \sqrt{81{{x}^{6}}{{y}^{4}}}[/latex]
Try It
Simplify. [latex] \sqrt{144{{x}^{14}}{{y}^{12}}}[/latex]
So far we have seen examples that have perfect squares under the radicals: the exponents have all been even. If we have an odd exponent then it is not a perfect square. For example, [latex]x^5[/latex] is not a perfect square. However, it contains perfect square factors of [latex]x^2[/latex] and [latex]x^4[/latex].
[latex]x^5=x\cdot x\cdot x\cdot x\cdot x=x^2\cdot x^2\cdot x=x^4\cdot x[/latex]
Each pair of factors makes a perfect square. This means that we can always write a variable with an odd exponent as the variable to one power less times the variable. For example, [latex]x^7=x^6\cdot x,\;y^9=y^8\cdot y,\; z^21=z^{20}\cdot z[/latex].
perfect square factors of powers of variables
[latex]x^n=x^{n-1}\cdot x[/latex]
This means that we can now simplify more radical expressions.
Example
Simplify. [latex] \sqrt{{{a}^{3}}{{b}^{5}}{{c}^{2}}}[/latex]
Solution
Factor to find variables with even exponents: [latex] \sqrt{{{a}^{2}}\cdot a\cdot {{b}^{4}}\cdot{b}\cdot{{c}^{2}}}[/latex]
Separate the perfect square factors into individual radicals: [latex] \sqrt{a^2}\cdot\sqrt{b^4}\cdot\sqrt{c^2}\cdot \sqrt{a\cdot b}[/latex]
Take the square root of each radical with a perfect square radicand. Remember that [latex] \sqrt{{{a}^{2}}}=\left| a \right|[/latex]: [latex] \left| a \right|\cdot \left |{b^2}\right |\cdot\left|{c}\right|\cdot\sqrt{a\cdot b}[/latex]
Simplify: [latex] \left| a\right |{b^2}\left | c \right|\sqrt{ab}[/latex]
Answer
[latex]\sqrt{{{a}^{3}}{{b}^{5}}{{c}^{2}}} \left| a\right |{b^2}\left | c \right|\sqrt{ab}[/latex]
Try It
Simplify. [latex] \sqrt{{{x}^{4}}{{y}^{9}}{{z}^{3}}}[/latex]
Try It
Simplify. [latex] \sqrt{{{m}^{9}}{{n}^{7}}{{p}^{11}}}[/latex]
