10.3: Multiply and divide radicals
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When we multiply radicals, we recall the product rule for radicals. As long as the roots of each radical in the product are the same, we can apply the product rule and then simplify as usual. At first, we will bring the radicals together under one radical, then simplify the radical by applying the product rule again.
If \(a\), \(b\) are any two positive real numbers, then \[\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\nonumber\]
In general, if \(a\), \(b\) are any two positive real numbers, then \[\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b},\nonumber\]
where \(n\) is a positive integer and \(n\geq 2\).
Simplify: \(-5\sqrt{14}\cdot 4\sqrt{6}\)
Solution
Notice both radicals are square roots and so, we can apply the product rule. Let’s rewrite the product so that coefficients are with coefficients and radicals are with radicals:
\[\begin{array}{rl}-5\sqrt{14}\cdot 4\sqrt{6}&\text{Rewrite} \\ -5\cdot 4\cdot\sqrt{14}\cdot\sqrt{16}&\text{Apply the product rule} \\ -5\cdot 4\cdot\sqrt{14\cdot 6}&\text{Multiply} \\ -20\sqrt{84} &\text{Simplify }\sqrt{84} \\ -20\sqrt{4\cdot 21}&\text{Apply the product rule} \\ -20\cdot 2\sqrt{21}&\text{Multiply coefficients} \\ -40\sqrt{21}&\text{Product}\end{array}\nonumber\]
Simplify: \(2\sqrt[3]{18}\cdot 6\sqrt[3]{15}\)
Solution
Notice both radicals are cube roots and so, we can apply the product rule. Let’s rewrite the product so that coefficients are with coefficients and radicals are with radicals:
\[\begin{array}{rl}2\sqrt[3]{18}\cdot 6\sqrt[3]{15}&\text{Rewrite} \\ 2\cdot 6\cdot\sqrt[3]{18}\cdot\sqrt[3]{15}&\text{Apply the product rule} \\ 2\cdot 6\cdot\sqrt[3]{18\cdot 15}&\text{Multiply} \\ 12\sqrt[3]{270}&\text{Simplify }\sqrt[3]{270} \\ 12\sqrt[3]{27\cdot 10}&\text{Apply the product rule} \\ 12\cdot 3\sqrt[3]{10}&\text{Multiply coefficients} \\ 36\sqrt[3]{10}&\text{Product}\end{array}\nonumber\]
Multiply Radicals with Monomials
Here we begin to multiply radicals with variables. In this section, we assume all variables to be positive.
Simplify: \(\sqrt[5]{8x^2}\cdot\sqrt[5]{4x^3}\)
Solution
Notice both radicals are fifth roots and so, we can apply the product rule.
\[\begin{array}{rl}\sqrt[5]{8x^2}\cdot\sqrt[5]{4x^3}&\text{Apply the product rule} \\ \sqrt[5]{8x^2\cdot 4x^3}&\text{Multiply} \\ \sqrt[5]{32x^5}&\text{Simplify} \\ \sqrt[5]{2^5\cdot x^5}&\text{Apply the product rule} \\ 2x&\text{Product}\end{array}\nonumber\]
Simplify: \(\sqrt{60x^4}\cdot\sqrt{6x^7}\)
Solution
Notice both radicals are square roots and so, we can apply the product rule.
\[\begin{array}{rl}\sqrt{60x^4}\cdot \sqrt{6x^7}&\text{Apply the product rule} \\ \sqrt{60x^4\cdot 6x^7}&\text{Multiply} \\ \sqrt{360x^{11}}&\text{Simplify} \\ \sqrt{36\cdot 10\cdot x^4\cdot x}&\text{Apply the product rule} \\ 6\cdot x^2\cdot\sqrt{10\cdot x}&\text{Rewrite} \\ 6x^2\sqrt{10x}&\text{Product}\end{array}\nonumber\]
Distribute with Radicals
When there is a factor in front of the parenthesis, we distribute that term to each inside the parenthesis. This method is applied to radicals. Recall, methods never change, just problems. Take the following example:
\[\begin{array}{cc}\color{blue}{2x}\color{black}{(}5y+3)&\color{blue}{2\sqrt{7}}\color{black}{(}5\sqrt{3}+3) \\ \color{blue}{2x}\color{black}{\:\cdot\;}5y+\color{blue}{2x}\color{black}{\:\cdot\:}3&\color{blue}{2\sqrt{7}}\color{black}{\:\cdot\:}5\sqrt{3}+\color{blue}{2\sqrt{7}}\color{black}{\:\cdot\:}3 \\ 10xy+6x&10\sqrt{21}+6\sqrt{7}\end{array}\nonumber\]
Notice, we distribute in the same sense as if we were in the polynomial chapter. Let’s take a look at more examples. Recall, we assume all variables are positive.
Simplify: \(7\sqrt{6}(3\sqrt{10}-5\sqrt{15})\)
Solution
\[\begin{array}{rl}7\sqrt{6}(3\sqrt{10}-5\sqrt{15})&\text{Distribute} \\ \color{blue}{7\sqrt{6}}\color{black}{\:\cdot\:}3\sqrt{10}-\color{blue}{7\sqrt{6}}\color{black}{\:\cdot\:}5\sqrt{15}&\text{Apply the product rule} \\ 21\sqrt{60}-35\sqrt{90}&\text{Simplify each term as usual} \\ 21\sqrt{4\cdot 15}-35\sqrt{9\cdot 10}&\text{Apply the product rule} \\ 21\cdot 2\sqrt{15} - 35\cdot 3\sqrt{10}&\text{Multiply coefficients} \\ 42\sqrt{15}-105\sqrt{10}&\text{Simplified expression}\end{array}\nonumber\]
Note, if the final expression had like radicals, then we would combine like radicals. Even though this resulted in unlike radicals, we continue to add or subtract radicals as usual.
Simplify: \(\sqrt{3}(7\sqrt{15x^3}+8x\sqrt{60x})\)
Solution
\[\begin{array}{rl}\sqrt{3}(7\sqrt{15x^3}+8x\sqrt{60x})&\text{Distribute} \\ \color{blue}{\sqrt{3}}\color{black}{\:\cdot\:}7\sqrt{15x^3}+\color{blue}{\sqrt{3}}\color{black}{\:\cdot\:}8x\sqrt{60x}&\text{Apply the product rule} \\ 7\sqrt{45x^3}+8x\sqrt{180x}&\text{Simplify each term as usual} \\ 7\sqrt{9\cdot 5\cdot x^2\cdot x}+8x\sqrt{36\cdot 5\cdot x}&\text{Apply the product rule} \\ 7\cdot 3x\sqrt{5x}+8x\cdot 6\sqrt{5x}&\text{Multiply coefficients} \\ 21x\color{blue}{\sqrt{5x}}\color{black}{\: +\:}48x\color{blue}{\sqrt{5x}}&\color{black}{\text{Combine like radicals}} \\ 69x\sqrt{5x}&\text{Simplified expression}\end{array}\nonumber\]
Multiple Radicals using FOIL
We can use the method of FOIL to multiply radicals that take the form of the “product of two binomials.” Even though the factors aren’t exactly two binomials, but the expression shares that form. Recall, we are only using the method of FOIL. Again, methods never change, just problems. We continue to assume all variables are positive.
Simplify: \((\sqrt{5}-2\sqrt{3})(4\sqrt{10}+6\sqrt{6})\)
Solution
\[\begin{array}{rl}(\sqrt{5}-2\sqrt{3})(4\sqrt{10}+6\sqrt{6})&\text{FOIL} \\ \underset{\text{F}}{\underbrace{\sqrt{5}\cdot 4\sqrt{10}}} + \underset{\text{O}}{\underbrace{\sqrt{5}\cdot 6\sqrt{6}}}-\underset{\text{I}}{\underbrace{2\sqrt{3}\cdot 4\sqrt{10}}}-\underset{\text{L}}{\underbrace{2\sqrt{3}\cdot 6\sqrt{6}}}&\text{Simplify and apply the product rule} \\ 4\sqrt{50}+6\sqrt{30}-8\sqrt{30}-12\sqrt{18}&\text{Simplify each term as usual} \\ 4\sqrt{25\cdot 2}+6\sqrt{30}-8\sqrt{30}-12\sqrt{9\cdot 2}&\text{Apply the product rule} \\ 4\cdot 5\sqrt{2}+6\sqrt{30}-8\sqrt{30}-12\cdot 3\sqrt{2}&\text{Multiply coefficients} \\ 20\color{blue}{\sqrt{2}}\color{black}{\:+\:}6\color{red}{\sqrt{30}}\color{black}{\: -\:}8\color{red}{\sqrt{30}}\color{black}{\: -\:}36\color{blue}{\sqrt{2}}&\color{black}{\text{Combine like radicals}} \\ -16\sqrt{2}-2\sqrt{30}&\text{Simplified expression} \end{array}\nonumber\]
Clay tablets have been discovered revealing much about Babylonian mathematics dating back from 1800 to 1600 BC. In one of the tables, there is an approximation of \(\sqrt{2}\) accurate to five decimal places: \(1.41421\).
Multiply Radicals with Special-Product Formulas
Simplify: \((5\sqrt{7}+\sqrt{2})^2\)
Solution
This should remind of you of a perfect square trinomial:
\[(a+b)^2=a^2+2ab+b^2\nonumber\]
Since this expression takes the form of a perfect square trinomial, we can apply the same method as we did in multiplying polynomials. Recall, we are only using the method of a perfect square trinomial.
\[\begin{array}{rl}(5\sqrt{7}+\sqrt{2})^2&\text{Apply perfect square trinomial formula} \\ (5\sqrt{7})^2+2(5\sqrt{7})(\sqrt{2})+(\sqrt{2})^2&\text{Simplify each term} \\ 25\cdot \sqrt{7^2}+10\sqrt{14}+\sqrt{2^2}&\text{Notice, }(\sqrt{7})^2=\sqrt{7^2}\text{ and }(\sqrt{2})^2=\sqrt{2^2} \\ 25\cdot 7+10\sqrt{14}+2&\text{Multiply} \\ 175+10\sqrt{14}+2&\text{Combine like terms} \\ 177+10\sqrt{14}&\text{Simplified expression}\end{array}\nonumber\]
Simplify: \((8-\sqrt{5})(8+\sqrt{5})\)
Solution
This should remind of you of a difference of two squares:
\[(a+b)(a-b)=a^2-b^2\nonumber\]
Since this expressions takes the form of a difference of two squares, we can apply the same method as we did in multiplying polynomials. Recall, we are only using the method of a difference of two squares.
\[\begin{array}{rl}(8-\sqrt{5})(8+\sqrt{5})&\text{Apply difference of two squares formula} \\ (8)^2-(\sqrt{5})^2&\text{Simplify each term} \\ 64-\sqrt{5^2}&\text{Notice, }(\sqrt{5})^2=\sqrt{5^2} \\ 64-5&\text{Subtract} \\ 59&\text{Simplified expression}\end{array}\nonumber\]
It’s interesting that the original expression contains radicals and the simplified expression contains no radicals. This displays that even though the original expression may contain radicals, in the process of simplifying, we may result in reducing out all radicals.
Simplify Quotient with Radicals
ls Division with radicals is very similar to multiplication. If we think about division as reducing fractions, we can reduce the coefficients outside the radicals and reduce the values inside the radicals.
Simplify: \(\dfrac{-3+\sqrt{27}}{3}\)
Solution
We simplify the \(\sqrt{27}\) and then try to reduce the fraction.
\[\begin{array}{rl}\dfrac{-3+\sqrt{27}}{3}&\text{Rewrite the radicand} \\ \dfrac{-3+\sqrt{9\cdot 3}}{3}&\text{Apply product rule to the numerator} \\ \dfrac{-3+3\sqrt{3}}{3}&\text{Factor the numerator} \\ \dfrac{3(-1+\sqrt{3})}{3}&\text{Reduce the fraction by a factor of }3 \\ \dfrac{\cancel{3}(-1+\sqrt{3})}{\cancel{3}}&\text{Simplify} \\ -1+\sqrt{3}&\text{Simplified expression}\end{array}\nonumber\]
If \(a\), \(b\) are any two positive real numbers, then \[\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\nonumber\]
In general, if \(a\), \(b\) are any two positive real numbers, then \[\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}},\nonumber\] where \(n\) is a positive integer and \(n\geq 2\).
Simplify: \(\dfrac{\sqrt{44y^6a^4}}{\sqrt{9y^2a^8}}\)
Solution
We apply the quotient rule of radicals and then simplify the radicand:
\[\begin{array}{rl}\dfrac{\sqrt{44y^6a^4}}{\sqrt{9y^2a^8}}&\text{Apply the quotient rule} \\ \sqrt{\dfrac{44y^6a^4}{9y^2a^8}}&\text{Reduce the radicand} \\ \sqrt{\dfrac{44\color{blue}{y^{\cancelto{4}{6}}}\color{black}{\cancel{a^4}}}{9\cancel{y^2}\color{blue}{a^{\cancelto{4}{8}}}}} &\text{Simplify} \\ \sqrt{\dfrac{44y^4}{9a^4}}&\text{Apply the quotient rule} \\ \dfrac{\sqrt{44y^4}}{\sqrt{9a^4}}&\text{Simplify the radicals} \\ \dfrac{\sqrt{4\cdot 11\cdot y^4}}{3a^2}&\text{Rewrite} \\ \dfrac{2y^2\sqrt{11}}{3a^2}&\text{Simplified expression} \end{array}\nonumber\]
Simplify: \(\dfrac{15\sqrt[3]{108}}{20\sqrt[3]{2}}\)
Solution
First we simplify the coefficients, then apply the quotient rule.
\[\begin{array}{rl}\dfrac{\cancelto{3}{15}\cdot\sqrt[3]{108}}{\cancelto{4}{20}\cdot \sqrt[3]{2}}&\text{Simplify coefficients} \\ \dfrac{3\sqrt[3]{108}}{4\sqrt[3]{2}}&\text{Apply quotient rule} \\ \dfrac{3}{4}\cdot\sqrt[3]{\dfrac{108}{2}}&\text{Reduce the radicand} \\ \dfrac{3}{4}\cdot\sqrt[3]{54}&\text{Rewrite the radicand} \\ \dfrac{3}{4}\cdot\sqrt[3]{27\cdot 2}&\text{Apply product rule} \\ \dfrac{3}{4}\cdot 3\cdot\sqrt[3]{2}&\text{Rewrite as one fraction} \\ \dfrac{3\cdot 3\sqrt[3]{2}}{4}&\text{Multiply coefficients} \\ \dfrac{9\sqrt[3]{2}}{4}&\text{Simplified expression}\end{array}\nonumber\]
Multiply and Divide Radicals Homework
Simplify.
\(3\sqrt{5}\cdot -4\sqrt{16}\)
\(\sqrt{12m}\cdot\sqrt{15m}\)
\(\sqrt[3]{4x^3}\cdot\sqrt[3]{2x^4}\)
\(\sqrt{6}(\sqrt{2}+2)\)
\(-5\sqrt{15}(3\sqrt{3}+2)\)
\(5\sqrt{10}(5n+\sqrt{2})\)
\((2+2\sqrt{2})(-3+\sqrt{2})\)
\((\sqrt{5}-5)(2\sqrt{5}-1)\)
\((\sqrt{2a}+2\sqrt{3a})(3\sqrt{2a}+\sqrt{5a})\)
\((-5-4\sqrt{3})(-3-4\sqrt{3})\)
\(\dfrac{\sqrt{12}}{5\sqrt{100}}\)
\(\dfrac{\sqrt{5}}{4\sqrt{125}}\)
\(\dfrac{\sqrt{10}}{\sqrt{8}}\)
\(\dfrac{2\sqrt{3}}{3\sqrt{4}}\)
\(\dfrac{5x^2}{4\sqrt{9x^4y^8}}\)
\(\dfrac{\sqrt{12p^2}}{\sqrt{3p}}\)
\(\dfrac{3\sqrt[3]{10}}{5\sqrt[3]{27}}\)
\(\dfrac{\sqrt[3]{5}}{4\sqrt[3]{625}}\)
\(\dfrac{5\sqrt[4]{5r^4}}{\sqrt[4]{80r^2}}\)
\(-5\sqrt{10}\cdot\sqrt{15}\)
\(\sqrt{5r^3}\cdot -5\sqrt{10r^2}\)
\(3\sqrt[3]{4a^4}\cdot\sqrt[3]{10a^3}\)
\(\sqrt{10}(\sqrt{5}+\sqrt{2})\)
\(5\sqrt{15}(3\sqrt{3}+2)\)
\(\sqrt{15}(\sqrt{5}-3\sqrt{3v})\)
\((-2+\sqrt{3})(-5+2\sqrt{3})\)
\((2\sqrt{3}+\sqrt{5})(5\sqrt{3}+2\sqrt{4})\)
\((-2\sqrt{2p}+5\sqrt{5})(\sqrt{5p}+\sqrt{5p})\)
\((5\sqrt{2}-1)(-\sqrt{2m}+5)\)
\(\dfrac{\sqrt{15}}{2\sqrt{4}}\)
\(\dfrac{\sqrt{12}}{\sqrt{3}}\)
\(\dfrac{\sqrt{2}}{3\sqrt{32}}\)
\(\dfrac{4\sqrt{30}}{\sqrt{15}}\)
\(\dfrac{4\sqrt{12xy^{10}}}{5\sqrt{3xy^4}}\)
\(\dfrac{\sqrt{8n^2}}{\sqrt{32n}}\)
\(\dfrac{\sqrt[3]{15}}{\sqrt[3]{64}}\)
\(\dfrac{\sqrt[4]{4}}{2\sqrt[4]{64}}\)
\(\dfrac{4m}{\sqrt[4]{81m^4n^4}}\)