# 9.3 Multiplication of square root expressions

 Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. The distinction between the principal square root of the number x and the secondary square root of the number x is made by explanation and by example. The simplification of the radical expressions that both involve and do not involve fractions is shown in many detailed examples; this is followed by an explanation of how and why radicals are eliminated from the denominator of a radical expression. Real-life applications of radical equations have been included, such as problems involving daily output, daily sales, electronic resonance frequency, and kinetic energy.Objectives of this module: be able to use the product property of square roots to multiply square roots.

## Overview

• The Product Property of Square Roots
• Multiplication Rule for Square Root Expressions

## The product property of square roots

In our work with simplifying square root expressions, we noted that

$\sqrt{xy}=\sqrt{x}\sqrt{y}$

Since this is an equation, we may write it as

$\sqrt{x}\sqrt{y}=\sqrt{xy}$

To multiply two square root expressions, we use the product property of square roots.

## The product property $\sqrt{x}\sqrt{y}=\sqrt{xy}$

$\sqrt{x}\sqrt{y}=\sqrt{xy}$

The product of the square roots is the square root of the product.

In practice, it is usually easier to simplify the square root expressions before actually performing the multiplication. To see this, consider the following product:

$\sqrt{8}\sqrt{48}$
We can multiply these square roots in either of two ways:

Simplify then multiply.

$\sqrt{4·2}\sqrt{16·3}=\left(2\sqrt{2}\right)\left(4\sqrt{3}\right)=2·4\sqrt{2·3}=8\sqrt{6}$

Multiply then simplify.

$\sqrt{8}\sqrt{48}=\sqrt{8·48}=\sqrt{384}=\sqrt{64·6}=8\sqrt{6}$

Notice that in the second method, the expanded term (the third expression, $\sqrt{384}$ ) may be difficult to factor into a perfect square and some other number.

## Multiplication rule for square root expressions

The preceding example suggests that the following rule for multiplying two square root expressions.

## Rule for multiplying square root expressions

1. Simplify each square root expression, if necessary.
2. Perform the multiplecation.
3. Simplify, if necessary.

## Sample set a

Find each of the following products.

$\sqrt{3}\sqrt{6}=\sqrt{3·6}=\sqrt{18}=\sqrt{9·2}=3\sqrt{2}$

$\sqrt{8}\sqrt{2}=2\sqrt{2}\sqrt{2}=2\sqrt{2·2}=2\sqrt{4}=2·2=4$

This product might be easier if we were to multiply first and then simplify.

$\sqrt{8}\sqrt{2}=\sqrt{8·2}=\sqrt{16}=4$

$\sqrt{20}\sqrt{7}=\sqrt{4}\sqrt{5}\sqrt{7}=2\sqrt{5·7}=2\sqrt{35}$

$\begin{array}{lll}\sqrt{5{a}^{3}}\sqrt{27{a}^{5}}=\left(a\sqrt{5a}\right)\left(3{a}^{2}\sqrt{3a}\right)\hfill & =\hfill & 3{a}^{3}\sqrt{15{a}^{2}}\hfill \\ \hfill & =\hfill & 3{a}^{3}·a\sqrt{15}\hfill \\ \hfill & =\hfill & 3{a}^{4}\sqrt{15}\hfill \end{array}$

$\begin{array}{lll}\sqrt{{\left(x+2\right)}^{7}}\sqrt{x-1}=\sqrt{{\left(x+2\right)}^{6}\left(x+2\right)}\sqrt{x-1}\hfill & =\hfill & {\left(x+2\right)}^{3}\sqrt{\left(x+2\right)}\sqrt{x-1}\hfill \\ \hfill & =\hfill & {\left(x+2\right)}^{3}\sqrt{\left(x+2\right)\left(x-1\right)}\hfill \\ \begin{array}{cccc}& & & \begin{array}{cccc}& & & \begin{array}{cccc}& & & \text{or}\end{array}\end{array}\end{array}\hfill & =\hfill & {\left(x+2\right)}^{3}\sqrt{{x}^{2}+x-2}\hfill \end{array}$

## Practice set a

Find each of the following products.

$\sqrt{5}\sqrt{6}$

$\sqrt{30}$

$\sqrt{32}\sqrt{2}$

8

$\sqrt{x+4}\sqrt{x+3}$

$\sqrt{\left(x+4\right)\left(x+3\right)}$

$\sqrt{8{m}^{5}n}\sqrt{20{m}^{2}n}$

$4{m}^{3}n\sqrt{10m}$

$\sqrt{9{\left(k-6\right)}^{3}}\sqrt{{k}^{2}-12k+36}$

$3{\left(k-6\right)}^{2}\sqrt{k-6}$

$\sqrt{3}\left(\sqrt{2}+\sqrt{5}\right)$

$\sqrt{6}+\sqrt{15}$

$\sqrt{2a}\left(\sqrt{5a}-\sqrt{8{a}^{3}}\right)$

$a\sqrt{10}-4{a}^{2}$

$\sqrt{32{m}^{5}{n}^{8}}\left(\sqrt{2m{n}^{2}}-\sqrt{10{n}^{7}}\right)$

$8{m}^{3}{n}^{2}\sqrt{n}-8{m}^{2}{n}^{5}\sqrt{5m}$

## Exercises

$\sqrt{2}\sqrt{10}$

$2\sqrt{5}$

$\sqrt{3}\sqrt{15}$

$\sqrt{7}\sqrt{8}$

$2\sqrt{14}$

$\sqrt{20}\sqrt{3}$

$\sqrt{32}\sqrt{27}$

$12\sqrt{6}$

$\sqrt{45}\sqrt{50}$

$\sqrt{5}\sqrt{5}$

5

$\sqrt{7}\sqrt{7}$

$\sqrt{8}\sqrt{8}$

8

$\sqrt{15}\sqrt{15}$

$\sqrt{48}\sqrt{27}$

36

$\sqrt{80}\sqrt{20}$

$\sqrt{5}\sqrt{m}$

$\sqrt{5m}$

$\sqrt{7}\sqrt{a}$

$\sqrt{6}\sqrt{m}$

$\sqrt{6m}$

$\sqrt{10}\sqrt{h}$

$\sqrt{20}\sqrt{a}$

$2\sqrt{5a}$

$\sqrt{48}\sqrt{x}$

$\sqrt{75}\sqrt{y}$

$5\sqrt{3y}$

$\sqrt{200}\sqrt{m}$

$\sqrt{a}\sqrt{a}$

$a$

$\sqrt{x}\sqrt{x}$

$\sqrt{y}\sqrt{y}$

$y$

$\sqrt{h}\sqrt{h}$

$\sqrt{3}\sqrt{3}$

3

$\sqrt{6}\sqrt{6}$

$\sqrt{k}\sqrt{k}$

$k$

$\sqrt{m}\sqrt{m}$

$\sqrt{{m}^{2}}\sqrt{m}$

$m\sqrt{m}$

$\sqrt{{a}^{2}}\sqrt{a}$

$\sqrt{{x}^{3}}\sqrt{x}$

${x}^{2}$

$\sqrt{{y}^{3}}\sqrt{y}$

$\sqrt{y}\sqrt{{y}^{4}}$

${y}^{2}\sqrt{y}$

$\sqrt{k}\sqrt{{k}^{6}}$

$\sqrt{{a}^{3}}\sqrt{{a}^{5}}$

${a}^{4}$

$\sqrt{{x}^{3}}\sqrt{{x}^{7}}$

$\sqrt{{x}^{9}}\sqrt{{x}^{3}}$

${x}^{6}$

$\sqrt{{y}^{7}}\sqrt{{y}^{9}}$

$\sqrt{{y}^{3}}\sqrt{{y}^{4}}$

${y}^{3}\sqrt{y}$

$\sqrt{{x}^{8}}\sqrt{{x}^{5}}$

$\sqrt{x+2}\sqrt{x-3}$

$\sqrt{\left(x+2\right)\left(x-3\right)}$

$\sqrt{a-6}\sqrt{a+1}$

$\sqrt{y+3}\sqrt{y-2}$

$\sqrt{\left(y+3\right)\left(y-2\right)}$

$\sqrt{h+1}\sqrt{h-1}$

$\sqrt{x+9}\sqrt{{\left(x+9\right)}^{2}}$

$\left(x+9\right)\sqrt{x+9}$

$\sqrt{y-3}\sqrt{{\left(y-3\right)}^{5}}$

$\sqrt{3{a}^{2}}\sqrt{15{a}^{3}}$

$3{a}^{2\text{\hspace{0.17em}}}\sqrt{5a}$

$\sqrt{2{m}^{4}{n}^{3}}\sqrt{14{m}^{5}n}$

$\sqrt{12{\left(p-q\right)}^{3}}\sqrt{3{\left(p-q\right)}^{5}}$

$6{\left(p-q\right)}^{4}$

$\sqrt{15{a}^{2}{\left(b+4\right)}^{4}}\sqrt{21{a}^{3}{\left(b+4\right)}^{5}}$

$\sqrt{125{m}^{5}{n}^{4}{r}^{8}}\sqrt{8{m}^{6}r}$

$10{m}^{5}{n}^{2}{r}^{4}\sqrt{10mr}$

$\sqrt{7{\left(2k-1\right)}^{11}{\left(k+1\right)}^{3}}\sqrt{14{\left(2k-1\right)}^{10}}$

$\sqrt{{y}^{3}}\sqrt{{y}^{5}}\sqrt{{y}^{2}}$

${y}^{5}$

$\sqrt{{x}^{6}}\sqrt{{x}^{2}}\sqrt{{x}^{9}}$

$\sqrt{2{a}^{4}}\sqrt{5{a}^{3}}\sqrt{2{a}^{7}}$

$2{a}^{7}\sqrt{5}$

$\sqrt{{x}^{n}}\sqrt{{x}^{n}}$

$\sqrt{{y}^{2}n}\sqrt{{y}^{4}n}$

${y}^{3n}$

$\sqrt{{a}^{2n+5}}\sqrt{{a}^{3}}$

$\sqrt{2{m}^{3n+1}}\sqrt{10{m}^{n+3}}$

$2{m}^{2n+2}\sqrt{5}$

$\sqrt{75{\left(a-2\right)}^{7}}\sqrt{48a-96}$

$\sqrt{2}\left(\sqrt{8}+\sqrt{6}\right)$

$2\left(2+\sqrt{3}\right)$

$\sqrt{5}\left(\sqrt{3}+\sqrt{7}\right)$

$\sqrt{3}\left(\sqrt{x}+\sqrt{2}\right)$

$\sqrt{3x}+\sqrt{6}$

$\sqrt{11}\left(\sqrt{y}+\sqrt{3}\right)$

$\sqrt{8}\left(\sqrt{a}-\sqrt{3a}\right)$

$2\sqrt{2a}-2\sqrt{6a}$

$\sqrt{x}\left(\sqrt{{x}^{3}}-\sqrt{2{x}^{4}}\right)$

$\sqrt{y}\left(\sqrt{{y}^{5}}+\sqrt{3{y}^{3}}\right)$

${y}^{2}\left(y+\sqrt{3}\right)$

$\sqrt{8{a}^{5}}\left(\sqrt{2a}-\sqrt{6{a}^{11}}\right)$

$\sqrt{12{m}^{3}}\left(\sqrt{6{m}^{7}}-\sqrt{3m}\right)$

$6{m}^{2}\left({m}^{3}\sqrt{2}-1\right)$

$\sqrt{5{x}^{4}{y}^{3}}\left(\sqrt{8xy}-5\sqrt{7x}\right)$

## Exercises for review

( [link] ) Factor ${a}^{4}{y}^{4}-25{w}^{2}.$

$\left({a}^{2}{y}^{2}+5w\right)\left({a}^{2}{y}^{2}-5w\right)$

( [link] ) Find the slope of the line that passes through the points $\left(-5,4\right)$ and $\left(-3,4\right).$

( [link] ) Perform the indicated operations:

$\frac{15{x}^{2}-20x}{6{x}^{2}+x-12}·\frac{8x+12}{{x}^{2}-2x-15}÷\frac{5{x}^{2}+15x}{{x}^{2}-25}$

$\frac{4\left(x+5\right)}{{\left(x+3\right)}^{2}}$

( [link] ) Simplify $\sqrt{{x}^{4}{y}^{2}{z}^{6}}$ by removing the radical sign.

( [link] ) Simplify $\sqrt{12{x}^{3}{y}^{5}{z}^{8}}.$

$2x{y}^{2}{z}^{4}\sqrt{3xy}$

a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
how did I we'll learn this
f(x)= 2|x+5| find f(-6)
f(n)= 2n + 1
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Got questions? Join the online conversation and get instant answers!