ORCCA Square Root Properties

Section 9.1 Square Root Properties

In this chapter, we will learn how to both simplify square roots and to do operations with square roots.

Definition 9.1.1. The Definition of the Square Root of a Number.

If \(y^2=x\) for a positive number \(y\text{,}\) then \(y\) is called the square root of \(x\text{,}\) and we write \(y=\sqrt{x}\text{,}\) where the \(\sqrt{\phantom{x}}\) symbol is called the radical or the root. We call expressions with a root symbol radical expressions. The number inside the radical is called the radicand.

For example, since \(4^2=16\text{,}\) then \(\sqrt{16}=4\text{.}\) Both \(\sqrt{2}\) and \(3\sqrt{2}\) are radical expressions. In both expressions, the number \(2\) is the radicand. You can review the square root basics in Section 1.4.

The word “radical” means something like “on the fringes” when used in politics, sports, and other places. It actually has that same meaning in math, when you consider a square with area \(A\) as in Figure 9.1.2.

A drawing of a square labeled with area A; the side lengths are labeled the square root of A — Figure 9.1.2. “Radical” means “off to the side.”

Subsection 9.1.1 Estimating Square Roots

When the radicand is a perfect square, its square root is a rational number. If the radicand is not a perfect square, the square root is irrational. We want to be able to estimate square roots without using a calculator.

To estimate \(\sqrt{10}\text{,}\) we can find the nearest perfect squares that are whole numbers on either side of \(10\text{.}\) Recall that the perfect squares are \(1, 4, 9, 16, 25, 36, 49, 64,\dots\) The perfect square that is just below \(10\) is \(9\) and the perfect square just above \(10\) is \(16\text{.}\) This tells us that \(\sqrt{10}\) is between \(\sqrt{9}\) and \(\sqrt{16}\text{,}\) or between \(3\) and \(4\text{.}\) We can also say that \(\sqrt{10}\) is much closer to \(3\) than \(4\) because \(10\) is closer to \(9\text{,}\) so we think \(3.1\) or \(3.2\) would be a good estimate.

To check our estimate, let's find \(\sqrt{10}\) with a calculator:

\begin{equation*} \sqrt{10}\approx3.162 \end{equation*}

The actual value is just above \(3\) as we estimated, and between \(3.1\) and \(3.2\text{.}\) Let's look at some more examples.

Checkpoint 9.1.3.

Example 9.1.4.

Estimate \(\sqrt{3.2}\) without a calculator.

Explanation

The radicand \(3.2\) is between \(1\) and \(4\text{,}\) so \(\sqrt{3.2}\) is between \(\sqrt{1}\) and \(\sqrt{4}\text{,}\) or between \(1\) and \(2\text{.}\)

To be more precise, we notice that \(3.2\) is much closer to \(4\) than \(1\text{.}\) We estimate \(\sqrt{3.2}\) to be about \(1.8\text{.}\)

We will check our estimate with a calculator:

\begin{equation*} \sqrt{3.2}\approx1.788 \end{equation*}

Subsection 9.1.2 Multiplication and Division Properties of Square Roots

Here is an example using perfect squares and the rules of exponents to show a relationship between the product of two square roots:

\begin{gather*} \sqrt{9\cdot16}=\sqrt{3^2\cdot 4^2}=\sqrt{(3\cdot4)^2}=3\cdot4=12\\ \end{gather*}

and

\begin{gather*} \sqrt{9}\cdot\sqrt{16}=\sqrt{3^2}\cdot\sqrt{4^2}=3\cdot4=12 \end{gather*}

Whether we multiply the radicands first or take the square roots first, we get the same result. This tells us that in multiplication with radicals, we can combine factors into a single radical or separate them as needed.

Now let's look at division. When we learned how to find the square root of a fraction in Section 1.4, we saw that the numerators and denominators could be simplified separately. We multiply the numerators and denominators independently. Here is an example of two different ways to simplify a fraction in a square root:

\begin{gather*} \sqrt{\frac{25}{9}}=\sqrt{\left(\frac{5}{3}\right)^2}=\frac{5}{3}\\ \end{gather*}

and

\begin{gather*} \frac{\sqrt{25}}{\sqrt{9}}=\frac{\sqrt{5^2}}{\sqrt{3^2}}=\frac{5}{3} \end{gather*}

Just like with multiplication, we can separate the numerators and denominators in a radical expression or combine them as needed. Note that we worked with expressions that were perfect squares, but these properties will work regardless of the number inside the radical. Let's summarize these properties.

Fact 9.1.5. Multiplication and Division Properties of Square Roots.

For any positive real numbers \(x\) and \(y\) we have the following properties:

Multiplication Property of Square Roots: \(\displaystyle \sqrt{x\cdot y}=\sqrt{x}\cdot\sqrt{y}\)
Division Property of Square Roots: \(\displaystyle \sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\)

Subsection 9.1.3 Simplifying Square Roots

We can use the Multiplication and Division Properties of Square Roots to simplify a radicand that is not a perfect square. Simplifying radicals is similar to simplifying fractions because we want the radicand to be as small as possible.

To understand why we can simplify radicals, let's use a calculator to compare \(\sqrt{12}\) and \(2\sqrt{3}\text{.}\)

\begin{align*} \sqrt{12}\amp=3.4641\ldots \amp\amp\text{and}\amp2\sqrt{3}\amp=3.4641\ldots \end{align*}

These are equivalent expressions so let's see how we can simplify \(\sqrt{12}\) to \(2\sqrt{3}\text{.}\)

First, we will make a table of factor pairs for the number \(12\text{,}\) as we did in Section 7.3.

\begin{align*} 1\amp\cdot12\\ 2\amp\cdot6\\ 3\amp\cdot4 \end{align*}

The factor pair with the largest perfect square is \(3\cdot4\text{.}\) We will use the property of multiplying radicals to separate the perfect square from the other factor. We write the perfect square first because it will end up in front of the radical.

\begin{align*} \sqrt{12}\amp=\sqrt{4}\cdot\sqrt{3}\\ \amp=2\cdot\sqrt{3}\\ \amp=2\sqrt{3} \end{align*}

This process can be used to simplify any square root, or to determine that it is fully simplified. Let's look at a few more examples.

Example 9.1.6.

Simplify \(\sqrt{72}\text{.}\)

Explanation

Here is a table of factor pairs for the number \(72\text{.}\)

\begin{align*} 1\amp\cdot72\amp4\amp\cdot18\\ 2\amp\cdot36\amp6\amp\cdot12\\ 3\amp\cdot24\amp8\amp\cdot9 \end{align*}

The largest perfect square is \(36\) so we will rewrite \(72\) as \(36\cdot2\text{.}\)

\begin{align*} \sqrt{72}\amp=\sqrt{36\cdot2}\\ \amp=\sqrt{36}\cdot\sqrt{2}\\ \amp=6\sqrt{2} \end{align*}

Notice that if we had chosen \(4\cdot18\) we could simplify the radical partially, but we would need to continue and find the perfect square of \(9\) in \(18\text{.}\)

Checkpoint 9.1.7.

Example 9.1.8.

Simplify \(\sqrt{30}\text{.}\)

Explanation

Here is a table of factor pairs for the number \(30\text{.}\)

\begin{align*} 1\amp\cdot30\amp3\amp\cdot10\\ 2\amp\cdot15\amp5\amp\cdot6 \end{align*}

The number \(30\) does not have any factors that are perfect squares so it cannot be simplified further.

We can also use the Division Property of Square Roots to simplify expressions.

Example 9.1.9.

Simplify \(\sqrt{\frac{9}{16}}\text{.}\)
Simplify \(\frac{\sqrt{50}}{\sqrt{2}}\text{.}\)

Explanation

For the first expression, we will use the Division Property of Square Roots:

\begin{align*} \sqrt{\frac{9}{16}}\amp=\frac{\sqrt{9}}{\sqrt{16}}\\ \amp=\frac{3}{4} \end{align*}
For the second expression, we use the same property in reverse: \(\frac{\sqrt{x}}{\sqrt{y}}=\sqrt{\frac{x}{y}}\text{:}\)

\begin{align*} \frac{\sqrt{50}}{\sqrt{2}}\amp=\sqrt{\frac{50}{2}}\\ \amp=\sqrt{25}\\ \amp=5 \end{align*}

Subsection 9.1.4 Multiplying Square Root Expressions

If we use the Multiplication Property and the Division Property of Square Roots in the reverse order as

\begin{align*} \sqrt{x}\cdot\sqrt{y}\amp=\sqrt{xy}\amp\amp \text{and}\amp \frac{\sqrt{x}}{\sqrt{y}}\amp=\sqrt{\frac{x}{y}}\text{,} \end{align*}

we can use these properties to multiply and divide square root expressions. We want to simplify each radical first to keep the radicands as small as possible. Let's look at a few examples.

Example 9.1.10.

Multiply \(\sqrt{8}\cdot\sqrt{54}\text{.}\)

Explanation

We will simplify each radical first, and then multiply them together. We do not want to multiply \(8\cdot54\) because we will end up with a larger number that is harder to factor.

\begin{align*} \sqrt{8}\cdot\sqrt{54}\amp=\sqrt{4\cdot2}\cdot\sqrt{9\cdot6}\\ \amp=2\sqrt{2}\cdot3\sqrt{6}\\ \amp=2\cdot3\sqrt{2\cdot6}\\ \amp=2\cdot3\sqrt{2\cdot2\cdot3}\\ \amp=6\cdot2\sqrt{3}\\ \amp=12\sqrt{3} \end{align*}

We could have multiplied \(2\cdot6\) inside the radical to get \(12\) and then factored \(12\) into \(4\cdot3\text{.}\) Whenever you find a pair of identical factors, this is a perfect square.

Checkpoint 9.1.11.

Example 9.1.12.

Multiply \(\sqrt{\frac{6}{5}}\cdot\sqrt{\frac{3}{5}}\text{.}\)

Explanation

\begin{align*} \sqrt{\frac{6}{5}}\cdot\sqrt{\frac{3}{5}}\amp=\sqrt{\frac{6}{5}\cdot\frac{3}{5}}\\ \amp=\sqrt{\frac{18}{25}}\\ \amp=\frac{\sqrt{18}}{\sqrt{25}}\\ \amp=\frac{\sqrt{9\cdot2}}{5}\\ \amp=\frac{3\sqrt{2}}{5} \end{align*}

Subsection 9.1.5 Adding and Subtracting Square Root Expressions

We learned the Multiplication Property previously and applied this to multiplication of square roots, but we cannot apply this property to the operations of addition or subtraction. Here are two examples to demonstrate this:

\begin{align*} \sqrt{9+16}\amp\stackrel{?}{=}\sqrt{9}+\sqrt{16}\amp\sqrt{169-25}\amp\stackrel{?}{=}\sqrt{169}-\sqrt{25}\\ \sqrt{25}\amp\stackrel{?}{=}3+4\amp\sqrt{144}\amp\stackrel{?}{=}13-5\\ 5\amp\stackrel{\text{no}}{=}7\amp12\amp\stackrel{\text{no}}{=}8 \end{align*}

We do not get the same result if we separate the radicals, so we must complete any additions and subtractions inside the radical first.

To add and subtract radical expressions, we will need to recognize that we can only add and subtract like terms. In this case, we will call them like radicals. In fact, adding like radicals will work just like adding like terms

\begin{align*} x+x\amp=2x\\ \end{align*}

and

\begin{align*} \sqrt{5}+\sqrt{5}\amp=2\sqrt{5} \end{align*}

We can verify that the second equation is true by replacing \(x\) with \(\sqrt{5}\) in the second equation. Let's look at a few more examples.

Example 9.1.13.

Simplify \(\sqrt{2}+\sqrt{8}\text{.}\)

Explanation

\begin{align*} \sqrt{2}+\sqrt{8}\amp=\sqrt{2}+\sqrt{4\cdot2}\\ \amp=\sqrt{2}+2\sqrt{2}\\ \amp=3\sqrt{2} \end{align*}

To help understand \(\sqrt{2}+2\sqrt{2}=3\sqrt{2}\text{,}\) think of \(x+2x=3x\) or “a taco plus two tacos is three tacos.”

Checkpoint 9.1.14.

Example 9.1.15.

Simplify \(\sqrt{2}+\sqrt{27}\text{.}\)

Explanation

\begin{align*} \sqrt{2}+\sqrt{27}\amp=\sqrt{2}+\sqrt{9\cdot3}\\ \amp=\sqrt{2}+3\sqrt{3} \end{align*}

We cannot simplify the expression further because \(\sqrt{2}\) and \(\sqrt{3}\) are not like radicals.

Example 9.1.16.

Simplify \(\sqrt{6}-\sqrt{18}\cdot\sqrt{12}\text{.}\)

Explanation

In this example, we should multiply the latter two square roots first and then see if we have like radicals.

\begin{align*} \sqrt{6}-\sqrt{18}\cdot\sqrt{12}\amp=\sqrt{6}-\sqrt{9\cdot2}\cdot\sqrt{4\cdot3}\\ \amp=\sqrt{6}-3\sqrt{2}\cdot2\sqrt{3}\\ \amp=\sqrt{6}-3\cdot2\cdot\sqrt{2}\cdot\sqrt{3}\\ \amp=\sqrt{6}-6\sqrt{6}\\ \amp=-5\sqrt{6} \end{align*}

Subsection 9.1.6 Rationalizing the Denominator

When simplifying square root expressions, we have seen that we need to make the radicand as small as possible. Another rule is that we do not leave any irrational numbers, such as \(\sqrt{3}\) or \(2\sqrt{5}\text{,}\) in the denominator of a fraction. In other words, we want the denominator to be rational. The process of dealing with such numbers in the denominator is called rationalizing the denominator.

Let's see how we can remove the square root symbol from the denominator in \(\frac{1}{\sqrt{5}}\text{.}\) If we multiply a radical by itself, the result is the radicand, by Definition 9.1.1. As an example:

\begin{equation*} \sqrt{5}\cdot\sqrt{5}=5 \end{equation*}

To write \(\frac{1}{\sqrt{5}}\) as an equivalent fraction, we must multiply both the numerator and denominator by the same number. If we multiply the numerator and denominator by \(\sqrt{5}\text{,}\) we have:

\begin{align*} \frac{1}{\sqrt{5}}\amp=\frac{1}{\sqrt{5}} \multiplyright{\frac{\sqrt{5}}{\sqrt{5}}}\\ \amp=\frac{1\cdot\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}\\ \amp=\frac{\sqrt{5}}{5} \end{align*}

We can use a calculator to verify that \(\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\text{.}\) They both are \(0.4472\ldots\text{.}\) Let's look at a few more examples.

Example 9.1.17.

Rationalize the denominator in \(\frac{6}{\sqrt{3}}\text{.}\)

Explanation

We will rationalize this denominator by multiplying the numerator and denominator by \(\sqrt{3}\text{:}\)

\begin{align*} \frac{6}{\sqrt{3}}\amp=\frac{6}{\sqrt{3}}\multiplyright{\frac{\sqrt{3}}{\sqrt{3}}}\\ \amp=\frac{6\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}\\ \amp=\frac{6\sqrt{3}}{3}\\ \amp=2\sqrt{3} \end{align*}

Note that we reduced any fractions that are outside the radical.

Checkpoint 9.1.18.

Example 9.1.19.

Rationalize the denominator in \(\sqrt{\frac{2}{7}}\text{.}\)

Explanation

\begin{align*} \sqrt{\frac{2}{7}}\amp=\frac{\sqrt{2}}{\sqrt{7}}\\ \amp=\frac{\sqrt{2}}{\sqrt{7}}\multiplyright{\frac{\sqrt{7}}{\sqrt{7}}}\\ \amp=\frac{\sqrt{2}\cdot\sqrt{7}}{\sqrt{7}\cdot\sqrt{7}}\\ \amp=\frac{\sqrt{14}}{7} \end{align*}

Subsection 9.1.7 More Complicated Square Root Operations

In Section 6.5, we learned how to multiply polynomials like \(2(x+3)\) and \((x+2)(x+3)\text{.}\) All the methods we learned apply when we multiply square root expressions. We will look at a few examples done with different methods.

Example 9.1.20.

Multiply \(\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)\text{.}\)

Explanation

We will use the distributive property to do this problem:

\begin{align*} \sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)\amp=\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{2}\\ \amp=\sqrt{15}-\sqrt{10} \end{align*}

Example 9.1.21.

Multiply \(\left(\sqrt{6}+\sqrt{12}\right)\left(\sqrt{3}-\sqrt{2}\right)\text{.}\)

Explanation

We will use the FOIL Method to do this problem:

\begin{align*} \left(\sqrt{6}+\sqrt{12}\right)\left(\sqrt{3}-\sqrt{2}\right)\amp=\sqrt{6}\sqrt{3}-\sqrt{6}\sqrt{2}+\sqrt{12}\sqrt{3}-\sqrt{12}\sqrt{2}\\ \amp=\sqrt{18}-\sqrt{12}+\sqrt{36}-\sqrt{24}\\ \amp=3\sqrt{2}-2\sqrt{3}+6-2\sqrt{6} \end{align*}

When simplifying radicals it is useful to keep in mind that for any \(x\ge 0\text{,}\)

\begin{equation*} \sqrt{x}\cdot\sqrt{x}=x\text{.} \end{equation*}

Example 9.1.22.

Expand \(\left(\sqrt{3}-\sqrt{2}\right)^2\text{.}\)

Explanation

We will use the FOIL method to expand this expression:

\begin{align*} \left(\sqrt{3}-\sqrt{2}\right)^2\amp=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)\\ \amp=\left(\sqrt{3}\right)^2-\sqrt{3}\sqrt{2}-\sqrt{2}\sqrt{3}+\left(\sqrt{2}\right)^2\\ \amp=3-\sqrt{6}-\sqrt{6}+2\\ \amp=5-2\sqrt{6} \end{align*}

Example 9.1.23.

Multiply \(\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)\text{.}\)

Explanation

We will again use the FOIL method to expand this expression, but will note that it is a special form \((a-b)(a+b)\) and will simplify to \(a^2-b^2\text{:}\)

\begin{align*} \left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)\amp=\left(\sqrt{5}\right)^2+\sqrt{5}\sqrt{7}-\sqrt{7}\sqrt{5}-\left(\sqrt{7}\right)^2\\ \amp=5+\sqrt{35}-\sqrt{35}-7\\ \amp=-2 \end{align*}

Exercises 9.1.8 Exercises

Review and Warmup

1.

Which of the following are square numbers? There may be more than one correct answer.

\(\displaystyle 108\)
\(\displaystyle 131\)
\(\displaystyle 37\)
\(\displaystyle 9\)
\(\displaystyle 36\)
\(\displaystyle 64\)

2.

Which of the following are square numbers? There may be more than one correct answer.

\(\displaystyle 87\)
\(\displaystyle 92\)
\(\displaystyle 100\)
\(\displaystyle 62\)
\(\displaystyle 81\)
\(\displaystyle 4\)

3.

Evaluate the following.

\(\displaystyle{ \sqrt{121} }\) =
\(\displaystyle{ \sqrt{81} }\) =
\(\displaystyle{ \sqrt{4} }\) =

4.

Evaluate the following.

\(\displaystyle{ \sqrt{144} }\) =
\(\displaystyle{ \sqrt{25} }\) =
\(\displaystyle{ \sqrt{49} }\) =

5.

Evaluate the following.

\(\displaystyle{ \sqrt{{{\frac{1}{9}}}} }\) =
\(\displaystyle{ \sqrt{{-{\frac{144}{25}}}} }\) =

6.

Evaluate the following.

\(\displaystyle{ \sqrt{{{\frac{9}{100}}}} }\) =
\(\displaystyle{ \sqrt{{-{\frac{25}{49}}}} }\) =

7.

Evaluate the following.

Do not use a calculator.

\(\displaystyle{ \sqrt{25} }\) =
\(\displaystyle{ \sqrt{0.25} }\) =
\(\displaystyle{ \sqrt{2500} }\) =

8.

Evaluate the following.

Do not use a calculator.

\(\displaystyle{ \sqrt{36} }\) =
\(\displaystyle{ \sqrt{0.36} }\) =
\(\displaystyle{ \sqrt{3600} }\) =

9.

Evaluate the following.

Do not use a calculator.

\(\displaystyle{ \sqrt{49} }\) =
\(\displaystyle{ \sqrt{4900} }\) =
\(\displaystyle{ \sqrt{490000} }\) =

10.

Evaluate the following.

Do not use a calculator.

\(\displaystyle{ \sqrt{64} }\) =
\(\displaystyle{ \sqrt{6400} }\) =
\(\displaystyle{ \sqrt{640000} }\) =

11.

Evaluate the following.

Do not use a calculator.

\(\displaystyle{ \sqrt{81} }\) =
\(\displaystyle{ \sqrt{0.81} }\) =
\(\displaystyle{ \sqrt{0.0081} }\) =

12.

Evaluate the following.

Do not use a calculator.

\(\displaystyle{ \sqrt{121} }\) =
\(\displaystyle{ \sqrt{1.21} }\) =
\(\displaystyle{ \sqrt{0.0121} }\) =

13.

Without using a calculator, estimate the value of \(\sqrt{99}\text{:}\)

10.05
9.05
9.95
10.95

14.

Without using a calculator, estimate the value of \(\sqrt{19}\text{:}\)

4.64
3.36
4.36
3.64

Simplify Radical Expressions

15.

Evaluate the following.

\(\displaystyle{\sqrt{{{\frac{9}{64}}}}={}}\).

16.

Evaluate the following.

\(\displaystyle{\sqrt{{{\frac{16}{49}}}}={}}\).

17.

Evaluate the following.

\(-\sqrt{36}={}\).

18.

Evaluate the following.

\(-\sqrt{49}={}\).

19.

Evaluate the following.

\(\sqrt{-64}=\).

20.

Evaluate the following.

\(\sqrt{-100}=\).

21.

Evaluate the following.

\(\displaystyle{\sqrt{-{{\frac{100}{121}}}}={}}\).

22.

Evaluate the following.

\(\displaystyle{\sqrt{-{{\frac{121}{144}}}}={}}\).

23.

Evaluate the following.

\(\displaystyle{-\sqrt{{{\frac{1}{121}}}}={}}\).

24.

Evaluate the following.

\(\displaystyle{-\sqrt{{{\frac{4}{49}}}}={}}\).

25.

Evaluate the following.

\(\displaystyle{\sqrt{25}-\sqrt{16}=}\)
\(\displaystyle{\sqrt{25-16}=}\)

26.

Evaluate the following.

\(\displaystyle{\sqrt{100}-\sqrt{36}=}\)
\(\displaystyle{\sqrt{100-36}=}\)

27.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ \frac{{\sqrt{64}}}{{\sqrt{4}}} =}\)

28.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ \frac{{\sqrt{50}}}{{\sqrt{2}}} =}\)

29.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ \frac{{\sqrt{5}}}{{\sqrt{125}}} =}\)

30.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ \frac{{\sqrt{4}}}{{\sqrt{144}}} =}\)

31.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ {\sqrt{52}} = }\)

32.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ {\sqrt{50}} = }\)

33.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ {\sqrt{300}} = }\)

34.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ {\sqrt{180}} = }\)

35.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ {\sqrt{231}} = }\)

36.

Simplify the radical expression or state that it is not a real number.

\(\displaystyle{ {\sqrt{70}} = }\)

Multiplying Square Root Expressions

37.

Simplify the expression.

\(6\sqrt{7} \cdot 9\sqrt{2}=\)

38.

Simplify the expression.

\(7\sqrt{13} \cdot 6\sqrt{11}=\)

39.

Simplify the expression.

\(8\sqrt{13} \cdot 4\sqrt{{4}} =\)

40.

Simplify the expression.

\(9\sqrt{3} \cdot 9\sqrt{{25}} =\)

41.

Simplify the expression.

\(\displaystyle{2\sqrt{10} \cdot 4\sqrt{45}=}\)

42.

Simplify the expression.

\(\displaystyle{2\sqrt{10} \cdot 2\sqrt{20}=}\)

43.

Simplify the expression.

\(\displaystyle{ {\sqrt{3}} \cdot {8\sqrt{12}} = }\)

44.

Simplify the expression.

\(\displaystyle{ {\sqrt{4}} \cdot {4\sqrt{9}} = }\)

45.

Simplify the expression.

\(\displaystyle{ \sqrt{\frac{1}{6}} \cdot \sqrt{\frac{5}{6}} =}\)

46.

Simplify the expression.

\(\displaystyle{ \sqrt{\frac{1}{7}} \cdot \sqrt{\frac{3}{7}} =}\)

47.

Simplify the expression.

\(\displaystyle{ {\sqrt{\frac{15}{13}}} \cdot {\sqrt{\frac{5}{13}}} =}\)

48.

Simplify the expression.

\(\displaystyle{ {\sqrt{\frac{12}{19}}} \cdot {\sqrt{\frac{6}{19}}} =}\)

Adding and Subtracting Square Root Expressions

49.

Simplify the expression.

\(\displaystyle{{20\sqrt{13}} - {21\sqrt{13}} =}\)

50.

Simplify the expression.

\(\displaystyle{{10\sqrt{7}} - {11\sqrt{7}} =}\)

51.

Simplify the expression.

\(\displaystyle{{11\sqrt{17}} - {11\sqrt{17}} + {10\sqrt{17}} =}\)

52.

Simplify the expression.

\(\displaystyle{{13\sqrt{21}} - {18\sqrt{21}} + {16\sqrt{21}} =}\)

53.

Simplify the expression.

\(\displaystyle{{\sqrt{48}} + {\sqrt{108}} =}\)

54.

Simplify the expression.

\(\displaystyle{{\sqrt{20}} + {\sqrt{45}} =}\)

55.

Simplify the expression.

\(\displaystyle{{\sqrt{252}} - {\sqrt{343}} =}\)

56.

Simplify the expression.

\(\displaystyle{{\sqrt{112}} - {\sqrt{63}} =}\)

57.

Simplify the expression.

\(\displaystyle{{\sqrt{175}} + {\sqrt{63}} + {\sqrt{32}} + {\sqrt{18}} =}\)

58.

Simplify the expression.

\(\displaystyle{{\sqrt{28}} + {\sqrt{112}} + {\sqrt{20}} + {\sqrt{80}} =}\)

59.

Simplify the expression.

\(\displaystyle{{\sqrt{98}} - {\sqrt{8}} - {\sqrt{48}} - {\sqrt{48}} =}\)

60.

Simplify the expression.

\(\displaystyle{{\sqrt{98}} - {\sqrt{18}} - {\sqrt{80}} - {\sqrt{20}} =}\)

Rationalizing the Denominator

61.

Evaluate the following.

\(\displaystyle{\frac{3}{\sqrt{{49}}}}\) = .