6.1 Square root expressions (Page 2/2)

Page 2 / 2

The number $\sqrt{50}$ is between what two whole numbers?

Since $7^{2} = 49, \sqrt{49} = 7.$

Since $8^{2} = 64, \sqrt{64} = 8.$ Thus,

$7 < \sqrt{50} < 8$

Thus, $\sqrt{50}$ is a number between 7 and 8.

A number line with arrows on each end, labeled at zero, seven and eight. Seven is also labeled as square root of forty-nine and eight is labeled as square root of sixty-four. There is a closed circle at square root of fifty and it is labeled as square root of fifty.

Practice set b

Write the principal and secondary square roots of each number.

100

$\sqrt{100} = 10$ and $- \sqrt{100} = - 10$

121

$\sqrt{121} = 11$ and $- \sqrt{121} = - 11$

$\sqrt{35}$ and $- \sqrt{35}$

Use a calculator to obtain a decimal approximation for the two square roots of 35. Round to two decimal places.

$5.92 and - 5.92$

Meaningful expressions

Since we know that the square of any real number is a positive number or zero, we can see that expressions such as $\sqrt{- 16}$ do not describe real numbers. There is no real number that can be squared that will produce −16. For $\sqrt{x}$ to be a real number, we must have $x \geq 0.$ In our study of algebra, we will assume that all variables and all expressions in radicands represent nonnegative numbers (numbers greater than or equal to zero).

Sample set c

Write the proper restrictions that must be placed on the variable so that each expression represents a real number.

For $\sqrt{x - 3}$ to be a real number, we must have

$\begin{array}{l} x - 3 \geq 0 & or & x \geq 3 \end{array}$

For $\sqrt{2 m + 7}$ to be a real number, we must have

$\begin{array}{l} 2 m + 7 \geq 0 & or & 2 m \geq - 7 & or & m \geq \frac{- 7}{2} \end{array}$

Practice set c

Write the proper restrictions that must be placed on the variable so that each expression represents a real number.

$\sqrt{x + 5}$

$x \geq - 5$

$\sqrt{y - 8}$

$y \geq 8$

$\sqrt{3 a + 2}$

$a \geq - \frac{2}{3}$

$\sqrt{5 m - 6}$

$m \geq \frac{6}{5}$

Simplifying square roots

When variables occur in the radicand, we can often simplify the expression by removing the radical sign. We can do so by keeping in mind that the radicand is the square of some other expression. We can simplify a radical by seeking an expression whose square is the radicand. The following observations will help us find the square root of a variable quantity.

Since ${(x^{3})}^{2} = x^{3} \cdot^{2} = x^{6}, x^{3}$ is a square root of $x^{6} .$ Also
Six divided by two is equal to three. There is an arrow pointing towards six that is labeled as

Since ${(x^{4})}^{2} = x^{4 \cdot 2} = x^{8}, x^{4}$ is a square root of $x^{8} .$ Also

Eight divided by two is equal to four. There is an arrow pointing towards eight that is labeled as

Since ${(x^{6})}^{2} = x^{6 \cdot 2} = x^{12}, x^{6}$ is a square root of $x^{12} .$ Also

Twelve divided by two is equal to six. There is an arrow pointing towards twelve that is labeled as

These examples suggest the following rule:

If a variable has an even exponent, its square root can be found by dividing that exponent by 2.

The examples of Sample Set B illustrate the use of this rule.

Sample set d

Simplify each expression by removing the radical sign. Assume each variable is nonnegative .

$\begin{array}{l} \sqrt{a^{2}} . & We seek an expression whose square is a^{2} . Since {(a)}^{2} = a^{2}, \\ \sqrt{a^{2}} = a & Notice that 2 \div 2 = 1. \end{array}$

$\begin{array}{l} \sqrt{y^{8}} . & We seek an expression whose square is y^{8} . Since {(y^{4})}^{2} = y^{8}, \\ \sqrt{y^{8}} = y^{4} & Notice that 8 \div 2 = 4. \end{array}$

$\begin{array}{l} \sqrt{25 m^{2} n^{6}} . & We seek an expression whose square is 25 m^{2} n^{6} . Since {(5 m n^{3})}^{2} = 25 m^{2} n^{6}, \\ \sqrt{25 m^{2} n^{6}} = 5 m n^{3} & Notice that 2 \div 2 = 1 and 6 \div 2 = 3. \end{array}$

$\begin{array}{l} - \sqrt{121 a^{10} {(b - 1)}^{4}} . & We seek an expression whose square is 121 a^{10} {(b - 1)}^{4} . Since \\ {[11 a^{5} {(b - 1)}^{2}]}^{2} & = & 121 a^{10} {(b - 1)}^{4}, \\ \sqrt{121 a^{10} {(b - 1)}^{4}} & = & 11 a^{5} {(b - 1)}^{2} \\ Then, - \sqrt{121 a^{10} {(b - 1)}^{4}} & = & - 11 a^{5} {(b - 1)}^{2} & Notice that 10 \div 2 = 5 and 4 \div 2 = 2. \end{array}$

Practice set d

Simplify each expression by removing the radical sign. Assume each variable is nonnegative.

$\sqrt{y^{8}}$

$y^{4}$

$\sqrt{16 a^{4}}$

$4 a^{2}$

$\sqrt{49 x^{4} y^{6}}$

$7 x^{2} y^{3}$

$- \sqrt{100 x^{8} y^{12} z^{2}}$

$- 10 x^{4} y^{6} z$

$- \sqrt{36 {(a + 5)}^{4}}$

$- 6 {(a + 5)}^{2}$

$\sqrt{225 w^{4} {(z^{2} - 1)}^{2}}$

$15 w^{2} (z^{2} - 1)$

$\sqrt{0.25 y^{6} z^{14}}$

$0.5 y^{3} z^{7}$

$\sqrt{x^{2 n}},$ where $n$ is a natural number.

$x^{n}$

$\sqrt{x^{4 n}},$ where $n$ is a natural number.

$x^{2 n}$

Exercises

How many square roots does every positive real number have?

two

The symbol $\sqrt{\begin{matrix} \end{matrix}}$ represents which square root of a number?

The symbol – $\sqrt{\begin{matrix} \end{matrix}}$ represents which square root of a number?

secondary

For the following problems, find the two square roots of the given number.

9 and $- 9$

121

11 and $- 11$

144

225

15 and $- 15$

10,000

$\frac{1}{16}$

$\frac{1}{4} and - \frac{1}{4}$

$\frac{1}{49}$

$\frac{25}{36}$

$\frac{5}{6} and - \frac{5}{6}$

$\frac{121}{225}$

$0.04$

$0.2 and - 0.2$

$0.16$

$1.21$

$1.1 and - 1.1$

For the following problems, evaluate each expression. If the expression does not represent a real number, write "not a real number."

$\sqrt{49}$

$\sqrt{64}$

$- \sqrt{36}$

$- \sqrt{100}$

$- 10$

$- \sqrt{169}$

$- \sqrt{\frac{36}{81}}$

$- \frac{2}{3}$

$- \sqrt{\frac{121}{169}}$

$\sqrt{- 225}$

not a real number

$\sqrt{- 36}$

$- \sqrt{- 1}$

not a real number

$- \sqrt{- 5}$

$- (- \sqrt{9})$

$- (- \sqrt{0.81})$

For the following problems, write the proper restrictions that must be placed on the variable so that the expression represents a real number.

$\sqrt{y + 10}$

$y \geq - 10$

$\sqrt{x + 4}$

$\sqrt{a - 16}$

$a \geq 16$

$\sqrt{h - 11}$

$\sqrt{2 k - 1}$

$k \geq \frac{1}{2}$

$\sqrt{7 x + 8}$

$\sqrt{- 2 x - 8}$

$x \leq - 4$

$\sqrt{- 5 y + 15}$

For the following problems, simplify each expression by removing the radical sign.

$\sqrt{m^{6}}$

$m^{3}$

$\sqrt{k^{10}}$

$\sqrt{a^{8}}$

$a^{4}$

$\sqrt{h^{16}}$

$\sqrt{x^{4} y^{10}}$

$x^{2} y^{5}$

$\sqrt{a^{6} b^{20}}$

$\sqrt{a^{4} b^{6}}$

$a^{2} b^{3}$

$\sqrt{x^{8} y^{14}}$

$\sqrt{81 a^{2} b^{2}}$

$9 a b$

$\sqrt{49 x^{6} y^{4}}$

$\sqrt{100 m^{8} n^{2}}$

$10 m^{4} n$

$\sqrt{225 p^{14} r^{16}}$

$\sqrt{36 x^{22} y^{44}}$

$6 x^{11} y^{22}$

$\sqrt{169 w^{4} z^{6} {(m - 1)}^{2}}$

$\sqrt{25 x^{12} {(y - 1)}^{4}}$

$5 x^{6} {(y - 1)}^{2}$

$\sqrt{64 a^{10} {(a + 4)}^{14}}$

$\sqrt{9 m^{6} n^{4} {(m + n)}^{18}}$

$3 m^{3} n^{2} {(m + n)}^{9}$

$\sqrt{25 m^{26} n^{42} r^{66} s^{84}}$

$\sqrt{{(f - 2)}^{2} {(g + 6)}^{4}}$

$(f - 2) {(g + 6)}^{4}$

$\sqrt{{(2 c - 3)}^{6} + {(5 c + 1)}^{2}}$

$- \sqrt{64 r^{4} s^{22}}$

$- 8 r^{2} s^{11}$

$- \sqrt{121 a^{6} {(a - 4)}^{8}}$

$- [- \sqrt{{(w + 6)}^{2}}]$

$w + 6$

$- [- \sqrt{4 a^{2} b^{2} {(c^{2} + 8)}^{2}}]$

$\sqrt{1.21 h^{4} k^{4}}$

$1.1 h^{2} k^{2}$

$\sqrt{2.25 m^{6} p^{6}}$

$- \sqrt{\frac{169 a^{2} b^{4} c^{6}}{196 x^{4} y^{6} z^{8}}}$

$- \frac{13 a b^{2} c^{3}}{14 x^{2} y^{3} z^{4}}$

$- [\sqrt{\frac{81 y^{4} {(z - 1)}^{2}}{225 x^{8} z^{4} w^{6}}}]$

Exercised for review

( [link] ) Find the quotient. $\frac{x^{2} - 1}{4 x^{2} - 1} \div \frac{x - 1}{2 x + 1} .$

$\frac{x + 1}{2 x - 1}$

( [link] ) Find the sum. $\frac{1}{x + 1} + \frac{3}{x + 1} + \frac{2}{x^{2} - 1} .$

( [link] ) Solve the equation, if possible: $\frac{1}{x - 2} = \frac{3}{x^{2} - x - 2} - \frac{3}{x + 1} .$

No solution $; x = 2$ is excluded.

( [link] ) Perform the division: $\frac{15 x^{3} - 5 x^{2} + 10 x}{5 x} .$

( [link] ) Perform the division: $\frac{x^{3} - 5 x^{2} + 13 x - 21}{x - 3} .$

$x^{2} - 2 x + 7$

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Source: OpenStax, Algebra ii for the community college. OpenStax CNX. Jul 03, 2014 Download for free at http://cnx.org/content/col11671/1.1

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