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Section A.6 Chapter 6 Powers and Roots

Subsection A.6.1 Integer Exponents

Subsubsection A.6.1.1 Evaluate powers with negative exponents

Remember that a negative exponent indicates a reciprocal, so for example
\begin{equation*} x^{-2} = \dfrac{1}{x^2} \end{equation*}
A negative exponent does not mean that the power is negative. So for example
\begin{equation*} 4^{-2} = \dfrac{1}{16}\text{;} \end{equation*}
\(4^{-2}\) does not mean \(-16\text{.}\)
Example A.6.1.
Write each expression without using negative exponents.
  1. \(\displaystyle 10^{-4}\)
  2. \(\displaystyle \left(\dfrac{x}{4}\right)^{-3}\)
Solution.
  1. \(10^{-4} = \dfrac{1}{10^4} = \dfrac{1}{10,000}~\text{,}\) or \(~0.0001\text{.}\)
  2. To compute a negative power of a fraction, we compute the corresponding positive power of its reciprocal. Thus,
    \begin{equation*} \left(\dfrac{x}{4}\right)^{-3} = \left(\dfrac{4}{x}\right)^3 = \dfrac{64}{x^3} \end{equation*}
Example A.6.2.
Write each expression using negative exponents.
  1. \(\displaystyle \dfrac{1}{3a^4a^2}\)
  2. \(\displaystyle \dfrac{8}{x^4}\)
Solution.
  1. \(\displaystyle \dfrac{1}{3a^4a^2} = 3^{-4}a^{-2}\)
  2. \(\displaystyle \dfrac{8}{x^4} = 8x^{-4}\)
Checkpoint A.6.3.
Write each expression using negative exponents and evaluate.
  1. \(\displaystyle (-6)^{-2}\)
  2. \(\displaystyle \left(\dfrac{3}{5}\right)^{-2}\)
Answer.
  1. \(\displaystyle \dfrac{1}{6^2} = \dfrac{1}{36}\)
  2. \(\displaystyle \dfrac{5^2}{3^2} = \dfrac{25}{9}\)
Checkpoint A.6.4.
Write each expression using negative exponents.
  1. \(\displaystyle 4t^{-2}\)
  2. \(\displaystyle (4t)^{-2}\)
Answer.
  1. \(\displaystyle \dfrac{4}{t^2}\)
  2. \(\displaystyle \dfrac{1}{16t^2}\)

Subsubsection A.6.1.2 Use the laws of exponents

Recall the five Laws of Exponents.
Laws of Exponents.
  1. \(\displaystyle \displaystyle{a^m\cdot a^n = a^{m+n}}\)
  2. \(\displaystyle \dfrac{a^m}{a^n}=\begin{cases} a^{m-n} \amp \text{if}~m \gt n\\ \dfrac{1}{a^{n-m}} \amp \text{if}~n \gt m \end{cases}\)
  3. \(\displaystyle \displaystyle{\left(a^m\right)^n=a^{mn}}\)
  4. \(\displaystyle \displaystyle{\left(ab\right)^n=a^n b^n}\)
  5. \(\displaystyle \displaystyle{\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}}\)
Example A.6.5.
Multiply \(~(2x^2y)(5x^4y^3)~\text{.}\)
Solution.
Rearrange the factors to group together the numerical coefficients and the powers of each base.
\begin{equation*} (2x^2y)(5x^4y^3) = (2)(5)x^2x^4yy^3 \end{equation*}
Multiply the coefficients together, and use the first law of exponents to find the products of the variable factors.
\begin{equation*} (2)(5)x^2x^4yy^3 = 10x^6y^4 ~~~~~~~~ \blert{\text{Add exponents on each base.}} \end{equation*}
Example A.6.6.
Divide \(~\dfrac{3x^2y^4}{6x^3y}\text{.}\)
Solution.
Consider the numerical coefficients and the powers of each base separately. Use the second law of exponents to simplify each quotient of powers.
\begin{align*} \dfrac{3x^2y^4}{6x^3y} \amp = \dfrac{3}{6} \cdot \dfrac{x^2}{x^3} \cdot \dfrac{y^4}{y} \amp\amp \blert{\text{Subtract exponents on each base.}}\\ \amp = \dfrac{1}{2} \cdot \dfrac{1}{x^{3-2}} \cdot y^{4-1} \amp\amp \blert{\text{Multiply factors.}}\\ \amp = \dfrac{1}{2} \cdot \dfrac{1}{x} \cdot \dfrac{y^3}{1} = \dfrac{y^3}{2x} \end{align*}
Example A.6.7.
Simplify \(~(5a^3b)^2\text{.}\)
Solution.
Apply the fourth law of exponents and square each factor.
\begin{equation*} (5a^3b)^2 = 5^2(a^3)^2b^2 = 25a^6b^2 ~~~~~~~~ \blert{\text{Apply the third law: multiply exponents.}} \end{equation*}
Example A.6.8.
Simplify \(~\left(\dfrac{2}{y^3}\right)^3\text{.}\)
Solution.
Apply the fifth law of exponents.
\begin{align*} \left(\dfrac{2}{y^3}\right)^3 \amp = \dfrac{2^3}{(y^3)^3} \amp\amp \blert{\text{Cube numerator and denominator.}}\\ \amp = \dfrac{2^3}{y^{3(3)}} = \dfrac{8}{y^9} \amp\amp \blert{\text{Apply the third law.}} \end{align*}
Checkpoint A.6.9.
Multiply \(~-3a^4b(-4a^3b)\text{.}\)
Answer.
\(12a^7b^2\)
Checkpoint A.6.10.
Divide \(~\dfrac{8x^2y}{12x^5y^3}\text{.}\)
Answer.
\(\dfrac{2}{3x^3y^2}\)
Checkpoint A.6.11.
Simplify \(~(6pq^4)^3\text{.}\)
Answer.
\(216p^3q^{12}\)
Checkpoint A.6.12.
Simplify \(~\left(\dfrac{n^3}{k^4}\right)^8\text{.}\)
Answer.
\(\dfrac{n^{24}}{k^{32}}\)

Subsubsection A.6.1.3 Use scientific notation

If we move the decimal point to the left, we are making a number smaller, so we must multiply by a positive power of 10 to compensate. If we move the decimal point to the right, we must multiply by a negative power of 10.
Example A.6.13.
Write each number in scientific notation.
  1. \(\displaystyle 62,000,000\)
  2. \(\displaystyle 0.000431\)
Solution.
  1. First, we position the decimal point so that there is just one nonzero digit to the left of the decimal.
    \begin{equation*} 62,000,000 = 6.2 \times \fillinmath{XXXXXX} \end{equation*}
    To recover \(62,000,000\) from \(6.2\text{,}\) we must move the decimal point seven places to the right. Therefore, we multiply \(6.2\) by \(10^7\text{.}\)
    \begin{equation*} 62,000,000 = 6.2 \times 10^7 \end{equation*}
  2. First, we position the decimal point so that there is just one nonzero digit to the left of the decimal.
    \begin{equation*} 0.000431 = 4.31 \times \fillinmath{XXXXXX} \end{equation*}
    To recover \(0.000431\) from \(4.31\text{,}\) we must move the decimal point seven places to the right. Therefore, we multiply \(4.31\) by \(10^{-4}\text{.}\)
    \begin{equation*} 0.000431 = 4.31 \times 10^{-4} \end{equation*}
Checkpoint A.6.14.
Write each number in scientific notation.
  1. The largest living animal is the blue whale, with an average weight of \(~120,000,000~\) grams.
  2. The smallest animal is the fairy fly beetle, which weighs about \(~0.000005~\) grams.
Answer.
  1. \(\displaystyle 1.2 \times 10^8\)
  2. \(\displaystyle 5 \times 10^{-6}\)

Subsection A.6.2 Roots and Radicals

Subsubsection A.6.2.1 Use the definition of root

Because \(~(\sqrt{a})(\sqrt{a})=a,~\) it is also true that \(~\dfrac{a}{\sqrt{a}}= \sqrt{a}\text{.}\)
Example A.6.15.
Simplify. Do not use a calculator!
  1. \(\displaystyle \left(\sqrt{7}\right)\left(\sqrt{7}\right)\)
  2. \(\displaystyle \sqrt{n}(\sqrt{n})\)
Solution.
By the definition of square root, \(\sqrt{a}\) is a number whose square is \(a\text{.}\)
  1. \(\displaystyle \left(\sqrt{7}\right)\left(\sqrt{7}\right)=7\)
  2. \(\displaystyle \sqrt{n}(\sqrt{n})=n\)
Example A.6.16.
Simplify. Do not use a calculator!
  1. \(\displaystyle \left(\sqrt[3]{5}\right)^3\)
  2. \(\displaystyle \left(\sqrt[3]{4}\right)\left(\sqrt[3]{4}\right)\left(\sqrt[3]{4}\right)\)
Solution.
By the definition of cube root, \(\sqrt[3]{a}\) is a number whose cube is \(a\text{.}\)
  1. \(\displaystyle \left(\sqrt[3]{5}\right)^3=5\)
  2. \(\displaystyle \left(\sqrt[3]{4}\right)\left(\sqrt[3]{4}\right)\left(\sqrt[3]{4}\right)=4\)
Example A.6.17.
Simplify. Do not use a calculator!
  1. \(\displaystyle \dfrac{3}{\sqrt{3}}\)
  2. \(\displaystyle \dfrac{p}{\sqrt{p}}\)
Solution.
  1. \(\displaystyle \dfrac{3}{\sqrt{3}} = \dfrac{\sqrt{3}\sqrt{3}}{\sqrt{3}} = \sqrt{3}\)
  2. \(\displaystyle \dfrac{p}{\sqrt{p}} = \dfrac{\sqrt{p}\sqrt{p}}{\sqrt{p}} = \sqrt{p}\)
Checkpoint A.6.18.
Simplify. Do not use a calculator!
  1. \(\displaystyle \sqrt{5}(\sqrt{5})\)
  2. \(\displaystyle \sqrt{x}(\sqrt{x})\)
Answer.
  1. \(\displaystyle 5\)
  2. \(\displaystyle x\)
Checkpoint A.6.19.
Simplify. Do not use a calculator!
  1. \(\displaystyle \left(\sqrt[3]{9}\right)\left(\sqrt[3]{9}\right)\left(\sqrt[3]{9}\right)\)
  2. \(\displaystyle \left(\sqrt[3]{20}\right)^3\)
Answer.
  1. \(\displaystyle 9\)
  2. \(\displaystyle 20\)
Checkpoint A.6.20.
Simplify. Do not use a calculator!
  1. \(\displaystyle \dfrac{10}{\sqrt{10}}\)
  2. \(\displaystyle \dfrac{H}{\sqrt{H}}\)
Answer.
  1. \(\displaystyle \sqrt{10}\)
  2. \(\displaystyle \sqrt{H}\)

Subsubsection A.6.2.2 Irrational numbers

Rational numbers are the integers and common fractions; we can represent them precisely in decimal form. But the best we can do for an irrational number is to write an approximate decimal form by rounding.
Example A.6.21.
Identify each number as rational or irrational.
  1. \(\displaystyle \sqrt{6}\)
  2. \(\displaystyle \dfrac{-5}{3}\)
  3. \(\displaystyle \sqrt{16}\)
  4. \(\displaystyle \sqrt{\dfrac{5}{9}}\)
Solution.
  1. Irrational: \(~\sqrt{6}~\) is not the quotient of two integers.
  2. Rational: \(~\dfrac{-5}{3}~\) is the quotient of two integers.
  3. Rational: \(~\sqrt{16} = 4~\) is an integer.
  4. Irrational: \(~\sqrt{\dfrac{5}{9}} = \dfrac{\sqrt{5}}{3}~\text{,}\) but \(\sqrt{5}\) is irrational.
Example A.6.22.
Give a decimal approximation rounded to thousandths.
  1. \(\displaystyle 5\sqrt{3}\)
  2. \(\displaystyle \dfrac{-2}{3}\sqrt{21}\)
  3. \(\displaystyle 2+\sqrt[3]{5}\)
Solution.
Use a calculator to evaluate.
  1. Enter \(~5~ \boxed{\sqrt{}}~3\) ENTER and round to three decimal places: \(~8.660 \)
  2. Enter (-) \(2\) \(\boxed{\sqrt{}}~\) \(21\) ) ÷ \(3\) ENTER and round to three decimal places: \(~-3.055\)
  3. Enter \(2\) + MATH 4 \(5\) ENTER and round to three decimal places: \(~3.710\)
Checkpoint A.6.23.
Identify each number as rational or irrational.
  1. \(\displaystyle \sqrt{250}\)
  2. \(\displaystyle \dfrac{\sqrt{3}}{2}\)
  3. \(\displaystyle \dfrac{\sqrt{81}}{4}\)
  4. \(\displaystyle \sqrt[3]{16}\)
Answer.
  1. Irrational
  2. Irrational
  3. Rational
  4. Irrational
Checkpoint A.6.24.
Give a decimal approximation rounded to thousandths.
  1. \(\displaystyle -6\sqrt[3]{5}\)
  2. \(\displaystyle \dfrac{3}{5}\sqrt{76}\)
  3. \(\displaystyle 7-\sqrt{19}\)
Answer.
  1. \(\displaystyle -10.260\)
  2. \(\displaystyle 5.231\)
  3. \(\displaystyle 2.641\)

Subsubsection A.6.2.3 Using exponents and roots

Be careful to avoid tempting but false operations with exponents and roots.
Example A.6.25.
Which equation is a correct application of the laws of exponents?
  1. \(20(1+r)^4 = 20+20r^4~~~~~\) or \(~~~~~(ab^t)^3 = a^3b^{3t}\)
  2. \(2^{t/5} = (2^{1/5})^t~~~~~\) or \(~~~~~6.8(10)^t = 68^t\)
Solution.
  1. The first statement is not correct. There is no law that says \((a+b)^n\) is equivalent to \(a^n+b^n\text{,}\) so \((1+r)^4\) is not equivalent to \(1^4+r^4\) or \(1+r^4\text{.}\)
    However, it is true that \((ab)^n = a^nb^n\text{,}\) so in particular the second statement is true:
    \begin{equation*} ~~(ab^t)^3 = a^3(b^t)^3 = a^3b^{3t} \end{equation*}
  2. The first statement is correct. If we start with \((2^{1/5})^t\text{,}\) we can apply the third law, \((a^m)^n = a^{mn}\text{,}\) to find
    \begin{equation*} (2^{1/5})^t = 2^{(1/5)t} = 2^{t/5}\text{.} \end{equation*}
    In the second statement, 6.8 is not raised to power \(t\text{,}\) so we cannot multiply 6.8 times 10.
Decide whether each equation is a correct application of the laws of exponents. Write a correct statement if possible.
Checkpoint A.6.26.
\(P(1-r)^6 \rightarrow P-Pr^6\)
Answer.
Not correct
Checkpoint A.6.27.
\(25(2^t) \cdot 4(2^t) \rightarrow 100 \cdot 2^{t^2}\)
Answer.
Not correct. \(~25(2^t) \cdot 4(2^t) = 100(2^{2t})\)
Checkpoint A.6.28.
\(a\left(b^{1/8}\right)^{2t} \rightarrow ab^{t/4}\)
Answer.
Correct
Checkpoint A.6.29.
\(N(0.94)^{1/8.3} \rightarrow \dfrac{N}{(0.94)^{8.3}}\)
Answer.
Not correct, but \(~N(0.94)^{-8.3} = \dfrac{N}{(0.94)^{8.3}}\)
Properties of Radicals.
Product Rule
\begin{equation*} \sqrt[n]{ab}=\sqrt[n]{a}~\sqrt[n]{b}~~~~~~~\text{for } a, b \ge 0 \end{equation*}
Quotient Rule
\begin{equation*} \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}~~~~~~~\text{for } a\ge 0,~~ b \gt 0 \end{equation*}
In general, it is not true that \(\sqrt[n]{a+b}\) is equivalent to \(\sqrt[n]{a}+\sqrt[n]{b}\text{,}\) or that \(\sqrt[n]{a-b}\) is equivalent to \(\sqrt[n]{a}-\sqrt[n]{b}\text{.}\)
Example A.6.30.
Which equation is a correct application of the properties of radicals?
  1. \(\sqrt{x^4+81} = x^2 + 9~~~~~\) or \(~~~~~\sqrt[3]{P^2}~\sqrt[3]{1+r} = \sqrt[3]{P^2(1+r)}\)
  2. \(\dfrac{\sqrt{x+y}}{\sqrt{x}} = \sqrt{y}~~~~~\) or \(~~~~~\dfrac{x+y}{\sqrt{x+y}} = \sqrt{x+y}\)
Solution.
  1. The first statement is incorrect. There is no property that says \(~\sqrt[n]{a+b} = \sqrt[n]{a} + \sqrt[n]{b}\text{.}\)
    However, it is true that \(~\sqrt[n]{a}~ \sqrt[n]{b} = \sqrt[n]{ab}~\text{,}\) so the second statement is correct.
  2. The first statement is incorrect, because \(\dfrac{x+y}{x}\) is not equivalent to \(y\text{.}\)
    The second statement is correct, because \(~\sqrt{x+y}~\sqrt{x+y} = x+y\text{.}\)
Decide whether each equation is a correct application of the properties of radicals. Write a correct statement if possible.
Checkpoint A.6.31.
\(\sqrt[4]{a^2-a^4} \rightarrow \sqrt[4]{a^2} - a\)
Answer.
Not correct
Checkpoint A.6.32.
\(\sqrt{b^4-16} \rightarrow \sqrt{b^2-4} \sqrt{b^2+4}\)
Answer.
Correct
Checkpoint A.6.33.
\(\sqrt[3]{t^4} + \sqrt[3]{t^4} \rightarrow \sqrt[3]{2t^4}\)
Answer.
Not correct.\(~\sqrt[3]{t^4} + \sqrt[3]{t^4} = 2\sqrt[3]{t^4}\)
Checkpoint A.6.34.
\(\dfrac{\sqrt{2p}}{\sqrt{4p+8p^2}} \rightarrow \dfrac{1}{\sqrt{2+4p}}\)
Answer.
Correct

Subsection A.6.3 Rational Exponents

Subsubsection A.6.3.1 Operations on fractions

When working with rational exponents, we often need to perform operations on fractions.
Example A.6.35.
Add \(~\dfrac{-3}{4}+\left(\dfrac{-5}{8}\right)\)
Solution.
The LCD for the fractions is 8, so we build the first fraction:
\begin{equation*} \dfrac{-3}{4} \cdot \alert{\dfrac{2}{2}} = \dfrac{-6}{8} \end{equation*}
Then we combine like fractions:
\begin{equation*} \dfrac{-6}{8}+\left(\dfrac{-5}{8}\right) = \dfrac{-6+(-5)}{8} = \dfrac{-11}{8} \end{equation*}
Example A.6.36.
Subtract \(~\dfrac{-5}{6}-\left(\dfrac{-3}{4}\right)\)
Solution.
The LCD for the fractions is 12, so we build each fraction:
\begin{equation*} \dfrac{-5}{6} \cdot \alert{\dfrac{2}{2}} = \dfrac{-10}{12};~~~~\dfrac{-3}{4} \cdot \alert{\dfrac{3}{3}} = \dfrac{-9}{12} \end{equation*}
Then we combine like fractions:
\begin{equation*} \dfrac{-10}{12}-\left(\dfrac{-9}{12}\right) = \dfrac{-10+9}{12} = \dfrac{-1}{12} \end{equation*}
Example A.6.37.
Multiply \(~\dfrac{-2}{3}\left(\dfrac{5}{4}\right)\)
Solution.
We multiply numerators together, and multiply denominators together:
\begin{equation*} \dfrac{-2}{3}\left(\dfrac{5}{4}\right) = \dfrac{-2 \cdot 5}{3 \cdot 4} = \dfrac{-10}{12} \end{equation*}
Then we reduce:
\begin{equation*} \dfrac{-10}{12} = \dfrac{-5 \cdot \cancel{2}}{6 \cdot \cancel{2}} = \dfrac{-5}{6} \end{equation*}
Checkpoint A.6.38.
Add \(~\dfrac{-3}{4}+\dfrac{1}{3}\)
Answer.
\(\dfrac{-5}{12}\)
Checkpoint A.6.39.
Subtract \(~\dfrac{3}{8}-\left(\dfrac{-1}{6}\right)\)
Answer.
\(\dfrac{13}{24}\)
Checkpoint A.6.40.
Multiply \(~\dfrac{3}{8} \cdot \left(\dfrac{-1}{6}\right)\)
Answer.
\(\dfrac{-1}{16}\)

Subsubsection A.6.3.2 Convert between fractions and decimals

Rational exponents may also be written in decimal form.
Example A.6.41.
Convert \(~0.016~\) to a common fraction.
Solution.
The numerator of the fraction is 016, or 16. The last digit, 6, is in the thousandths place, so the denominator of the fraction is 1000. Thus, \(0.016=\dfrac{16}{1000}\text{.}\) We can reduce this fraction by dividing top and bottom by 8:
\begin{equation*} ~ \dfrac{16}{1000} = \dfrac{\cancel{8} \cdot 2}{\cancel{8} \cdot 125} = \dfrac{2}{125} \end{equation*}
Example A.6.42.
Convert \(~\dfrac{5}{16}~\) to a decimal fraction.
Solution.
Using a calculator, divide 5 by 16:
\(\qquad\qquad 5\) ÷ \(16 = 0.3125\)
Example A.6.43.
Convert \(~\dfrac{5}{11}~\) to a decimal fraction.
Solution.
Using a calculator, divide 5 by 11:
\(\qquad\qquad 5\) ÷ \(11= 0.45454545 ... \)
This is a nonterminating decimal, which we indicate by a repeater bar:
\begin{equation*} \dfrac{5}{11}= 0.45454545 ... = 0.\overline{45} \end{equation*}
Checkpoint A.6.44.
Convert \(~0.1062~\) to a common fraction.
Answer.
\(\dfrac{531}{5000}\)
Checkpoint A.6.45.
Convert \(~2.08~\) to a common fraction.
Answer.
\(\dfrac{52}{25}\)
Checkpoint A.6.46.
Convert \(~\dfrac{4}{15}~\) to a decimal fraction.
Answer.
\(0.2\overline{6}\)

Subsubsection A.6.3.3 Solve equations

To solve an equation of the form \(~x^n = k\text{,}\) we can raise both sides to the reciprocal of the exponent:
\begin{align*} (x^n)^{1/n} \amp = k^{1/n}\\ x \amp = k^{1/n} \end{align*}
because \(~(x^n)^{1/n} = x^{n(1/n)} = x^1\text{.}\)
Example A.6.47.
Solve \(~0.6x^4 = 578\text{.}\) Round your answer to hundredths.
Solution.
First, we isolate the power.
\begin{align*} 0.6x^4 \amp = 578 \amp\amp \blert{\text{Divide both sides by 0.6.}}\\ x^4 \amp = 963.\overline{3} \end{align*}
We raise both sides to the reciprocal of the power.
\begin{align*} (x^4)^{1/4} \amp = (963.\overline{3})^{1/4} \amp\amp \blert{\text{By the third law of exponents,}~ (x^4)^{1/4}=x.}\\ x \amp = 5.57 \end{align*}
To evaluate \((963.\overline{3})^{1/4}\text{,}\) enter \(~~\text{ANS}\)^ \(.25\) ENTER
Example A.6.48.
Solve \(~x^{2/3}-4=60\text{.}\)
Solution.
First, we isolate the power.
\begin{align*} x^{2/3}-4 \amp = 60 \amp\amp \blert{\text{Add 4 to both sides.}}\\ x^{2/3} \amp = 64 \end{align*}
We raise both sides to the reciprocal of the power.
\begin{align*} \left(x^{2/3}\right)^{3/2} \amp = 64^{3/2} \amp\amp \blert{64^{3/2}=\left(64^{1/2}\right)^3=8^3}\\ x \amp = 512 \end{align*}
Or we can evaluate \(~64^{3/2}~\) by entering \(~~64\) ^ \(1.5\) ENTER
Example A.6.49.
Solve \(~18x^{0.24} = 6.5\text{.}\) Round your answer to thousandths.
Solution.
First, we isolate the power.
\begin{align*} 18x^{0.24} \amp = 6.5 \amp\amp \blert{\text{Divide both sides by 18.}}\\ x^{0.24} \amp = 0.36\overline{1} \end{align*}
We raise both sides to the reciprocal of the power.
\begin{align*} \left(x^{0.24}\right)^{1/0.24} \amp = (0.36\overline{1})^{1/0.24}\\ x \amp = 0.014 \end{align*}
We evaluate \((0.36\overline{1})^{1/0.24}\) by entering \(~~\text{ANS}\) ^ ( \(1\) ÷ \(.24\) ) ENTER
Checkpoint A.6.50.
Solve \(~4x^5 = 1825~\text{.}\) Round your answer to thousandths.
Answer.
\(3.403\)
Checkpoint A.6.51.
Solve \(~\dfrac{3}{4}x^{3/4} = 36~\text{.}\) Round your answer to thousandths.
Answer.
\(174.444\)
Checkpoint A.6.52.
Solve \(~0.2x^{1.4}+1.8=12.3~\text{.}\) Round your answer to thousandths.
Answer.
\(16.931\)

Subsection A.6.4 Working with Radicals

Subsubsection A.6.4.1 Perfect squares

To simplify a radical, we factor out the largest perfect square.
Example A.6.53.
Find the missiing factor.
  1. \(60x^9 = 3x^3 \cdot~\) ?
  2. \(9x^3 - 3x^9 \cdot~\) ?
Solution.
  1. We mentally divide \(60x^9\) by \(3x^3\) to find \(\dfrac{60x^9}{3x^3} = 20x^6\text{.}\) The missing factor is \(20x^6\text{.}\)
  2. We mentally divide each term by \(3x^3\) to find \(\dfrac{9x^3}{3x^3} = 3\) and \(\dfrac{3x^9}{3x^3}=x^6\text{.}\) The missing factor is \(3-x^6\text{.}\)
Example A.6.54.
Factor out the largest perfect square.
  1. \(\displaystyle 108a^5b^2\)
  2. \(\displaystyle \dfrac{a^2+4a^4}{8}\)
Solution.
  1. By trial and error, we find that 36 is the largest square that divides 108. From each power, we can factor out the power with the largest possible even exponent, namely \(a^4\) and \(b^2\text{.}\) Thus, we factor out \(36a^4b^2\) to find \(108a^5b^2 =36a^4b^2 \cdot 3a\text{.}\)
  2. The largest even power that divides into both \(a^2\) and \(a^4\) is \(a^2\text{,}\) so we factor \(a^2\) from the numerator:
    \begin{equation*} a^2+a^4=a^2(1+4a^2) \end{equation*}
    The largest perfect square that divides into the denominator is 4. Thus, we factor out \(\dfrac{a^2}{4}\) from the fraction to find
    \begin{equation*} \dfrac{a^2+4a^4}{8} = \dfrac{a^2}{4} \cdot \dfrac{1+4a^2}{2} \end{equation*}
Checkpoint A.6.55.
Find the missiing factor.
  1. \(16z^{16}+4z^6 = 4z^4 \cdot~\) ?
  2. \(\dfrac{20}{7}m^7 = 4m^6 \cdot~\) ?
Answer.
  1. \(\displaystyle 4z^{12}+z^2\)
  2. \(\displaystyle \dfrac{5}{7}m\)
Checkpoint A.6.56.
Factor out the largest perfect square.
  1. \(\displaystyle \dfrac{5k^5}{9n}\)
  2. \(\displaystyle 32a^{10}-48a^9\)
Answer.
  1. \(\displaystyle \dfrac{k^4}{9} \cdot \dfrac{5k}{n}\)
  2. \(\displaystyle 16a^8(2a^2-3a)\)

Subsubsection A.6.4.2 Apply properties of radicals

We have a Product Rule and a Quotient Rule for radicals.
Rules for Radicals.
  • \(\displaystyle \sqrt{ab} =\sqrt{a}~\sqrt{b}~~~~~~\text{if}~a,~b \ge 0\)
  • \(\displaystyle \sqrt{\dfrac{a}{b}} =\dfrac{\sqrt{a}}{\sqrt{b}}~~~~~~\text{if}~a \ge 0,~b \gt 0\)
Example A.6.57.
Decide whether each statement is true or false. Then use a calculator to verify your answer.
  1. \(\displaystyle \sqrt{6} = \sqrt{2}~\sqrt{3}\)
  2. \(\displaystyle \sqrt{6} = \sqrt{2}+\sqrt{4}\)
Solution.
  1. Yes: we can multiply (or divide) radicals together, if they have the same index. You can check that \(\sqrt{6} \approx 2.4495\text{,}\) and
    \begin{equation*} \sqrt{2}~\sqrt{3} \approx (1.4142)(1.7321) = 2.4495 \end{equation*}
    rounded to four decimal places.
  2. No: we cannot combine radicals with addition or subtraction. You can check that \(\sqrt{6} \approx 2.4495\text{,}\) but \(\sqrt{2}+\sqrt{4} \approx 1.4142+2=3.4142\text{.}\)
Example A.6.58.
Find and correct the error in each calculation.
  1. \(\displaystyle \sqrt{36+64} \rightarrow 6+8\)
  2. \(\displaystyle \sqrt{3}+\sqrt{3} \rightarrow \sqrt{6}\)
Solution.
  1. We cannot split radicals with addition or subtraction; we must follow the order of operations:
    \begin{equation*} = \sqrt{36+64} = \sqrt{100} = 10 \end{equation*}
  2. We cannot combine radicals with addition or subtraction. However, we can add like terms:
    \begin{equation*} \sqrt{3}+\sqrt{3} = 2\sqrt{3} \end{equation*}
Checkpoint A.6.59.
Decide whether each statement is true or false. Then use a calculator to verify your answer.
  1. \(\displaystyle \sqrt{16} = \sqrt{18} - \sqrt{2}\)
  2. \(\displaystyle \sqrt{8} = \dfrac{\sqrt{72}}{\sqrt{9}}\)
  3. \(\displaystyle \sqrt{5} + \sqrt{5} = \sqrt{10}\)
  4. \(\displaystyle \sqrt{2}~\sqrt{9} = \sqrt{18}\)
Answer.
  1. False
  2. True
  3. False
  4. True
Checkpoint A.6.60.
Find and correct the error in each calculation.
  1. \(\displaystyle \sqrt{25+5} \rightarrow 5+\sqrt{5}\)
  2. \(\displaystyle \sqrt{10}+\sqrt{15} \rightarrow 5\)
  3. \(\displaystyle \sqrt{9+x^2} \rightarrow 3+x\)
  4. \(\displaystyle \sqrt{a^2-b^2}\rightarrow a-b\)
Answer.
  1. \(\displaystyle \sqrt{30}\)
  2. cannot be simplified
  3. cannot be simplified
  4. cannot be simplified

Subsubsection A.6.4.3 Simplify radicals

We simplify square roots by factoring out any perfect squares.
Example A.6.61.
Simplify \(~\sqrt{45}\)
Solution.
The largest perfect square that divides evenly into 45 is 9, so we factor 45 as \(9 \cdot 5\text{.}\) We use the product rule to write
\begin{equation*} \sqrt{45} = \sqrt{9 \cdot 5} = \sqrt{9}~\sqrt{5} \end{equation*}
Finally, we simplify to get
\begin{equation*} \sqrt{45} = \sqrt{9}~\sqrt{5} = 3\sqrt{5} \end{equation*}
Example A.6.62.
Simplify \(~\sqrt{20x^2y^3}\)
Solution.
The largest perfect square that divides 20 is 4. We write the radicand as the product of two factors, one containing the perfect square and the largest possible even powers of the variables. That is,
\begin{equation*} 20x^2y^3 = 4x^2y^2 \cdot 5y \end{equation*}
Then we write the radical as a product.
\begin{equation*} \sqrt{20x^2y^3} = \sqrt{4x^2y^2 \cdot 5y} = \sqrt{4x^2y^2}~\sqrt{5y} \end{equation*}
Finally, we simplify the first of the two factors to find
\begin{equation*} \sqrt{20x^2y^3}=\blert{\sqrt{4x^2y^2}}~\sqrt{5y} = \blert{2xy}~\sqrt{5y} \end{equation*}
Checkpoint A.6.63.
Simplify \(~\sqrt{75}\)
Answer.
\(3\sqrt{5}\)
Checkpoint A.6.64.
Simplify \(~\sqrt{72u^6v^9}\)
Answer.
\(6u^3v^4\sqrt{2v}\)

Subsubsection A.6.4.4 Rationalize the denominator

Rationalizing the denominator of a fraction helps maintain accuracy.
Example A.6.65.
Simplify \(~\dfrac{3\sqrt{2}}{\sqrt{3}}\)
Solution.
We can rationalize the denominator by multiplying mumerator and denominator by \(\blert{\sqrt{3}}\text{:}\)
\begin{equation*} \dfrac{3\sqrt{2}}{\sqrt{3}} \cdot \blert{\dfrac{\sqrt{3}}{\sqrt{3}}} = \dfrac{3\sqrt{6}}{3} = \sqrt{6} \end{equation*}
or we can divide 3 by \(\sqrt{3}\) to get \(\sqrt{3}\text{.}\) (Remember that \(\sqrt{3}~\sqrt{3} = 3\text{.}\))
\begin{equation*} \dfrac{\blert{3}\sqrt{2}}{\blert{\sqrt{3}}} = \blert{\sqrt{3}}~\sqrt{2} = \sqrt{6} \end{equation*}
Example A.6.66.
Combine \(~\dfrac{3}{\sqrt{2}} + \dfrac{5}{2}\)
Solution.
The LCD for the two fractions is 2, and the building factor for the first fraction is \(\blert{\sqrt{2}}\text{.}\)
\begin{align*} \dfrac{3}{\sqrt{2}} + \dfrac{5}{2} \amp = \dfrac{3}{\sqrt{2}} \cdot \blert{\dfrac{\sqrt{2}}{\sqrt{2}}} + \dfrac{5}{2}\\ \amp = \dfrac{3\sqrt{2}}{2} + \dfrac{5}{2} = \dfrac{3\sqrt{2}+5}{2} \end{align*}
Checkpoint A.6.67.
Simplify \(~\dfrac{8}{2\sqrt{2}}\)
Answer.
\(2\sqrt{2}\)
Checkpoint A.6.68.
Simplify \(~\sqrt{6} \cdot \dfrac{\sqrt{3}}{2\sqrt{2}}\)
Answer.
\(\dfrac{3}{2}\)
Checkpoint A.6.69.
Simplify \(~\dfrac{\sqrt{2}}{6} - \dfrac{2}{\sqrt{3}}\)
Answer.
\(\dfrac{\sqrt{2}-4\sqrt{3}}{6}\)
Checkpoint A.6.70.
Simplify \(~\dfrac{1}{\sqrt{6}} + \dfrac{3}{\sqrt{2}}\)
Answer.
\(\dfrac{\sqrt{6}+9\sqrt{2}}{6}\)

Subsection A.6.5 Radical Equations

Subsubsection A.6.5.1 Solve radical equations

To solve a simple radical equation, we raise both sides to the index of the radical.
Example A.6.71.
Solve \(~\sqrt{x-3} = 4\)
Solution.
We square both sides of the equation to produce an equation without radicals.
\begin{align*} \left(\sqrt{x-3}\right)^2 \amp = 4^2\\ x-3 \amp = 16\\ x \amp = 19 \end{align*}
You can check that \(x=19\) satisfies the original equation.
Example A.6.72.
Solve \(~-2\sqrt[3]{x-4} = -6\)
Solution.
We first isolate the cube root.
\begin{align*} -2\sqrt[3]{x-4} \amp = -6 \amp \amp \blert{\text{Divide both sides by}~-2.}\\ \sqrt[3]{x-4} \amp = 3 \end{align*}
Next, we undo the cube root by cubing both sides of the equation.
\begin{align*} \left(\sqrt[3]{x-4}\right)^3 \amp = 3^3\\ x-4 \amp = 27 \end{align*}
Finally, we add 4 to both sides to find the solution, \(x=31\text{.}\) We do not have to check for extraneous solutions when we cube both sides of an equation, but it is a good idea to check the solution for accuracy anyway.
Check: We substitute \(\alert{31}\) for \(x\) into the left side of the equation.
\begin{equation*} 2\sqrt[3]{\alert{31}-4} = -2\sqrt[3]{27} = -2(3) = -6 \end{equation*}
The solutions checks.
Checkpoint A.6.73.
Solve \(~\sqrt{x-6} = 2\)
Answer.
\(x=10\)
Checkpoint A.6.74.
Solve \(~3\sqrt[3]{4x-1} = -15\)
Answer.
\(x=-31\)

Subsubsection A.6.5.2 Square binomials containing radicals

We may encounter binomials when squaring both sides of an equation.
Example A.6.75.
Expand \(~(\sqrt{x}-3)^2\)
Solution.
\((\sqrt{x}-3)^2 = (\sqrt{x}-3)(\sqrt{x}-3)\text{,}\) so we apply "FOIL" to get
\begin{align*} (\sqrt{x}-3)(\sqrt{x}-3) \amp = \sqrt{x}\sqrt{x}-3\sqrt{x}-3\sqrt{x}-3(-3) \amp \amp \blert{\text{Simplify.}}\\ \amp = x-6\sqrt{x}+9 \end{align*}
Example A.6.76.
Expand \(~(8+\sqrt{t-2})^2\)
Solution.
We multiply \(~(8+\sqrt{t-2})(8+\sqrt{t-2})~\) to get
\begin{align*} 8\cdot 8+8 \amp \sqrt{t-2}+8\sqrt{t-2}+\left(\sqrt{t-2}\right)\left(\sqrt{t-2}\right)\\ \amp = 64+\sqrt{t-2}+(t+2)\\ \amp = 66+16\sqrt{t-2}+t \end{align*}
Checkpoint A.6.77.
Expand \(~\left(6-\sqrt{3a+1}\right)^2\)
Answer.
\(37-12\sqrt{3a+1}+3a\)
Checkpoint A.6.78.
Expand \(~\left(2\sqrt{z+4}-5\right)^2\)
Answer.
\(4z+41-20\sqrt{z+4}\)

Subsubsection A.6.5.3 Use absolute value

Example A.6.79.
Explain why \(\sqrt{x^2}=x\) is not true for all values of \(x\text{.}\)
Solution.
Recall that the symbol \(\sqrt{a}\) means the non-negative square root of \(a\text{.}\) If \(x\) is a negative number, for example \(x=-6\text{,}\) then \(x^2=(-6)^2=36\text{,}\) and not \(\sqrt{x^2}=\sqrt{36}=6\text{,}\) not \(-6\text{.}\) So if \(x\) is a negative number, \(\sqrt{x^2}\not= x\) In fact, \(\sqrt{x^2}= \abs{x}\text{.}\)
Example A.6.80.
For what values of \(x\) is \(~\sqrt{(x-5)^2} = x-5~\text{?}\)
Solution.
\(~\sqrt{(x-5)^2} = x-5~\) when \(x-5\) is positive or zero, that is for \(x \ge 5\text{.}\) If \(x \lt 5\text{,}\) then \(x-5\) is negative. But the \(\sqrt{}\) symbol returns only the positive root, so we use absolute value bars to indicate that the root is positive:
\begin{equation*} \sqrt{(x-5)^2} = \abs{x-5} \end{equation*}
Checkpoint A.6.81.
For what values of \(x\) is \(~\sqrt{(2x+8)^2} = 2x+8~\text{?}\)
Answer.
\(x \ge -4\)
Checkpoint A.6.82.
For what values of \(x\) is \(~\sqrt{(x-9)^2} = 9-x~\text{?}\)
Answer.
\(x \le 9\)
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