1.3 Radicals and rational exponents  (Page 5/11)

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Simplifying rational exponents

Simplify:

1. $5\left(2{x}^{\frac{3}{4}}\right)\left(3{x}^{\frac{1}{5}}\right)$
2. ${\left(\frac{16}{9}\right)}^{-\frac{1}{2}}$

Simplify $\text{\hspace{0.17em}}{\left(8x\right)}^{\frac{1}{3}}\left(14{x}^{\frac{6}{5}}\right).$

$28{x}^{\frac{23}{15}}$

Access these online resources for additional instruction and practice with radicals and rational exponents.

Key concepts

• The principal square root of a number $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is the nonnegative number that when multiplied by itself equals $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ See [link] .
• If $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ are nonnegative, the square root of the product $\text{\hspace{0.17em}}ab\text{\hspace{0.17em}}$ is equal to the product of the square roots of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ See [link] and [link] .
• If $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ are nonnegative, the square root of the quotient $\text{\hspace{0.17em}}\frac{a}{b}\text{\hspace{0.17em}}$ is equal to the quotient of the square roots of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ See [link] and [link] .
• Radical expressions written in simplest form do not contain a radical in the denominator. To eliminate the square root radical from the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. See [link] and [link] .
• The principal n th root of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is the number with the same sign as $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ that when raised to the n th power equals $\text{\hspace{0.17em}}a.\text{\hspace{0.17em}}$ These roots have the same properties as square roots. See [link] .
• The properties of exponents apply to rational exponents. See [link] .

Verbal

What does it mean when a radical does not have an index? Is the expression equal to the radicand? Explain.

When there is no index, it is assumed to be 2 or the square root. The expression would only be equal to the radicand if the index were 1.

Where would radicals come in the order of operations? Explain why.

Every number will have two square roots. What is the principal square root?

The principal square root is the nonnegative root of the number.

Can a radical with a negative radicand have a real square root? Why or why not?

Numeric

For the following exercises, simplify each expression.

$\sqrt{256}$

16

$\sqrt{\sqrt{256}}$

$\sqrt{4\left(9+16\right)}$

10

$\sqrt{289}-\sqrt{121}$

$\sqrt{196}$

14

$\sqrt{1}$

$\sqrt{98}$

$7\sqrt{2}$

$\sqrt{\frac{27}{64}}$

$\sqrt{\frac{81}{5}}$

$\frac{9\sqrt{5}}{5}$

$\sqrt{800}$

$\sqrt{169}+\sqrt{144}$

25

$\sqrt{\frac{8}{50}}$

$\frac{18}{\sqrt{162}}$

$\sqrt{2}$

$\sqrt{192}$

$14\sqrt{6}-6\sqrt{24}$

$2\sqrt{6}$

$15\sqrt{5}+7\sqrt{45}$

$\sqrt{150}$

$5\sqrt{6}$

$\sqrt{\frac{96}{100}}$

$\left(\sqrt{42}\right)\left(\sqrt{30}\right)$

$6\sqrt{35}$

$12\sqrt{3}-4\sqrt{75}$

$\sqrt{\frac{4}{225}}$

$\frac{2}{15}$

$\sqrt{\frac{405}{324}}$

$\sqrt{\frac{360}{361}}$

$\frac{6\sqrt{10}}{19}$

$\frac{5}{1+\sqrt{3}}$

$\frac{8}{1-\sqrt{17}}$

$-\frac{1+\sqrt{17}}{2}$

$\sqrt[4]{16}$

$\sqrt[3]{128}+3\sqrt[3]{2}$

$7\sqrt[3]{2}$

$\sqrt[5]{\frac{-32}{243}}$

$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$

$15\sqrt{5}$

$3\sqrt[3]{-432}+\sqrt[3]{16}$

Algebraic

For the following exercises, simplify each expression.

$\sqrt{400{x}^{4}}$

$20{x}^{2}$

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

$\sqrt{49p}$

$7\sqrt{p}$

${\left(144{p}^{2}{q}^{6}\right)}^{\frac{1}{2}}$

${m}^{\frac{5}{2}}\sqrt{289}$

$17{m}^{2}\sqrt{m}$

$9\sqrt{3{m}^{2}}+\sqrt{27}$

$3\sqrt{a{b}^{2}}-b\sqrt{a}$

$2b\sqrt{a}$

$\frac{4\sqrt{2n}}{\sqrt{16{n}^{4}}}$

$\sqrt{\frac{225{x}^{3}}{49x}}$

$\frac{15x}{7}$

$3\sqrt{44z}+\sqrt{99z}$

$\sqrt{50{y}^{8}}$

$5{y}^{4}\sqrt{2}$

$\sqrt{490b{c}^{2}}$

$\sqrt{\frac{32}{14d}}$

$\frac{4\sqrt{7d}}{7d}$

${q}^{\frac{3}{2}}\sqrt{63p}$

$\frac{\sqrt{8}}{1-\sqrt{3x}}$

$\frac{2\sqrt{2}+2\sqrt{6x}}{1-3x}$

$\sqrt{\frac{20}{121{d}^{4}}}$

${w}^{\frac{3}{2}}\sqrt{32}-{w}^{\frac{3}{2}}\sqrt{50}$

$-w\sqrt{2w}$

$\sqrt{108{x}^{4}}+\sqrt{27{x}^{4}}$

$\frac{\sqrt{12x}}{2+2\sqrt{3}}$

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

$\sqrt{147{k}^{3}}$

$\sqrt{125{n}^{10}}$

$5{n}^{5}\sqrt{5}$

$\sqrt{\frac{42q}{36{q}^{3}}}$

$\sqrt{\frac{81m}{361{m}^{2}}}$

$\frac{9\sqrt{m}}{19m}$

$\sqrt{72c}-2\sqrt{2c}$

$\sqrt{\frac{144}{324{d}^{2}}}$

$\frac{2}{3d}$

$\sqrt[3]{24{x}^{6}}+\sqrt[3]{81{x}^{6}}$

$\sqrt[4]{\frac{162{x}^{6}}{16{x}^{4}}}$

$\frac{3\sqrt[4]{2{x}^{2}}}{2}$

$\sqrt[3]{64y}$

$\sqrt[3]{128{z}^{3}}-\sqrt[3]{-16{z}^{3}}$

$6z\sqrt[3]{2}$

$\sqrt[5]{1,024{c}^{10}}$

Real-world applications

A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So the length of the guy wire can be found by evaluating $\text{\hspace{0.17em}}\sqrt{90,000+160,000}.\text{\hspace{0.17em}}$ What is the length of the guy wire?

500 feet

A car accelerates at a rate of where t is the time in seconds after the car moves from rest. Simplify the expression.

Extensions

For the following exercises, simplify each expression.

$\frac{\sqrt{8}-\sqrt{16}}{4-\sqrt{2}}-{2}^{\frac{1}{2}}$

$\frac{-5\sqrt{2}-6}{7}$

$\frac{{4}^{\frac{3}{2}}-{16}^{\frac{3}{2}}}{{8}^{\frac{1}{3}}}$

$\frac{\sqrt{m{n}^{3}}}{{a}^{2}\sqrt{{c}^{-3}}}\cdot \frac{{a}^{-7}{n}^{-2}}{\sqrt{{m}^{2}{c}^{4}}}$

$\frac{\sqrt{mnc}}{{a}^{9}cmn}$

$\frac{a}{a-\sqrt{c}}$

$\frac{x\sqrt{64y}+4\sqrt{y}}{\sqrt{128y}}$

$\frac{2\sqrt{2}x+\sqrt{2}}{4}$

$\left(\frac{\sqrt{250{x}^{2}}}{\sqrt{100{b}^{3}}}\right)\left(\frac{7\sqrt{b}}{\sqrt{125x}}\right)$

$\sqrt{\frac{\sqrt[3]{64}+\sqrt[4]{256}}{\sqrt{64}+\sqrt{256}}}$

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

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