Radicals and Rational Exponents: Algebra 2 Math Worksheet Answer Key – PDF Download

Algebra 2 Radicals Worksheet Answer Key: problems 1–5

Problem 1: Evaluate nth roots

1. \(13^{2}=169\), so \(13\).
2. \((-4)^{3}=-64\), so \(-4\).
3. \(2^{4}=16\), so \(2\).
4. \((-2)^{5}=-32\), so \(-2\).

Problem 2: Real or not real

1. An even root of a negative number: not a real number.
2. \((-7)^{3}=-343\), so \(-7\).
3. Even index, negative radicand: not a real number.
4. \((-1)^{5}=-1\), so \(-1\).

Problem 3: Radical to exponent form

1. \(x^{1/3}\)
2. The index \(5\) is the denominator: \(y^{2/5}\).
3. \(z^{3/4}\)
4. \(7^{3/2}\)

Problem 4: Exponent to radical form

1. \(\sqrt{5}\)
2. \(\sqrt[3]{x^{2}}\)
3. \(\sqrt[4]{(3a)^{3}}\)
4. Only \(m\) carries the exponent, so the factor \(2\) stays outside: \(2\sqrt[3]{m}\).

Problem 5: Evaluate rational exponents

1. \(\sqrt{25}=\) \(5\).
2. \(\sqrt[3]{64}=\) \(4\).
3. \((\sqrt[4]{16})^{3}=2^{3}=\) \(8\).
4. \((\sqrt[3]{8})^{2}=2^{2}=\) \(4\).
5. \((\sqrt[5]{32})^{3}=2^{3}=\) \(8\).

Algebra 2 Radicals Worksheet Answer Key: problems 6–10

Problem 6: Negative rational exponents

1. \(\frac{1}{9^{1/2}}=\frac{1}{3}\). \(\frac{1}{3}\)
2. \(27^{2/3}=3^{2}=9\), so \(\frac{1}{9}\).
3. \(16^{3/4}=2^{3}=8\), so \(\frac{1}{8}\).
4. Flip the base: \(4^{3/2}=(\sqrt{4})^{3}=2^{3}=\) \(8\).

Problem 7: Simplify square roots

1. \(\sqrt{16\cdot 3}=4\sqrt{3}\). \(4\sqrt{3}\)
2. \(\sqrt{9\cdot 10}=\) \(3\sqrt{10}\).
3. \(\sqrt{100\cdot 2}=\) \(10\sqrt{2}\).
4. \(\sqrt{25\cdot 3}=\) \(5\sqrt{3}\).

Problem 8: Simplify higher roots

1. \(\sqrt[3]{8\cdot 3}=\) \(2\sqrt[3]{3}\).
2. \(\sqrt[3]{125\cdot 2}=\) \(5\sqrt[3]{2}\).
3. \(\sqrt[3]{-27\cdot 2}=\) \(-3\sqrt[3]{2}\).
4. \(\sqrt[4]{16\cdot 3}=\) \(2\sqrt[4]{3}\).

Problem 9: Radicals with variables

1. \(\sqrt{9x^{4}\cdot 2x}=\) \(3x^{2}\sqrt{2x}\).
2. \(\sqrt[3]{8y^{6}\cdot 2y}=\) \(2y^{2}\sqrt[3]{2y}\).
3. \(\sqrt{25a^{2}b^{4}\cdot 2a}=\) \(5ab^{2}\sqrt{2a}\).
4. \(\sqrt[3]{27m^{3}n^{9}\cdot 2m^{2}}=\) \(3mn^{3}\sqrt[3]{2m^{2}}\).

Problem 10: Multiply radicals

1. \(\sqrt{60}=\sqrt{4\cdot 15}=\) \(2\sqrt{15}\).
2. \(\sqrt[3]{24}=\) \(2\sqrt[3]{3}\).
3. \(15\sqrt{28}=15\cdot 2\sqrt{7}=\) \(30\sqrt{7}\).
4. \(2\sqrt{18}=2\cdot 3\sqrt{2}=\) \(6\sqrt{2}\).

Algebra 2 Radicals Worksheet Answer Key: problems 11–15

Problem 11: Add and subtract radicals

1. \(6\sqrt{3}+3\sqrt{3}=\) \(9\sqrt{3}\).
2. \(10\sqrt{2}-6\sqrt{2}=\) \(4\sqrt{2}\).
3. \(2\sqrt[3]{2}+3\sqrt[3]{2}=\) \(5\sqrt[3]{2}\).
4. \(2\sqrt{5}-3\sqrt{5}+\sqrt{5}=\) \(0\).

Problem 12: Rationalize denominators

1. \(\frac{6\sqrt{3}}{3}=\) \(2\sqrt{3}\).
2. \(\frac{5\sqrt{10}}{10}=\) \(\frac{\sqrt{10}}{2}\).
3. Multiply by \(\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}\): the denominator becomes \(7-3=4\), so the result is \(\sqrt{7}+\sqrt{3}\).
4. Multiply by \(\frac{\sqrt[3]{4}}{\sqrt[3]{4}}\): \(\frac{3\sqrt[3]{4}}{\sqrt[3]{8}}=\) \(\frac{3\sqrt[3]{4}}{2}\).

Problem 13: Exponent properties

1. \(\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\): \(x^{5/6}\).
2. \(\frac{2}{3}\cdot\frac{3}{4}=\frac{1}{2}\): \(y^{1/2}\).
3. \(\frac{5}{4}-\frac{1}{4}=1\): \(x\).
4. \(8^{2/3}a^{2}=\) \(4a^{2}\).

Problem 14: Change the index

1. \(x^{4/6}=x^{2/3}=\) \(\sqrt[3]{x^{2}}\).
2. \(9^{1/4}=(3^{2})^{1/4}=3^{1/2}=\) \(\sqrt{3}\).
3. \(64^{1/6}=\) \(2\), since \(2^{6}=64\).
4. \(x^{12/3}=x^{4}\), then \((x^{4})^{1/2}=\) \(x^{2}\).

Problem 15: One-step radical equations

1. Square both sides: \(x=49\).
2. Cube both sides: \(x=-27\).
3. \(x-4=25\), so \(x=29\). Check: \(\sqrt{25}=5\).
4. \(2x+1=81\), so \(x=40\). Check: \(\sqrt{81}=9\).

Algebra 2 Radicals Worksheet Answer Key: problems 16–20

Problem 16: Check for extraneous solutions

Square both sides: \(3x+4=x^{2}\), so \(x^{2}-3x-4=0\) and \((x-4)(x+1)=0\). The candidates are \(x=4\) and \(x=-1\).

Check \(x=4\): \(\sqrt{16}=4\), true. Check \(x=-1\): \(\sqrt{1}=1\), but \(x=-1\), false. The solution is \(x=4\); \(x=-1\) is extraneous.

Problem 17: Radical on the right

Square both sides: \(x^{2}-4x+4=x+4\), so \(x^{2}-5x=0\) and \(x(x-5)=0\). The candidates are \(x=0\) and \(x=5\).

Check \(x=0\): \(0-2=-2\), but \(\sqrt{4}=2\), false. Check \(x=5\): \(5-2=3\) and \(\sqrt{9}=3\), true. The solution is \(x=5\); \(x=0\) is extraneous.

Problem 18: Isolate the radical first

1. \(\sqrt{x+5}=7\), so \(x+5=49\) and \(x=44\).
2. \(2\sqrt{x-1}=8\), \(\sqrt{x-1}=4\), \(x-1=16\): \(x=17\).
3. \(\sqrt[3]{2x-5}=3\), \(2x-5=27\), \(2x=32\): \(x=16\).
4. \(\sqrt{x+7}=-2\). A principal square root is never negative: no solution.

Problem 19: Rational exponent equations

1. Raise both sides to \(\frac{2}{3}\): \(x=27^{2/3}=9\). Check: \(9^{3/2}=27\). \(x=9\)
2. The numerator is even, so there are two solutions: \(x=\pm 16^{3/2}=\pm 64\). Check: \(64^{2/3}=16\) and \((-64)^{2/3}=(-4)^{2}=16\). \(x=\pm 64\)
3. Raise both sides to the fourth power: \(x=81\).
4. \(x-1=\pm 4^{3/2}=\pm 8\), so \(x=9\) or \(x=-7\). Check: \(8^{2/3}=4\) and \((-8)^{2/3}=4\). \(x=9\) or \(x=-7\)

Problem 20: Equations with two radicals

1. \(2x+3=x+7\), so \(x=4\). Check: both sides equal \(\sqrt{11}\).
   

2. \(x+4=3x-10\), so \(2x=14\) and \(x=7\). Check: both sides equal \(\sqrt[3]{11}\).
3. \(5x-1=2x+8\), so \(3x=9\) and \(x=3\). Check: both sides equal \(\sqrt{14}\).

Algebra 2 Radicals Worksheet Answer Key: problems 21–25

Problem 21: Pendulum period

1. \(T=2\pi\sqrt{\frac{8}{32}}=2\pi\cdot\frac{1}{2}=\pi\). \(\pi\approx 3.14\) seconds.
2. \(\sqrt{\frac{L}{32}}=1\), so \(\frac{L}{32}=1\). \(L=32\) feet.
3. \(\sqrt{\frac{L}{32}}=2\), so \(\frac{L}{32}=4\). \(L=128\) feet.

Problem 22: Free fall time

1. \(t=\sqrt{\frac{144}{16}}=\sqrt{9}=\) \(3\) seconds.
2. \(\frac{h}{16}=2.5^{2}=6.25\), so \(h=100\). \(100\) feet.
3. \(t=\sqrt{25}=\) \(5\) seconds.

Problem 23: Skid marks

1. \(s=\sqrt{30(0.75)(40)}=\sqrt{900}=\) \(30\) mph.
2. \(45=\sqrt{30(0.6)d}\), so \(2025=18d\). \(d=112.5\) feet.
3. \(s=\sqrt{720}=\sqrt{144\cdot 5}=12\sqrt{5}\). \(12\sqrt{5}\approx 26.83\) mph.

Problem 24: Distance to the horizon

1. \(d=\sqrt{225}=\) \(15\) miles.
2. \(441=1.5h\), so \(h=294\) feet.
3. \(d=\sqrt{9}=\) \(3\) miles.

Problem 25: Cube-shaped box

1. \(e=\sqrt[3]{250}=\sqrt[3]{125\cdot 2}=\) \(5\sqrt[3]{2}\) inches.
2. \(5(1.2599)\approx\) \(6.30\) inches.
3. \(e^{2}=(5\sqrt[3]{2})^{2}=25\sqrt[3]{4}\approx 25(1.5874)\). \(25\sqrt[3]{4}\approx 39.69\text{ in}^{2}\).

Algebra 2 Radicals Worksheet Answer Key: problems 26–26

Problem 26: Bike value

1. \(V=900(0.64)^{1/2}=900(0.8)=\) \(\$720\).
2. \(0.64^{3/2}=0.8^{3}=0.512\), so \(\$460.80\).
3. \((0.64)^{t/2}=\frac{576}{900}=0.64\), so \(\frac{t}{2}=1\). \(t=2\) years.
4. \(0.64^{5/2}=0.8^{5}=0.32768\), so \(V=\$294.912\). \(\$294.91\)

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