Algebra 1 · Unit 6
Write square roots as exponent 1/2 and cube roots as exponent 1/3. Justify equivalent expressions.
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Square root is exponent 12
√a = \(a^{\frac{1}{2}}\). Cube root is exponent 13: ∛a = \(a^{\frac{1}{3}}\).
A power inside
∛8³ = \(8^{\frac{3}{3}}\) = 8. The cube and the cube root undo each other.
Rewrite, then simplify
Change the radical to a rational exponent, or the exponent to a radical, then use exponent laws or perfect powers.
Only 12 and 13
This page stays with square roots and cube roots.
Rewrite one radical as a rational exponent, or one exponent as a radical.
√5 = \(5^{\frac{1}{2}}\)
Answer: \(5^{\frac{1}{2}}\)
∛7 = \(7^{\frac{1}{3}}\)
Answer: \(7^{\frac{1}{3}}\)
\(9^{\frac{1}{2}}\) = √9 = 3
Answer: 3
\(8^{\frac{1}{3}}\) = ∛8 = 2
Answer: 2
√16 = \(16^{\frac{1}{2}}\) = 4
Answer: 4
∛27 = \(27^{\frac{1}{3}}\) = 3
Answer: 3
A power inside a radical, or a rational exponent of a perfect power.
∛8³ = \(8^{\frac{3}{3}}\)
The cube and the cube root undo.
8
Answer: 8
√5² = 5, since 5 ≥ 0
Answer: 5
\(32^{\frac{1}{2}}\) = √32 = 4√2
Answer: 4√2
∛8 = 2
Answer: 2
\(x^{\frac{2}{2}}\) = x¹ = x
Answer: x
∛125 = 5
Answer: 5
Justify a chain of equivalent expressions.
∛8³ = \(8^{\frac{3}{3}}\)
The numerator 3 and the denominator 3 cancel.
8
Answer: 8
9³ = 729
√729 = 27
Or √9³ = 9¹ × √9 = 9 × 3 = 27
Answer: 27
\(8^{\frac{1}{2}}\)
√8 = 2√2
Answer: \(8^{\frac{1}{2}}\) = 2√2
27² = 729
∛729 = 9
Or 27 = 3³, so ∛3⁶ = 3² = 9
Answer: 9
True. The cube root of −8 is −2.
\((-8)^{\frac{1}{2}}\) is not real, but the exponent here is 13.
Answer: true
\(50^{\frac{1}{2}}\)
5√2
Answer: \(50^{\frac{1}{2}}\) = 5√2
∛8³ undoes the cube, leaving 8.
∛8 = ∛2³ = 2.
Cubing and taking a cube root are inverse operations when the exponent matches the index.
Answer: 8, and 2
-3² = 9
√9 = 3
The principal square root is nonnegative. √a² = |a|.
Answer: 3
√16 = 4
\(16^{\frac{1}{2}}\) = 4
∛4³ = 4
Answer: √16, 16^(1/2), ∛(4³)
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