← Back to the Rational Exponents worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Each solution shows the conversion or check that matters most. When an equation is squared, the final answer is only the candidate that survives substitution into the original equation.
Standards: N-RN.A.1, N-RN.A.2, A-REI.A.2 · Level: Grade 11 Algebra 2
Problem 1 Evaluating roots
a) (−5)3 = −125, so ∛(−125) = −5.
b) 34 = 81, so the fourth root is 3.
c) √49 = 7 and √64 = 8, so 78.
d) 25 = 32, so 2.
Problem 2 Evaluating rational exponents
a) √25 = 5.
b) (∛8)2 = 22 = 4.
c) (4√(16))3 = 23 = 8.
d) (√49)3 = 73 = 343.
Problem 3 Radical form to exponent form
a) x1/2.
b) x2/3.
c) x3/4.
d) y2/5.
Problem 4 Exponent form to radical form
a) ∛x.
b) √(y5) or (√y)5.
c) 4√(73) or (4√(7))3.
Problem 5 Simplifying numeric radicals
a) 162 = 81 · 2, so 9√2.
b) 40 = 8 · 5, so 2∛5.
c) 1,176 = 196 · 6, so 14√6.
Problem 6 Reading a root graph
The graph passes through (9, 3) and (4, 2), so 91/2 = 3 and 41/2 = 2.
At x = 12 the curve is between 3 (at 9) and 4 (at 16). Since 32 = 9 < 12 < 16 = 42, 121/2 lies between 3 and 4 (about 3.46).
Problem 7 Properties of rational exponents
a) 1/2 + 1/3 = 5/6: x5/6.
b) (2/3)(3/4) = 1/2: x1/2.
c) 3/4 − 1/4 = 1/2: x1/2.
Problem 8 Radicals with variables
a) 18x5 = 9x4 · 2x, so 3x2√(2x).
b) 54a4 = 27a3 · 2a, so 3a∛(2a).
Problem 9 Combining like radicals
√12 = 2√3, √27 = 3√3, √75 = 5√3.
10√3 − 6√3 + 5√3 = 9√3.
Problem 10 Multiplying radical expressions
a) 9 − 5 = 4.
b) 4 · 3 + 4√3 + 1 = 13 + 4√3.
Problem 11 Rationalizing denominators
a) √8 = 2√2, so 62√2 = 3√2 = 3√22. 3√22
b) Multiply by the conjugate: 5(√7 + 2)(√7)2 − 4 = 5(√7 + 2)3. 5√7 + 103
Problem 12 Solving with a graph
Square both sides: x + 6 = x2, so x2 − x − 6 = 0 and (x − 3)(x + 2) = 0. Candidates: 3 and −2.
Check 3: √9 = 3 ✓. Check −2: √4 = 2 ≠ −2 ✗.
x = 3 (and −2 is extraneous)
Problem 13 Single radical equations
a) 3x − 5 = 16, so x = 7. Check: √16 = 4 ✓. x = 7
b) 2x + 1 = −27, so x = −14. Check: ∛(−27) = −3 ✓. x = −14
Problem 14 An extraneous solution
Square: 2x + 9 = x2 + 6x + 9, so x2 + 4x = 0 and x(x + 4) = 0.
Check 0: √9 = 3 and 0 + 3 = 3 ✓. Check −4: √1 = 1 but −4 + 3 = −1 ✗.
x = 0 only
Problem 15 Isolating a radical first
Isolate the radical: √(x + 14) = x + 2. Square: x + 14 = x2 + 4x + 4, so x2 + 3x − 10 = 0, which factors as (x + 5)(x − 2) = 0.
Check 2: √16 = 4 = 2 + 2 ✓. Check −5: √9 = 3 but −5 + 2 = −3 ✗.
x = 2
Problem 16 Two radicals
Isolate one radical: √(x + 5) = 5 − √x. Square: x + 5 = 25 − 10√x + x.
So 10√x = 20, √x = 2, and x = 4. Check: √9 + √4 = 3 + 2 = 5 ✓.
x = 4
Problem 17 A rational exponent equation
Raise both sides to the power 3/2: |x − 4| = 93/2 = 27, so x − 4 = 27 or x − 4 = −27.
Check 31: 272/3 = 9 ✓. Check −23: (−27)2/3 = (∛(−27))2 = 9 ✓.
x = 31 or x = −23
Problem 18 Skid marks and speed
Set √(21d) = 42: 21d = 1,764, so d = 84.
For d = 147: s = √(21 · 147) = √3,087 = √(441 · 7) = 21√7 ≈ 55.6.
84 feet; 21√7 mph (about 55.6 mph)
Problem 19 Distance to the horizon
12 = √(1.5h), so 144 = 1.5h and h = 96.
For h = 6: d = √9 = 3.
The tower is 96 feet tall; the horizon is 3 miles away
Problem 20 Side of a cube
s = 2,7441/3 = 14, because 143 = 2,744.
Surface area = 6s2 = 6 · 196 = 1,176.
Side 14 inches; surface area 1,176 square inches