8th Grade Square Roots and Cube Roots Worksheet Solutions (PDF)

8th Grade math worksheet solutions: Square Roots and Cube Roots – Maths PDF

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This square cube roots answers for 8th Grade covers square roots and cube roots (Common Core 8.EE.A.2), with a free PDF to print.

This answer key gives a full worked solution for each of the 21 problems on the square roots and cube roots worksheet (Common Core 8.EE.A.2). Compare each step with your own work, not only the final answer, and note any step where your method differed.

Grade 8 · Common Core State Standards: 8.EE.A.2

Square cube roots answers: exercises 1 to 10

Exercise 1 – Square root

\(9 \cdot 9 = 81\).

\(9\)

Exercise 2 – Cube root

\(4 \cdot 4 \cdot 4 = 64\).

\(4\)

Exercise 3 – Solve x squared

Take both roots: \(x = \pm\sqrt{49}\).

\(x = 7 \text{ or } x = -7\)

Exercise 4 – Solve x cubed

\(4^{3} = 64\), and a cube has only one real root.

\(x = 4\)

Exercise 5 – Larger perfect square

\(14 \cdot 14 = 196\).

\(14\)

Exercise 6 – Larger cube root

\(6^{3} = 216\).

\(6\)

Exercise 7 – Bigger equation

\(15^{2} = 225\), so \(x = \pm 15\).

\(x = 15 \text{ or } x = -15\)

Exercise 8 – Fraction cube

\(\sqrt[3]{27} = 3\) and \(\sqrt[3]{64} = 4\).

\(x = \frac{3}{4}\)

Exercise 9 – Fraction square

\(\sqrt{49} = 7\) and \(\sqrt{81} = 9\), so \(x = \pm\frac{7}{9}\).

\(x = \frac{7}{9} \text{ or } x = -\frac{7}{9}\)

Exercise 10 – Combine roots

\(6 + 2 = 8\).

\(8\)

Exercises 11 and up

Exercise 11 – Estimate a cube root

\(3^{3} = 27\) and \(4^{3} = 64\), and \(27 < 30 < 64\).

\(3 < \sqrt[3]{30} < 4\)

Exercise 12 – Not a perfect square

\(x = \pm\sqrt{20}\). Since \(4.47^{2} = 19.98\) and \(4.48^{2} = 20.07\), \(\sqrt{20} \approx 4.47\).

\(x = \pm\sqrt{20} \approx \pm 4.47\)

Exercise 13 – Two-step equation

Divide by 2: \(x^{2} = 49\).

\(x = \pm 7\)

Exercise 14 – Cube with a coefficient

Divide by 4: \(x^{3} = 27\), so \(x = 3\).

Check: \(4 \cdot 27 = 108\).

\(x = 3\)

Exercise 15 – Square tile

Side = 13 in.

Side \(= \sqrt{169} = 13\).

13 inches

Exercise 16 – Shipping cube

Edge = 7 in.

Edge \(= \sqrt[3]{343} = 7\), because \(7 \cdot 7 \cdot 7 = 343\).

7 inches

Exercise 17 – Garden fence

Side \(= \sqrt{196} = 14\) ft.

Perimeter \(= 4 \cdot 14 = 56\).

56 feet

Exercise 18 – Gift box surface

Edge = 6 in.

Edge \(= \sqrt[3]{216} = 6\) in.

One face: \(6^{2} = 36\). Six faces: \(6 \cdot 36 = 216\).

Edge 6 in; 216 square inches of paper

Exercise 19 – Jogging lap

Side \(= \sqrt{2500} = 50\) yd.

One lap: \(4 \cdot 50 = 200\) yd. Three laps: \(3 \cdot 200 = 600\).

600 yards

Exercise 20 – Compare

\(\sqrt{90} \approx 9.5\) because \(9.5^{2} = 90.25\).

\(\sqrt[3]{90} \approx 4.5\) because \(4.5^{3} = 91.125\) and \(4.4^{3} = 85.18\).

The square: about 9.5 cm versus about 4.5 cm

Exercise 21 – Explain

No. The symbol \(\sqrt{36}\) means 6, but \((-6)^{2}\) is also 36.

\(x = 6 \text{ or } x = -6\)

Key facts behind this square cube roots answers

Key facts

\(\sqrt{81} = 9\) because \(9^{2} = 81\); \(\sqrt[3]{125} = 5\) because \(5^{3} = 125\).
\(x^{2} = p\) with \(p > 0\) has two solutions: \(x = \pm\sqrt{p}\).
\(x^{3} = p\) has one real solution: \(x = \sqrt[3]{p}\).
Know the perfect squares up to \(15^{2}\) and the perfect cubes up to \(5^{3}\).
Roots of non-perfect squares, like \(\sqrt{20}\), are irrational and can be estimated.

How to use this square cube roots answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

All Grade 8 chapters: Grade 8 math.

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