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Number - Squares, cubes, and roots

Grade 11IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of Square Numbers: The result of multiplying an integer by itself (e.g., 1,4,9,16,25,...1, 4, 9, 16, 25, ...).

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Definition of Cube Numbers: The result of multiplying an integer by itself twice (e.g., 1,8,27,64,125,...1, 8, 27, 64, 125, ...).

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Square Roots: The inverse operation of squaring. A positive number xx has two square roots, x\sqrt{x} and −x-\sqrt{x}, though the symbol \sqrt{} usually refers to the principal (positive) root.

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Cube Roots: The inverse operation of cubing. Unlike square roots, the cube root of a negative number is defined and remains negative.

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Order of Operations (BIDMAS/BODMAS): Powers and roots are processed after Brackets but before Division/Multiplication.

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Estimating Roots: Finding the approximate value of a non-perfect square by identifying the two closest perfect squares (e.g., 10\sqrt{10} is between 9\sqrt{9} and 16\sqrt{16}).

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Fractional Indices: Understanding that x1/2x^{1/2} is equivalent to x\sqrt{x} and x1/3x^{1/3} is equivalent to x3\sqrt[3]{x}.

📐Formulae

x2=x×xx^2 = x \times x

x3=x×x×xx^3 = x \times x \times x

x=x12\sqrt{x} = x^{\frac{1}{2}}

x3=x13\sqrt[3]{x} = x^{\frac{1}{3}}

a×b=a×b\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}

ab=ab\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

💡Examples

Problem 1:

Evaluate (−4)2(-4)^2 and (−4)3(-4)^3.

Solution:

(−4)2=16(-4)^2 = 16; (−4)3=−64(-4)^3 = -64.

Explanation:

Squaring a negative number always results in a positive value because a negative times a negative is positive. Cubing a negative number results in a negative value because negative times negative is positive, and positive times negative is negative.

Problem 2:

Find the value of 2163\sqrt[3]{216}.

Solution:

6

Explanation:

We look for a number that, when multiplied by itself twice, equals 216. Since 6×6=366 \times 6 = 36 and 36×6=21636 \times 6 = 216, the cube root is 6.

Problem 3:

Estimate the value of 55\sqrt{55} to one decimal place.

Solution:

Approximately 7.4

Explanation:

Identify the nearest perfect squares: 72=497^2 = 49 and 82=648^2 = 64. Since 55 is between 49 and 64, the root is between 7 and 8. Because 55 is slightly less than the midpoint between 49 and 64, 7.4 is a reasonable estimate.

Problem 4:

Simplify 144+−273\sqrt{144} + \sqrt[3]{-27}.

Solution:

12+(−3)=912 + (-3) = 9

Explanation:

The square root of 144 is 12. The cube root of -27 is -3 (since −3×−3×−3=−27-3 \times -3 \times -3 = -27). Adding them together gives 12−3=912 - 3 = 9.