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Power Play - The Other Side of Powers

Grade 8CBSE

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

🔑Concepts

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The operation of finding the square root is the inverse operation of squaring. If x2=yx^2 = y, then y=x\sqrt{y} = x.

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A natural number is called a perfect square if it is the square of some natural number. For example, 8181 is a perfect square because 92=819^2 = 81.

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Properties of Square Numbers: Square numbers can only end with 0,1,4,5,6,90, 1, 4, 5, 6, 9 at the unit's place. Numbers ending in 2,3,7,2, 3, 7, or 88 are never perfect squares.

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The cube root of a number xx is the number yy such that y3=xy^3 = x, denoted as x3=y\sqrt[3]{x} = y.

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Finding square roots can be done through Prime Factorization (for perfect squares) or Long Division (for any number).

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Negative exponents represent the reciprocal of the base raised to the positive power: a−n=1ana^{-n} = \frac{1}{a^n}.

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The 'Other Side' of powers refers to fractional exponents: x12x^{\frac{1}{2}} is the square root x\sqrt{x} and x13x^{\frac{1}{3}} is the cube root x3\sqrt[3]{x}.

📐Formulae

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

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

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

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

ab3=a3×b3\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}

a−m=1ama^{-m} = \frac{1}{a^m}

(am)n=am×n(a^m)^n = a^{m \times n}

💡Examples

Problem 1:

Find the square root of 729729 using the Prime Factorization method.

Solution:

729=3×3×3×3×3×3729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 729=(3×3)×(3×3)×(3×3)\sqrt{729} = \sqrt{(3 \times 3) \times (3 \times 3) \times (3 \times 3)} 729=3×3×3=27\sqrt{729} = 3 \times 3 \times 3 = 27

Explanation:

We first resolve the number into its prime factors. Then we pair the identical factors and take one factor from each pair to find the square root.

Problem 2:

Evaluate 5123\sqrt[3]{512}.

Solution:

512=2×2×2×2×2×2×2×2×2512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 5123=(2×2×2)×(2×2×2)×(2×2×2)3\sqrt[3]{512} = \sqrt[3]{(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)} 5123=2×2×2=8\sqrt[3]{512} = 2 \times 2 \times 2 = 8

Explanation:

To find the cube root, we factorize the number and form groups of three identical factors (triplets). We then take one factor from each triplet.

Problem 3:

Subtract the square of 1515 from the square of 2020.

Solution:

202=40020^2 = 400 152=22515^2 = 225 400−225175\begin{array}{r} 400 \\ - 225 \\ \hline 175 \end{array}

Explanation:

First, calculate the squares of both numbers: 20×20=40020 \times 20 = 400 and 15×15=22515 \times 15 = 225. Then perform vertical subtraction.

Problem 4:

Simplify (64)−13(64)^{-\frac{1}{3}}.

Solution:

(64)−13=16413(64)^{-\frac{1}{3}} = \frac{1}{64^{\frac{1}{3}}} 64=4364 = 4^3 1(43)13=143×13=141=14\frac{1}{(4^3)^{\frac{1}{3}}} = \frac{1}{4^{3 \times \frac{1}{3}}} = \frac{1}{4^1} = \frac{1}{4}

Explanation:

First, convert the negative exponent to a positive one by taking the reciprocal. Then, express 6464 as 434^3 to simplify the fractional exponent.