Solved Examples Square Roots and Cube Roots - Quantitative Aptitude for SSC CGL

Square Root

The square root of a number y is a number x such that x² = y. It is written as √y = x. For example, √9 = 3, √16 = 4.

Cube Root

The cube root of a number y is a number x such that x³ = y. It is written as ∛y = x. For example, ∛8 = 2, ∛27 = 3.

Important Formulae and Identities

• √(ab) = √a × √b (both a and b must be non-negative)
• √(a/b) = √a / √b (b ≠ 0, a and b non-negative)
• √(a²) = |a|
• √(a + b) ≠ √a + √b in general (care required)
• ∛(ab) = ∛a × ∛b
• ∛(a/b) = ∛a / ∛b (b ≠ 0)
• To rationalise a denominator containing a square root, multiply numerator and denominator by the conjugate when needed.

Methods and Concepts

Prime-factor method for perfect square and perfect cube

Write the number as a product of prime powers. For a perfect square each prime exponent must be even; for a perfect cube each prime exponent must be a multiple of three. To make the number a perfect square/cube, multiply by the smallest integer that makes every prime exponent satisfy the required multiple.

Rationalisation of denominator (square-root case)

To remove a surd from the denominator of a fraction such as (√a + √b)/(√a - √b), multiply numerator and denominator by the conjugate (√a + √b) to use the identity (√a - √b)(√a + √b) = a - b.

Simplifying expressions with decimals under square root

Use place-value factors: √(m / 10^k) = √m / 10^(k/2). Convert to whole-number square roots when possible.

Solved Examples

Example 1: What is the least number required to multiply to 9720 to make a perfact cube?
(a) 55
(b) 65
(c) 75
(d) 85
Ans: (c)

Factorise 9720 into primes.

\(9720 = 2^3 \times 3^5 \times 5^1\)

To make a perfect cube, every prime exponent must be a multiple of 3.

\(2^3\) already has exponent a multiple of 3.

\(3^5\) needs 1 more power of 3 to become \(3^6\).

\(5^1\) needs 2 more powers of 5 to become \(5^3\).

Therefore the smallest multiplier is:

\(3^1 \times 5^2 = 3 \times 25 = 75\)

Hence the required number is 75.

Example 2: If √ 15 = 3.8729 then what is (√5 + √3)/(√5 - √3)?
(a) 7.8729
(b) 6.8729
(c) 5.8729
(d) 4.8729
Ans: (a)

Rationalise the denominator by multiplying numerator and denominator by the conjugate.

\(\dfrac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3} \times \dfrac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}\)

\(\dfrac{(\sqrt5+\sqrt3)^2}{(\sqrt5)^2 - (\sqrt3)^2}\)

\(\dfrac{5 + 3 + 2\sqrt{15}}{5 - 3}\)

\(\dfrac{8 + 2\sqrt{15}}{2}\)

\(4 + \sqrt{15}\)

Using \(\sqrt{15}=3.8729\) gives \(4 + 3.8729 = 7.8729\).

Example 3: If √ 1369 = 37 then what is √13.69 + √0.1369 + √ 0.001369 + √ 0.00001369?
(a) 4.0021
(b) 4.1107
(c) 3.1232
(d) 2.1323
Ans: (b)

Express each decimal under a common whole number by factoring powers of 10.

\(\sqrt{13.69} = \sqrt{\dfrac{1369}{100}} = \dfrac{\sqrt{1369}}{\sqrt{100}} = \dfrac{37}{10}\)

\(\sqrt{0.1369} = \sqrt{\dfrac{1369}{10000}} = \dfrac{37}{100}\)

\(\sqrt{0.001369} = \sqrt{\dfrac{1369}{1000000}} = \dfrac{37}{1000}\)

\(\sqrt{0.00001369} = \sqrt{\dfrac{1369}{100000000}} = \dfrac{37}{10000}\)

Sum the values:

\(\dfrac{37}{10} + \dfrac{37}{100} + \dfrac{37}{1000} + \dfrac{37}{10000} = 3.7 + 0.37 + 0.037 + 0.0037 = 4.1107\)

Example 4: √(25/16) = ?
(a) 3/4
(b) 5/4
(c) 4
(d) 4/5
Ans: (b)

Separate numerator and denominator under square root.

\(\sqrt{\dfrac{25}{16}} = \dfrac{\sqrt{25}}{\sqrt{16}} = \dfrac{5}{4}\)

Example 5: √ 54 x √ 6 = ?
(a) 24
(b) 15
(c) 18
(d) 13
Ans: (c)

Combine the radicals by multiplication.

\(\sqrt{54} \times \sqrt{6} = \sqrt{54 \times 6}\)

\(\sqrt{324}\)

\(18\)

Hence the value is 18.

Example 6: ∛9261 = ?
(a) 21
(b) 17
(c) 29
(d) 23
Ans: (a)

Factorise 9261 into primes.

\(9261 = 3^3 \times 7^3\)

Take cube root:

\(\sqrt[3]{9261} = \sqrt[3]{3^3 \times 7^3} = 3 \times 7 = 21\)

Example 7: If √15 = 3.88. What is √(5/3)?
(a) 1.213
(b) 1.293
(c) 1.321
(d) 1.432
Ans: (b)

Express √(5/3) in terms of √15.

\(\sqrt{\dfrac{5}{3}} = \dfrac{\sqrt5}{\sqrt3} = \dfrac{\sqrt5}{\sqrt3} \times \dfrac{\sqrt3}{\sqrt3} = \dfrac{\sqrt{15}}{3}\)

Using \(\sqrt{15} = 3.88\) gives:

\(\dfrac{3.88}{3} = 1.293\)

Example 8: √(248 + √(51 + √169)) = ?
(a) 15
(b) 12
(c) 13
(d) 16
Ans: (d)

Evaluate the innermost square root first.

\(\sqrt{169} = 13\)

\(\sqrt{51 + 13} = \sqrt{64} = 8\)

\(\sqrt{248 + 8} = \sqrt{256} = 16\)

Example 9: √ 2025 = ?
(a) 45
(b) 35
(c) 34
(d) 30
Ans: (a)

Factorise 2025 into prime powers.

\(2025 = 3^4 \times 5^2\)

Take square root:

\(\sqrt{2025} = \sqrt{3^4 \times 5^2} = 3^2 \times 5 = 9 \times 5 = 45\)

Example 10: √64009 = ?
(a) 803
(b) 363
(c) 253
(d) 347
Ans: (c)

Recognise or test the nearest integer square. Compute:

\(253^2 = (250 + 3)^2 = 250^2 + 2 \times 250 \times 3 + 3^2 = 62500 + 1500 + 9 = 64009\)

Therefore \(\sqrt{64009} = 253\).

Additional Tips and Shortcuts

• To check if a number is a perfect square or cube quickly, use its prime exponents: all even exponents ⇒ perfect square; all exponents multiples of 3 ⇒ perfect cube.
• When rationalising denominators containing two-term surds, use the conjugate; for cube roots with sums, use algebraic identities if applicable.
• For decimals under a square root, shift decimal point by multiplying/dividing by powers of 10 to convert to a whole number square root.
• Memorise small squares up to 30² and cubes up to 20³ for quick recognition in competitive tests.

Summary

Understanding prime factorisation, properties of surds and rationalisation are the essential tools for solving square-root and cube-root problems efficiently. Use the stepwise methods above to simplify, evaluate nested radicals and to find the smallest multiplier to reach perfect powers.