Important Formulas Square Roots and Cube Roots - Quantitative Aptitude for SSC CGL

Square Root

If x² = y, the square root of y is x and we write √y = x.

Examples: √4 = 2, √9 = 3, √196 = 14.

Definition and basic facts

• Principal square root: For a non-negative real number y, √y denotes the non-negative number whose square is y.
• Notation: √y and y^1/2 denote the square root of y.
• Perfect square: An integer n is a perfect square if n = k² for some integer k. Examples: 1, 4, 9, 16, 25, 36, ...
• Prime factor method: To check whether a number is a perfect square, write its prime factorisation. All prime exponents must be even for the number to be a perfect square.
• Square of last digits (unit-digit rule): The unit digit of a perfect square can be only 0, 1, 4, 5, 6 or 9.

Properties of square roots

1. √(xy) = √x × √y for x ≥ 0, y ≥ 0.
2. √(x/y) = √x / √y for x ≥ 0, y > 0.
3. (√x)² = x for x ≥ 0.
4. √(x²) = |x| for all real x.

Cube Root

The cube root of a number x is the number whose cube equals x. It is denoted by ∛x or x^1/3.

Examples: ∛8 = 2 because 2 × 2 × 2 = 8; ∛343 = 7 because 7 × 7 × 7 = 343.

Properties of cube roots

• ∛(abc) = ∛a × ∛b × ∛c for real a, b, c when signs are handled correctly.
• ∛(x³) = x for all real x.
• Prime factor method: to test a perfect cube, write prime factorisation; all prime exponents must be multiples of 3.

Examples

Example 1: (√625 x 14 x 11 is equal to: 11 √25 √196
(a) 5
(b) 6
(c) 8
(d) 11
Ans: (a)
Given Expression

Example 2: How many two-digit numbers satisfy this property. The last digit (unit's digit) of the square of the two-digit number is 8?
(a) 1
(b) 2
(c) 3
(d) None of these
Ans: (d)

Explanation:

Squares of integers end with digits 0, 1, 4, 5, 6 or 9 only. No square ends in 8.

To check the unit digits explicitly:

\(0^2 \to 0\)

\(1^2 \to 1\)

\(2^2 \to 4\)

\(3^2 \to 9\)

\(4^2 \to 6\)

\(5^2 \to 5\)

\(6^2 \to 6\)

\(7^2 \to 9\)

\(8^2 \to 4\)

\(9^2 \to 1\)

None yield 8; therefore no two-digit number has a square ending in 8.

Example 3:
(a) 10
(b) 100
(c) 1000
(d) None of these
Ans: (b)

Solution:

\(0.0169x = (1.3)^2\)

\((1.3)^2 = 1.69\)

\(x = \dfrac{1.69}{0.0169}\)

\(x = 100\)

Example 4: If √5 = 2.236;
Then the value of

Explanation:

The question statement and the expression to evaluate are shown in the image(s) above.

Use the given approximation √5 = 2.236 and substitute into the expression shown to obtain the numerical value. The computed value matches option (b) 7.826.

Example 5: The square root of (7 + 3√5) (7 - 3√5) is
(a) 2
(b) 4
(c) √5
(d) 3√5
Ans: (a)

Solution:

\((7 + 3\sqrt{5})(7 - 3\sqrt{5}) = 7^2 - (3\sqrt{5})^2\)

\(7^2 - (3\sqrt{5})^2 = 49 - 9 \times 5\)

\(49 - 45 = 4\)

\(\sqrt{4} = 2\)

Example 6: What should come in place of both x in the equation:

(a) 12
(b) 14
(c) 144
(d) 196
Ans: (a)

Solution:

The expression simplifies to \(8 \times 6 \times 3\).

\(8 \times 6 \times 3 = 144\)

\(x = \sqrt{144} = 12\)

Example 7: The cube root of .000216 is:
(a) .6
(b) .06
(c) 77
(d) 87
Ans: (b)

Solution:

\(0.06 = \dfrac{6}{100} = 6 \times 10^{-2}\)

\((0.06)^3 = 6^3 \times 10^{-6} = 216 \times 10^{-6} = 0.000216\)

Therefore \(\sqrt[3]{0.000216} = 0.06\).

Example 8: The least perfect square, which is divisible by each of 21, 36 and 66 is:
(a) 213444
(b) 214344
(c) 214434
(d) 231444
Ans: (a)

Solution:

Find LCM of 21, 36 and 66.

\(21 = 3 \times 7\)

\(36 = 2^2 \times 3^2\)

\(66 = 2 \times 3 \times 11\)

LCM uses highest powers: \(2^2, 3^2, 7, 11\).

\(\text{LCM} = 2^2 \times 3^2 \times 7 \times 11 = 2772\)

Prime exponents in 2772 are: \(2^2, 3^2, 7^1, 11^1\). To make a perfect square, every exponent must be even.

Multiply by \(7 \times 11\) to make exponents even (become \(7^2, 11^2\)).

Required number \(= 2^2 \times 3^2 \times 7^2 \times 11^2\)

\(= 4 \times 9 \times 49 \times 121 = 213444\)

Example 9: If a = 0.1039, then the value of

Solution:

\((1 - 2a) + 3a = 1 + a\)

Substitute \(a = 0.1039\).

\(1 + a = 1 + 0.1039 = 1.1039\)

Example 10: A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to ₹ 59.29, the number of the member is the group is:
(a) 57
(b) 67
(c) 77
(d) 87
Ans: (c)

Solution:

Convert rupees to paise: \(59.29 \text{ rupees} = 59.29 \times 100 = 5929 \text{ paise}\)

The collection rule implies the total amount in paise equals \(n^2\) when n members give n paise each.

So \(n^2 = 5929\).

\(\sqrt{5929} = 77\)

Therefore the number of members is 77.

Useful methods and shortcuts

• Perfect square by prime factors: Factor the number. All prime exponents must be even; to make a non-square into the least perfect square multiple, multiply by missing primes to make exponents even.
• Perfect cube by prime factors: All prime exponents must be multiples of 3.
• Decimal roots: For square root of a decimal, count digits after decimal and group digits in pairs from the decimal point outward to use digit-by-digit methods; for cube root, group digits in threes.
• Working with surds: Use conjugates to simplify expressions of the form √a ± √b and rationalise denominators when required.
• Checking last digits: Use unit-digit patterns for squares and cubes to eliminate impossible options quickly.

Concluding remarks

Mastery of square roots and cube roots requires facility with prime factorisation, recognition of perfect squares/cubes, and comfort with surd arithmetic and decimal approximations. Practice evaluating expressions, using conjugates where needed, and applying unit-digit tests for quick eliminations in competitive questions.