SSC CGL Maths · Surds, Indices and Simplification

Indices and Square Roots

The laws of indices count repeated factors; the square root of a perfect square is found from its last digit and the nearest square of a round number; a root inside a root is opened from the inside.

All Surds, Indices and Simplification notes

Why this matters

Twelve questions in the bank, all EASY, and ten of them come from the 2025 shifts. Six are square roots, three of them of four-digit perfect squares that must be found in seconds, so the last-digit method on this page is worth practising until it is automatic.

Concept 1 of 3: The laws of indices

A power is repeated multiplication:

a^3 = a \times a \times a

. Every law of indices is just a way of counting how many factors of

a

there are.

Definition

Same base, multiply: add the indices. Same base, divide: subtract them. A power of a power: multiply them.
$a^0 = 1$ for any $a \ne 0$ , and $a^{-n} = \frac{1}{a^n}$ .
$(ab)^n = a^n b^n$ , and $a^{1/2} = \sqrt{a}$ .
Take out the common power: $3^{x + 2} - 3^x = 3^x(3^2 - 1) = 8 \times 3^x$ . When every term above and below carries the same $a^x$ , it cancels.
Fast route: if the answer cannot depend on the letter (the options are plain numbers), put the letter equal to 0 and work with ordinary numbers.
Powers of ten: $0.00052 = 5.2 \times 10^{-4}$ . Move the point and count the places.

Laws of indices

a^m a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn} \qquad a^0 = 1 \qquad a^{-n} = \frac{1}{a^n}

athe base, not zero
m, nthe indices (powers)

Worked example

Simplify

\frac{2^{n + 5} - 3 \times 2^{n + 2}}{5 \times 2^{n} - 2^{n}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 24 Jul 2023, Shift 1 · Q64Easy

Example 1 · Surds, Indices and Simplification · Indices and Square Roots

What is the value of the given expression?

\frac{4^{a + 4} - 5 \times 4^{a + 2}}{15 \times 4^a - 2^2 \times 4^a}

Adding indices for different bases

2^3 \times 3^2

8 \times 9 = 72

, not

6^5

. The add-the-indices law works only when the bases are the same.

Multiplying indices in a product

a^3 \times a^4 = a^7

, not

a^{12}

. Indices multiply only for a power of a power:

(a^3)^4 = a^{12}

Reading a zero index as zero

7^0 = 1

, not 0. Any non-zero number to the power 0 is 1.

Drill 2 more on the laws of indices

Concept 2 of 3: Square roots of perfect squares

SSC prints four-digit perfect squares and expects the root at once. The last digit narrows the root to two choices, and the square of the number ending in 5 between them picks the right one.

Definition

Last digit of the square gives the last digit of the root: 1 gives 1 or 9; 4 gives 2 or 8; 9 gives 3 or 7; 6 gives 4 or 6; 5 gives 5; 0 gives 0.
Tens digit of the root: cover the last two digits of the square and find the largest square at or below what is left. For 3969, 39 lies between $6^2 = 36$ and $7^2 = 49$ , so the root is in the sixties.
Choose between the two: compare with the square of the number ending in 5. 3969 ends in 9, so the root is 63 or 67; $65^2 = 4225$ , and 3969 is less, so the root is the smaller choice, 63.
A number ending in 5: if $n$ is its tens part, its square is $n(n + 1)$ followed by 25. So $65^2 = 4225$ , since $6 \times 7 = 42$ .
Inside a root, finish the BODMAS first: $\sqrt{3 \times 12} = \sqrt{36} = 6$ .

Square of a number ending in 5

(10n + 5)^2 = 100\,n(n + 1) + 25

nthe tens part of the number

Worked example

Find

\sqrt{3364}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 18 Sep 2025, Shift 1 · Q52Easy

Example 2 · Surds, Indices and Simplification · Indices and Square Roots

Solve:

\left(\sqrt{7225} - \sqrt{1225}\right) \times \left(\frac{1}{5}\right)

Trusting the last digit alone

A square ending in 6 can have a root ending in 4 or 6:

\sqrt{1296} = 36

but

\sqrt{1156} = 34

. Compare with

35^2 = 1225

to choose.

Taking the root of each term

\sqrt{64 + 36}

\sqrt{100} = 10

, not

8 + 6 = 14

. The root of a sum is not the sum of the roots.

Drill 5 more on square roots of perfect squares

Concept 3 of 3: Nested square roots

A root inside a root is opened from the inside, like brackets. When the roots go on for ever, the whole expression equals one of its own parts, and that gives a simple equation.

Definition

Finite: start at the innermost root. In SSC questions each result makes the next number a perfect square.
Infinite: if $x = \sqrt{a + \sqrt{a + \cdots}}$ , the part under the first root is $a + x$ , so $x = \sqrt{a + x}$ and $x^2 - x - a = 0$ . Keep the positive root.
Fast route: write $a$ as a product of two consecutive numbers, $n(n - 1)$ . Then $\sqrt{a + \sqrt{a + \cdots}} = n$ , and with minus signs $\sqrt{a - \sqrt{a - \cdots}} = n - 1$ . For 20 = 5 × 4 the values are 5 and 4.
The fast route is safe only when $a$ really is such a product. Otherwise solve the quadratic.

Infinite nested roots

\sqrt{n(n - 1) + \sqrt{n(n - 1) + \cdots}} = n \qquad \sqrt{n(n - 1) - \sqrt{n(n - 1) - \cdots}} = n - 1

Worked example

Find

\sqrt{30 + \sqrt{30 + \sqrt{30 + \cdots}}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 17 Sep 2025, Shift 3 · Q53Easy

Example 3 · Surds, Indices and Simplification · Indices and Square Roots

Evaluate

\sqrt{12 + \sqrt{12 + \sqrt{12 + \cdots}}}

Keeping the negative root

x^2 - x - 20 = 0

gives

x = 5

x = -4

. A square root is never negative, so

\sqrt{20 + \sqrt{20 + \cdots}} = 5

Using the shortcut on the wrong number

\sqrt{10 + \sqrt{10 + \cdots}}

is not a whole number, because 10 is not a product of two consecutive numbers. Solving

x^2 - x - 10 = 0

gives

x = \frac{1 + \sqrt{41}}{2}

, about 3.7.

Drill 2 more on nested square roots

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The laws of indices
Laws of indices
$a^m a^n = a^{m+n} \qquad \frac{a^m}{a^n} = a^{m-n} \qquad (a^m)^n = a^{mn} \qquad a^0 = 1 \qquad a^{-n} = \frac{1}{a^n}$
Square roots of perfect squares
Square of a number ending in 5
$(10n + 5)^2 = 100\,n(n + 1) + 25$
Nested square roots
Infinite nested roots
$\sqrt{n(n - 1) + \sqrt{n(n - 1) + \cdots}} = n \qquad \sqrt{n(n - 1) - \sqrt{n(n - 1) - \cdots}} = n - 1$

Watch out for (7)

Adding indices for different bases
→ The laws of indices
Multiplying indices in a product
→ The laws of indices
Reading a zero index as zero
→ The laws of indices
Trusting the last digit alone
→ Square roots of perfect squares
Taking the root of each term
→ Square roots of perfect squares
Keeping the negative root
→ Nested square roots
Using the shortcut on the wrong number
→ Nested square roots

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