SSC CGL Maths · Surds, Indices and Simplification

Surds and Rationalisation

A surd is a root that is not a whole number, such as √2; surds are simplified by taking out square factors, combined only when the roots match, cleared from a denominator with the conjugate, and compared by squaring.

All Surds, Indices and Simplification notes

Why this matters

Twenty questions in the bank, 14 EASY and 6 MODERATE, and 18 of them come from the 2025 shifts. Three moves cover all of them: simplify and expand, rationalise with the conjugate, and compare two surd sums by squaring. Knowing √2, √3, √5, √6 and √7 to three decimal places turns several of them into a quick estimate.

Concept 1 of 3: Simplifying, adding and expanding surds

Pull the largest square factor out of each root. Once the numbers left under the roots match, the surds behave like like terms in algebra:

3\sqrt{2}

and

5\sqrt{2}

add just as

3x

and

5x

do.

Definition

Simplify: $\sqrt{ab} = \sqrt{a}\,\sqrt{b}$ . Take out the largest square factor: $\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$ .
Like surds have the same number under the root after simplifying: $4\sqrt{3} + 2\sqrt{3} = 6\sqrt{3}$ . $\sqrt{2} + \sqrt{3}$ cannot be combined.
Expand like algebra: $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$ and $(\sqrt{a} + c)^2 = a + c^2 + 2c\sqrt{a}$ .
$(x + y)^2 - (x - y)^2 = 4xy$ , so $(\sqrt{a} + \sqrt{b})^2 - (\sqrt{a} - \sqrt{b})^2 = 4\sqrt{ab}$ .
Values to know: $\sqrt{2} \approx 1.414$ , $\sqrt{3} \approx 1.732$ , $\sqrt{5} \approx 2.236$ , $\sqrt{6} \approx 2.449$ , $\sqrt{7} \approx 2.646$ . A quick decimal check confirms an answer or removes options.

Surd rules

\sqrt{ab} = \sqrt{a}\,\sqrt{b} \qquad (\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab} \qquad (\sqrt{a} + \sqrt{b})^2 - (\sqrt{a} - \sqrt{b})^2 = 4\sqrt{ab}

Worked example

Simplify

\sqrt{75} - \sqrt{27} + \sqrt{12}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 15 Sep 2025, Shift 2 · Q71Easy

Example 1 · Surds, Indices and Simplification · Surds and Rationalisation

Simplify:

\sqrt{98} - \sqrt{32} + \sqrt{50}

Adding the numbers under the roots

\sqrt{9} + \sqrt{16} = 3 + 4 = 7

, but

\sqrt{25} = 5

. Roots add only when they are like surds, and then you add the numbers in front of them.

Dropping the middle term of a surd square

(\sqrt{7} + 2)^2 = 7 + 4 + 4\sqrt{7} = 11 + 4\sqrt{7}

, not

7 + 4 = 11

. The middle term

2ab

carries the surd.

Drill 8 more on simplifying, adding and expanding surds

Concept 2 of 3: Rationalising with the conjugate

A root in the denominator makes a fraction hard to use. Multiply the top and bottom by the conjugate, the same two terms with the sign between them changed. The bottom becomes a difference of squares, and the root disappears.

Definition

$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b$ , with no root left.
So $\frac{1}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{a} + \sqrt{b}}{a - b}$ .
When the two squares differ by 1, the reciprocal is just the conjugate: $\frac{1}{\sqrt{10} - 3} = \sqrt{10} + 3$ . This makes long alternating chains collapse, each root cancelling the next.
A surd plus its conjugate has no root: $\frac{1}{\sqrt{a} - \sqrt{b}} + \frac{1}{\sqrt{a} + \sqrt{b}} = \frac{2\sqrt{a}}{a - b}$ .
A number and its reciprocal: for $x = 4 + \sqrt{15}$ , $\frac{1}{x} = 4 - \sqrt{15}$ (since $16 - 15 = 1$ ), so $x + \frac{1}{x} = 8$ and $x \cdot \frac{1}{x} = 1$ . Questions on $x^2 + \frac{1}{x^2}$ or $x^3 + \frac{1}{x^3}$ then use the identities.
Sign check: if the denominator is negative, as in $\sqrt{2} - \sqrt{3}$ , the answer is negative.

Conjugate

(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b \qquad \frac{1}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{a} + \sqrt{b}}{a - b}

Worked example

Rationalise

\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 3 · Q71Easy

Example 2 · Surds, Indices and Simplification · Surds and Rationalisation

Simplify:

\frac{1}{\sqrt{7}-\sqrt{3}} + \frac{1}{\sqrt{7}+\sqrt{3}}

Forgetting to divide by a minus b

\frac{1}{\sqrt{7} - \sqrt{2}} = \frac{\sqrt{7} + \sqrt{2}}{5}

, not

\sqrt{7} + \sqrt{2}

. The reciprocal is the bare conjugate only when the two squares differ by 1.

Missing a negative denominator

\frac{\sqrt{3} + \sqrt{5}}{\sqrt{3} - \sqrt{5}}

is negative, because

\sqrt{3} < \sqrt{5}

. It equals

\frac{8 + 2\sqrt{15}}{-2} = -(4 + \sqrt{15})

, and the same value without its minus sign is the wrong option SSC likes to print.

Drill 6 more on rationalising with the conjugate

Concept 3 of 3: Comparing sums and differences of surds

Two surd sums can look almost equal. Since both are positive, square them: the whole numbers match and only the surd parts are left to compare. A difference of surds is easier after rationalising.

Definition

Sums: $(\sqrt{a} + \sqrt{b})^2 = (a + b) + 2\sqrt{ab}$ . When $a + b = c + d$ , the sum with the larger product is larger, so the pair whose numbers are closer together wins.
Differences: $\sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$ . With the same $a - b$ , the difference of the larger numbers is the smaller one: $\sqrt{11} - \sqrt{10} < \sqrt{3} - \sqrt{2}$ .
Fast route: use the decimal values of the roots. Three decimal places separate the options unless two sums agree to two places.
Squaring keeps the order only when both sides are positive.

Comparing surd sums and differences

(\sqrt{a} + \sqrt{b})^2 = (a + b) + 2\sqrt{ab} \qquad \sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}}

Worked example

Which is larger:

\sqrt{10} + \sqrt{3}

\sqrt{8} + \sqrt{5}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 12 Sep 2025, Shift 2 · Q51Moderate

Example 3 · Surds, Indices and Simplification · Surds and Rationalisation

Let

x = \sqrt{3} + \sqrt{5}

and

y = \sqrt{8} + \sqrt{2}

. Which is greater?

Calling two surd sums equal because the numbers under the roots add up the same

\sqrt{7} + \sqrt{1}

and

\sqrt{4} + \sqrt{4}

both have a total of 8 under the roots, but they are not equal: about 3.646 against 4. Equal totals say nothing; compare the products, 7 against 16.

Squaring when a side may be negative

Squaring keeps the order of positive numbers only:

-3 < 2

but

9 > 4

. Check that both sides are positive before squaring them.

Drill 3 more on comparing sums and differences of surds

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)

Simplifying, adding and expanding surds
Surd rules
$\sqrt{ab} = \sqrt{a}\,\sqrt{b} \qquad (\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab} \qquad (\sqrt{a} + \sqrt{b})^2 - (\sqrt{a} - \sqrt{b})^2 = 4\sqrt{ab}$
Rationalising with the conjugate
Conjugate
$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b \qquad \frac{1}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{a} + \sqrt{b}}{a - b}$
Comparing sums and differences of surds
Comparing surd sums and differences
$(\sqrt{a} + \sqrt{b})^2 = (a + b) + 2\sqrt{ab} \qquad \sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$

Watch out for (6)

Adding the numbers under the roots
→ Simplifying, adding and expanding surds
Dropping the middle term of a surd square
→ Simplifying, adding and expanding surds
Forgetting to divide by a minus b
→ Rationalising with the conjugate
Missing a negative denominator
→ Rationalising with the conjugate
Calling two surd sums equal because the numbers under the roots add up the same
→ Comparing sums and differences of surds
Squaring when a side may be negative
→ Comparing sums and differences of surds

Test yourself on Surds, Indices and Simplification

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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