SSC CGL Maths · Surds, Indices and Simplification

Square Roots of Surds and Surd Equations

A number of the form a + 2√b is the square of √m + √n when m + n = a and mn = b; an equation with two roots is solved through their sum and difference, or by componendo and dividendo.

All Surds, Indices and Simplification notes

Why this matters

Ten questions in the bank, 4 EASY and 6 MODERATE, all from the 2025 shifts. Six ask for the square root of a whole number plus a surd, and four are equations with two square roots. Each has a two-line method, and each can also be settled by putting the options back in.

Concept 1 of 3: The square root of a + 2√b

Squaring

\sqrt{m} + \sqrt{n}

gives

m + n + 2\sqrt{mn}

: a whole number plus twice a root. So to undo it, look for two numbers that add up to the whole number and multiply to the number under the root.

Definition

First write the surd part as 2 times a root: $4\sqrt{5} = 2\sqrt{20}$ , $6\sqrt{3} = 2\sqrt{27}$ .
Find $m$ and $n$ with $m + n = a$ and $mn = b$ , taking $m > n$ .
Then $\sqrt{a + 2\sqrt{b}} = \sqrt{m} + \sqrt{n}$ and $\sqrt{a - 2\sqrt{b}} = \sqrt{m} - \sqrt{n}$ . The larger root goes first, because a square root is never negative.
The two results multiply to $m - n$ . When $m - n = 1$ they are reciprocals, so $\frac{1}{\sqrt{m} - \sqrt{n}} = \sqrt{m} + \sqrt{n}$ .
Fast route: square each option and keep the one that gives the number back.

Square root of a surd

\sqrt{a \pm 2\sqrt{b}} = \sqrt{m} \pm \sqrt{n} \quad \text{where } m + n = a,\ mn = b,\ m > n

Worked example

Find

\sqrt{8 + 2\sqrt{15}}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 1 · Q51Easy

Example 1 · Surds, Indices and Simplification · Square Roots of Surds and Surd Equations

Simplify the expression:

\sqrt{(9 + 4\sqrt{5})}

Not writing the surd part as 2√b first

11 + 6\sqrt{2}

the surd part is

2\sqrt{18}

, so look for

m + n = 11

and

mn = 18

: 9 and 2, giving

3 + \sqrt{2}

. Using

mn = 2

mn = 6

finds no pair.

Putting the smaller root first after a minus

\sqrt{7 - 2\sqrt{12}}

2 - \sqrt{3}

, not

\sqrt{3} - 2

. A square root is never negative, so the larger root goes first.

Drill 5 more on the square root of a + 2√b

Concept 2 of 3: Equations with a sum and a difference of square roots

The product of a sum and a difference of roots has no roots in it:

(\sqrt{A} + \sqrt{B})(\sqrt{A} - \sqrt{B}) = A - B

. So if you know the sum and can see

A - B

, you also know the difference, and adding the two gives

\sqrt{A}

Definition

Given $\sqrt{A} + \sqrt{B} = s$ and $\sqrt{A} - \sqrt{B} = d$ : add to get $2\sqrt{A} = s + d$ , subtract to get $2\sqrt{B} = s - d$ .
Given only the sum, as in $\sqrt{x + p} + \sqrt{x - q} = s$ : the two numbers under the roots differ by $p + q$ , so the difference of the roots is $\frac{p + q}{s}$ . Then add and subtract as above.
Fast route: try the options; each needs two quick square roots.
Always check the answer in the original equation.

Sum times difference of roots

(\sqrt{A} + \sqrt{B})(\sqrt{A} - \sqrt{B}) = A - B

Worked example

Solve

\sqrt{x + 7} + \sqrt{x - 5} = 6

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 · Q69Moderate

Example 2 · Surds, Indices and Simplification · Square Roots of Surds and Surd Equations

Given,

\sqrt{x + 3} + \sqrt{x - 2} = 5

, then find the value of

x

Squaring a sum of roots term by term

\sqrt{x + 7} + \sqrt{x - 5} = 6

does not give

(x + 7) + (x - 5) = 36

. Squaring a sum adds the middle term

2\sqrt{(x + 7)(x - 5)}

, which is why the sum-and-difference route is quicker.

Drill 1 more on equations with a sum and a difference of square roots

Concept 3 of 3: Componendo and dividendo

When a ratio has a sum on top and the matching difference below, adding and subtracting its top and bottom pulls the two parts apart. With roots, one step leaves the ratio of the two roots.

Definition

If $\frac{p}{q} = \frac{a}{b}$ , then $\frac{p + q}{p - q} = \frac{a + b}{a - b}$ . The rule also runs backwards.
So $\frac{\sqrt{A} + \sqrt{B}}{\sqrt{A} - \sqrt{B}} = k$ gives $\frac{\sqrt{A}}{\sqrt{B}} = \frac{k + 1}{k - 1}$ .
Square it: $\frac{A}{B} = \left(\frac{k + 1}{k - 1}\right)^2$ , then solve the plain equation in $x$ .
Fast route: test the options, starting with the ones that make both numbers under the roots perfect squares.

Componendo and dividendo

\frac{p}{q} = \frac{a}{b} \;\Rightarrow\; \frac{p + q}{p - q} = \frac{a + b}{a - b} \qquad \frac{\sqrt{A} + \sqrt{B}}{\sqrt{A} - \sqrt{B}} = k \;\Rightarrow\; \frac{A}{B} = \left(\frac{k + 1}{k - 1}\right)^2

Worked example

Solve

\frac{\sqrt{x + 4} + \sqrt{x - 4}}{\sqrt{x + 4} - \sqrt{x - 4}} = 3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 20 Sep 2025, Shift 1 · Q51Moderate

Example 3 · Surds, Indices and Simplification · Square Roots of Surds and Surd Equations

Find the value of x in given equation:

\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}} = 2

Forgetting to square the ratio of the roots

From

\frac{\sqrt{A}}{\sqrt{B}} = 3

the ratio

\frac{A}{B}

is 9, not 3. Componendo and dividendo gives the ratio of the roots; square it before solving for

x

Drill 1 more on componendo and dividendo

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 square root of a + 2√b
Square root of a surd
$\sqrt{a \pm 2\sqrt{b}} = \sqrt{m} \pm \sqrt{n} \quad \text{where } m + n = a,\ mn = b,\ m > n$
Equations with a sum and a difference of square roots
Sum times difference of roots
$(\sqrt{A} + \sqrt{B})(\sqrt{A} - \sqrt{B}) = A - B$
Componendo and dividendo
Componendo and dividendo
$\frac{p}{q} = \frac{a}{b} \;\Rightarrow\; \frac{p + q}{p - q} = \frac{a + b}{a - b} \qquad \frac{\sqrt{A} + \sqrt{B}}{\sqrt{A} - \sqrt{B}} = k \;\Rightarrow\; \frac{A}{B} = \left(\frac{k + 1}{k - 1}\right)^2$

Watch out for (4)

Not writing the surd part as 2√b first
→ The square root of a + 2√b
Putting the smaller root first after a minus
→ The square root of a + 2√b
Squaring a sum of roots term by term
→ Equations with a sum and a difference of square roots
Forgetting to square the ratio of the roots
→ Componendo and dividendo

Test yourself on Surds, Indices and Simplification

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