Surds and Rationalisation

What a Surd Is

A surd is one particular kind of irrational number: a root of a positive rational number that does not come out exact. $\sqrt{2}$, $\sqrt{7}$, $\sqrt[3]{5}$ and $\sqrt[4]{10}$ are surds. $\sqrt{49} = 7$ and $\sqrt[3]{27} = 3$ are not, because those roots come out exact. $\pi$ is irrational, but it is not a surd either, because it is not the root of any rational number.

The words that come up in questions.

• Order. In $\sqrt[n]{a}$, the small $n$ is the order. A square root $\sqrt{a}$ has order 2 and a cube root $\sqrt[3]{a}$ has order 3.
• Radicand. The number under the root sign. In $5\sqrt{3}$ the radicand is 3.
• Pure surd. Everything sits under the root, like $\sqrt{50}$ or $\sqrt[3]{24}$.
• Mixed surd. A rational number times a surd, like $5\sqrt{2}$ or $2\sqrt[3]{3}$.

Mixed to pure. Raise the outside number to the order of the root, then take it inside.

$$5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50} \qquad 2\sqrt[3]{3} = \sqrt[3]{8 \times 3} = \sqrt[3]{24}$$

Pure to simplest form. Split the radicand so that one part is the biggest perfect square you can find (a perfect cube for a cube root), and take its root outside.

$$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \qquad \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \qquad \sqrt[3]{54} = \sqrt[3]{27 \times 2} = 3\sqrt[3]{2}$$

If you only spot a smaller square, you still get there, just in more steps: $\sqrt{72} = 2\sqrt{18} = 2 \times 3\sqrt{2} = 6\sqrt{2}$. The squares worth testing are 4, 9, 16, 25, 36, 49, 64, 81, 100, 121 and 144.

Like and unlike surds. Two surds are like when, after simplifying, they have the same order and the same radicand. $\sqrt{12} = 2\sqrt{3}$ and $\sqrt{75} = 5\sqrt{3}$ are like surds. $\sqrt{8} = 2\sqrt{2}$ and $\sqrt{12} = 2\sqrt{3}$ are unlike, and so are $\sqrt{2}$ and $\sqrt[3]{2}$, because their orders differ.

[SSC Tip] Never judge two surds by how they look. $\sqrt{18}$ and $\sqrt{50}$ seem unrelated until you write them as $3\sqrt{2}$ and $5\sqrt{2}$. Simplify every surd in a question before you do anything else with it.

A surd can also be written as a power, such as $\sqrt[3]{5} = 5^{1/3}$ (see Indices and Powers). SSC usually writes surd questions in root form, so that is the form used here.

Adding, Multiplying and Dividing Surds

Adding and subtracting. You can only combine like surds, the same way you combine $3x + 7x = 10x$.

$$3\sqrt{5} + 7\sqrt{5} = 10\sqrt{5} \qquad \sqrt{48} + \sqrt{27} - \sqrt{12} = 4\sqrt{3} + 3\sqrt{3} - 2\sqrt{3} = 5\sqrt{3}$$

Unlike surds stay as they are. $2\sqrt{3} + 4\sqrt{2}$ cannot be squeezed into one term.

The biggest trap in this chapter: $\sqrt{a} + \sqrt{b}$ is not $\sqrt{a + b}$. Try it with perfect squares and the mistake shows at once: $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$, but $\sqrt{25} = 5$. SSC puts $\sqrt{a + b}$ among the options for exactly this reason.

Multiplying and dividing (same order). Multiply or divide under the root, and multiply the outside numbers separately.

$$\sqrt{a} \times \sqrt{b} = \sqrt{ab} \qquad \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \qquad \sqrt{a} \times \sqrt{a} = a$$

$$2\sqrt{6} \times 3\sqrt{10} = 6\sqrt{60} = 6 \times 2\sqrt{15} = 12\sqrt{15} \qquad \frac{\sqrt{72}}{\sqrt{8}} = \sqrt{9} = 3$$

Brackets open the same way as in algebra. Just remember that $\sqrt{3} \times \sqrt{3} = 3$:

$$(2 + \sqrt{3})(4 - \sqrt{3}) = 8 - 2\sqrt{3} + 4\sqrt{3} - 3 = 5 + 2\sqrt{3}$$

Surds of different orders. You can't multiply $\sqrt{2}$ by $\sqrt[3]{3}$ directly. First bring both to the same order, which is the LCM of the two orders. The rule: multiply the order by some number $k$ and raise the radicand to the power $k$.

$$\sqrt[n]{a} = \sqrt[nk]{a^k}$$

For orders 2 and 3 the LCM is 6, so $\sqrt{2} = \sqrt[6]{2^3} = \sqrt[6]{8}$ and $\sqrt[3]{3} = \sqrt[6]{3^2} = \sqrt[6]{9}$. Then

$$\sqrt{2} \times \sqrt[3]{3} = \sqrt[6]{8 \times 9} = \sqrt[6]{72}$$

One more: $\sqrt[3]{4} \times \sqrt{2} = \sqrt[6]{16} \times \sqrt[6]{8} = \sqrt[6]{128} = \sqrt[6]{64 \times 2} = 2\sqrt[6]{2}$.

The same trick of making the orders equal is how you compare roots like $\sqrt{3}$ and $\sqrt[3]{5}$. That comparison is done in full in the Indices and Powers section.

Options usually show surds in simplest form. If you get $\sqrt{108}$ and the options say $6\sqrt{3}$, they are the same number. Simplify your answer before you decide that none of the options match.

Rationalising the Denominator

Rationalising means rewriting a fraction so that no surd is left in the denominator. You multiply the top and the bottom by the same thing, so the value doesn't change. That "same thing" is called the rationalising factor.

One surd at the bottom. Multiply by that surd.

$$\frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \qquad \frac{15}{\sqrt{5}} = \frac{15\sqrt{5}}{5} = 3\sqrt{5}$$

Simplify the surd first and the numbers stay small: $\frac{6}{\sqrt{12}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$. The smallest rationalising factor of $\sqrt{12}$ is $\sqrt{3}$, not $\sqrt{12}$, because $\sqrt{12} \times \sqrt{3} = 6$ is already rational.

For a cube root, multiply by whatever completes a perfect cube. $\sqrt[3]{2} \times \sqrt[3]{4} = \sqrt[3]{8} = 2$, so $\frac{1}{\sqrt[3]{2}} = \frac{\sqrt[3]{4}}{2}$. The rationalising factor of $\sqrt[3]{4}$ is $\sqrt[3]{2}$, and of $\sqrt[3]{9}$ it is $\sqrt[3]{3}$.

Two terms at the bottom. Multiply by the conjugate, which is the same two terms with the middle sign flipped. The product is always rational:

$$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b \qquad (p + \sqrt{b})(p - \sqrt{b}) = p^2 - b$$

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

When the top is the same pair with a plus, the top becomes a square:

$$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{a + b + 2\sqrt{ab}}{a - b} \qquad \text{e.g. } \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \frac{8 + 2\sqrt{12}}{4} = 2 + \sqrt{3}$$

When the bottom becomes 1. If $a - b = 1$ or $p^2 - b = 1$, the denominator disappears completely:

$$\frac{1}{\sqrt{6} - \sqrt{5}} = \sqrt{6} + \sqrt{5} \qquad \frac{1}{3 - 2\sqrt{2}} = 3 + 2\sqrt{2}$$

(The second works because $(2\sqrt{2})^2 = 8$ and 9 − 8 = 1.) Learn to spot these pairs, because they drive the telescoping and conjugate questions later on: $2 \pm \sqrt{3}$, $3 \pm 2\sqrt{2}$, $5 \pm 2\sqrt{6}$, $4 \pm \sqrt{15}$ and $\sqrt{n + 1} \pm \sqrt{n}$.

Three terms at the bottom. Group two of them, treat the group as one term, and rationalise twice. Pick the grouping that makes the plain numbers cancel.

$$\frac{1}{1 + \sqrt{2} + \sqrt{3}} \times \frac{(1 + \sqrt{2}) - \sqrt{3}}{(1 + \sqrt{2}) - \sqrt{3}} = \frac{1 + \sqrt{2} - \sqrt{3}}{(1 + \sqrt{2})^2 - 3} = \frac{1 + \sqrt{2} - \sqrt{3}}{2\sqrt{2}}$$

Here $(1 + \sqrt{2})^2 = 3 + 2\sqrt{2}$, and its 3 cancels the 3 from $(\sqrt{3})^2$. One more step, multiplying by $\sqrt{2}$, gives $\frac{\sqrt{2} + 2 - \sqrt{6}}{4}$.

[Tier 2] Cube roots at the bottom. Use $(a - b)(a^2 + ab + b^2) = a^3 - b^3$ and $(a + b)(a^2 - ab + b^2) = a^3 + b^3$. For example $(\sqrt[3]{2} - 1)(\sqrt[3]{4} + \sqrt[3]{2} + 1) = 2 - 1 = 1$, so $\frac{1}{\sqrt[3]{4} + \sqrt[3]{2} + 1} = \sqrt[3]{2} - 1$.

Conjugate Pairs and Surd Values of x

For a surd like $x = p + \sqrt{q}$, the conjugate is $y = p - \sqrt{q}$. A number and its conjugate have a rational sum and a rational product, and most questions that start "if $x$ equals this surd, find that" run on these two facts:

$$x + y = 2p \qquad xy = p^2 - q \qquad x - y = 2\sqrt{q}$$

When the product is 1. For pairs like $5 \pm 2\sqrt{6}$ (25 − 24 = 1), the conjugate is also the reciprocal. If $x = 5 + 2\sqrt{6}$, then $\frac{1}{x} = 5 - 2\sqrt{6}$, so

$$x + \frac{1}{x} = 10 \qquad x - \frac{1}{x} = 4\sqrt{6} \qquad x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = 98$$

If the product isn't 1, rationalise to get $\frac{1}{x}$. For $x = 3 + \sqrt{5}$, $\frac{1}{x} = \frac{3 - \sqrt{5}}{9 - 5} = \frac{3 - \sqrt{5}}{4}$.

Two fractions that are reciprocals. A favourite SSC setup is $x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$ and $y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$. Rationalise $x$: $\frac{(\sqrt{5} + \sqrt{3})^2}{5 - 3} = \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$. Then $y$ is its reciprocal, $4 - \sqrt{15}$. Everything now follows from $x + y = 8$ and $xy = 1$:

Asked	Working	Value
$x^2 + y^2$	$(x + y)^2 - 2xy = 64 - 2$	62
$x^2 + xy + y^2$	$62 + 1$	63
$x^2 - xy + y^2$	$62 - 1$	61
$x^3 + y^3$	$(x + y)^3 - 3xy(x + y) = 512 - 24$	488
$x - y$	$(4 + \sqrt{15}) - (4 - \sqrt{15})$	$2\sqrt{15}$

Two shortcuts for this pair of fractions, worth remembering as they are:

$$\frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} + \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{2(a + b)}{a - b} \qquad \frac{\sqrt{a} + \sqrt{b}}{\sqrt{a} - \sqrt{b}} - \frac{\sqrt{a} - \sqrt{b}}{\sqrt{a} + \sqrt{b}} = \frac{4\sqrt{ab}}{a - b}$$

With $a = 5$ and $b = 3$ these give 2 × 8 ÷ 2 = 8 and $\frac{4\sqrt{15}}{2} = 2\sqrt{15}$, matching the table.

An expression in a surd $x$. If $x = 3 + \sqrt{5}$, don't substitute and expand. Get the surd alone and square:

$$x - 3 = \sqrt{5} \;\Rightarrow\; x^2 - 6x + 9 = 5 \;\Rightarrow\; x^2 - 6x + 4 = 0$$

Now anything built from $x^2 - 6x$ comes straight out. For example, $x^2 - 6x + 7 = -4 + 7 = 3$.

Matching rational and irrational parts. If $p + q\sqrt{r} = s + t\sqrt{r}$, where $p, q, s, t$ are rational and $\sqrt{r}$ is irrational, then $p = s$ and $q = t$. So if $(3 + \sqrt{2})^2 = a + b\sqrt{2}$, expanding gives $11 + 6\sqrt{2}$, and $a = 11$, $b = 6$.

The general identities for $x + \frac{1}{x}$, such as $x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)$, are done properly in the Algebra chapter. For surd questions the only new idea is spotting that $\frac{1}{x}$ is the conjugate.

Square Roots of $a \pm 2\sqrt{b}$ and $a \pm \sqrt{b}$

Start from the square of two surds added or subtracted:

$$(\sqrt{x} + \sqrt{y})^2 = x + y + 2\sqrt{xy} \qquad (\sqrt{x} - \sqrt{y})^2 = x + y - 2\sqrt{xy}$$

So to find $\sqrt{a + 2\sqrt{b}}$, look for two numbers that add up to $a$ and multiply to $b$. If they are $x$ and $y$, the answer is $\sqrt{x} + \sqrt{y}$. For $\sqrt{a - 2\sqrt{b}}$ the answer is $\sqrt{x} - \sqrt{y}$ with the bigger number first, because a square root is never negative.

$$\sqrt{7 + 2\sqrt{10}} = \sqrt{5} + \sqrt{2} \qquad (5 + 2 = 7,\ 5 \times 2 = 10)$$

$$\sqrt{9 - 2\sqrt{20}} = \sqrt{5} - \sqrt{4} = \sqrt{5} - 2 \qquad (5 + 4 = 9,\ 5 \times 4 = 20)$$

Make the 2 appear first. The method needs exactly "2 times a root". Rewrite whatever is there:

• No 2 at all: $\sqrt{48} = \sqrt{4 \times 12} = 2\sqrt{12}$.
• Some other number in front: $4\sqrt{5} = 2\sqrt{20}$, $6\sqrt{2} = 2\sqrt{18}$, $8\sqrt{3} = 2\sqrt{48}$. Halve the outside number and take the half inside as its square.

So $\sqrt{8 + \sqrt{48}} = \sqrt{8 + 2\sqrt{12}} = \sqrt{6} + \sqrt{2}$, and $\sqrt{19 - 8\sqrt{3}} = \sqrt{19 - 2\sqrt{48}} = \sqrt{16} - \sqrt{3} = 4 - \sqrt{3}$.

When the 2 won't come out whole: the $a \pm \sqrt{b}$ form. In $\sqrt{4 + \sqrt{15}}$ you can't write $\sqrt{15}$ as 2 times the root of a whole number. Multiply and divide inside by 2:

$$\sqrt{4 + \sqrt{15}} = \sqrt{\frac{8 + 2\sqrt{15}}{2}} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{2}} = \frac{\sqrt{10} + \sqrt{6}}{2}$$

Adding or subtracting the pair. $\sqrt{a + 2\sqrt{b}} + \sqrt{a - 2\sqrt{b}} = 2\sqrt{x}$ and the difference is $2\sqrt{y}$. For example $\sqrt{7 + 2\sqrt{10}} - \sqrt{7 - 2\sqrt{10}} = (\sqrt{5} + \sqrt{2}) - (\sqrt{5} - \sqrt{2}) = 2\sqrt{2}$.

Squares and cubes of surd pairs.

Expression	Value
$(\sqrt{a} + \sqrt{b})^2$	$a + b + 2\sqrt{ab}$
$(\sqrt{a} - \sqrt{b})^2$	$a + b - 2\sqrt{ab}$
$(\sqrt{a} + \sqrt{b})^2 + (\sqrt{a} - \sqrt{b})^2$	$2(a + b)$
$(\sqrt{a} + \sqrt{b})^2 - (\sqrt{a} - \sqrt{b})^2$	$4\sqrt{ab}$
$(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b})$	$a - b$

For a cube, use $(p + q)^3 = p^3 + 3p^2 q + 3pq^2 + q^3$. With $p = \sqrt{2}$ and $q = 1$: $(\sqrt{2} + 1)^3 = 2\sqrt{2} + 6 + 3\sqrt{2} + 1 = 7 + 5\sqrt{2}$.

[SSC Tip] If the options already show square roots, don't work forwards at all. Square each option and see which one gives the number under the root. Squaring $\sqrt{6} + \sqrt{2}$ takes five seconds.

Comparing and Ordering Surds

"Which of these is the largest?" came up often in the 2025 papers. There are five situations, and each has its own quick test.

1. Surds with numbers in front. Take each outside number back inside and compare what is under the root. $3\sqrt{7} = \sqrt{63}$, $4\sqrt{5} = \sqrt{80}$, $2\sqrt{17} = \sqrt{68}$ and $6\sqrt{2} = \sqrt{72}$, so the order is

$$3\sqrt{7} < 2\sqrt{17} < 6\sqrt{2} < 4\sqrt{5}$$

2. Sums: $\sqrt{a} + \sqrt{b}$ against $\sqrt{c} + \sqrt{d}$. Square both. You get $a + b + 2\sqrt{ab}$ against $c + d + 2\sqrt{cd}$.

When $a + b = c + d$, and SSC nearly always sets it up this way, just compare the products $ab$ and $cd$. The pair whose numbers are closer together has the bigger product, so it gives the bigger sum. For $\sqrt{10} + \sqrt{4}$ against $\sqrt{8} + \sqrt{6}$, both pairs add to 14, and 8 × 6 = 48 beats 10 × 4 = 40, so $\sqrt{8} + \sqrt{6}$ is bigger.

When the totals differ, squaring still works but you must compare the full results. $(\sqrt{5} + \sqrt{7})^2 = 12 + 2\sqrt{35}$ and $(\sqrt{3} + \sqrt{10})^2 = 13 + 2\sqrt{30}$. Here the bigger total wins, but only just (about 4.882 against 4.894), so don't assume it always will.

3. Differences with the same gap. Rationalise:

$$\sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$$

If $a - b$ is the same in every option, the option with the smallest numbers has the smallest bottom, so it is the biggest difference: $\sqrt{3} - \sqrt{2} > \sqrt{6} - \sqrt{5} > \sqrt{10} - 3$. Watch for whole numbers in disguise, like 3 for $\sqrt{9}$ or 4 for $\sqrt{16}$.

4. Differences with different gaps. The shortcut breaks, so use rough values. $\sqrt{12} - \sqrt{7} \approx 3.46 - 2.65 = 0.81$ and $\sqrt{10} - \sqrt{6} \approx 3.16 - 2.45 = 0.71$, so the first is bigger. Two decimal places are almost always enough.

5. Different orders. Make the orders equal first: $\sqrt{3} = \sqrt[6]{27}$ and $\sqrt[3]{5} = \sqrt[6]{25}$, so $\sqrt{3}$ is bigger. The Indices and Powers section covers this type fully.

What you are comparing	Quick test
$a\sqrt{b}$ forms	take the number inside, compare radicands
$\sqrt{a} + \sqrt{b}$, same total $a + b$	bigger product $ab$ wins
$\sqrt{a} - \sqrt{b}$, same gap $a - b$	smaller numbers win
different totals or gaps	square, or use rough decimal values
different orders	make the orders equal

Telescoping Surd Sums

Sums like $\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots$ look long, but almost everything in them cancels.

The key step. When the two numbers under the roots differ by 1, rationalising leaves a denominator of 1:

$$\frac{1}{\sqrt{n} + \sqrt{n + 1}} = \frac{\sqrt{n + 1} - \sqrt{n}}{(n + 1) - n} = \sqrt{n + 1} - \sqrt{n}$$

So each term turns into a difference, and neighbouring differences cancel:

$$(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \cdots + (\sqrt{n} - \sqrt{n - 1}) = \sqrt{n} - 1$$

Only the last root and the first root survive. For example

$$\frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \cdots + \frac{1}{\sqrt{35} + \sqrt{36}} = \sqrt{36} - \sqrt{4} = 6 - 2 = 4$$

When the step is $k$. If the numbers under the roots go up in steps of $k$, each term has $k$ at the bottom after rationalising:

$$\frac{1}{\sqrt{n} + \sqrt{n + k}} = \frac{\sqrt{n + k} - \sqrt{n}}{k}$$

So the whole sum is (last root − first root) ÷ $k$. For $\frac{1}{\sqrt{2} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{6}} + \cdots + \frac{1}{\sqrt{48} + \sqrt{50}}$, the answer is $\frac{\sqrt{50} - \sqrt{2}}{2} = \frac{5\sqrt{2} - \sqrt{2}}{2} = 2\sqrt{2}$.

Minus signs inside, alternating signs outside. Now each term becomes a sum:

$$\frac{1}{\sqrt{n + 1} - \sqrt{n}} = \sqrt{n + 1} + \sqrt{n}$$

On their own these wouldn't cancel, so SSC pairs them with alternating + and − signs. Then each middle root appears once with + and once with −, and only the two ends are left:

$$\frac{1}{\sqrt{3} - \sqrt{2}} - \frac{1}{\sqrt{4} - \sqrt{3}} + \frac{1}{\sqrt{5} - \sqrt{4}} = (\sqrt{3} + \sqrt{2}) - (2 + \sqrt{3}) + (\sqrt{5} + 2) = \sqrt{5} + \sqrt{2}$$

With an odd number of terms (starting with +), both end roots come out with a + sign. With an even number, the last end comes out with a − sign. If you are unsure, write the first two and the last two terms in full and watch what cancels.

Read the pattern before you use the rule.

• Do the numbers under the roots go up by the same step every time?
• Is the step 1? If not, divide by the step.
• Where does the chain start and end? A sum from $\sqrt{9}$ to $\sqrt{100}$ is not the same as one from $\sqrt{1}$.

The chain is often written as $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \cdots + \frac{1}{\sqrt{99} + \sqrt{100}}$. The first 1 is just $\sqrt{1}$, so the answer is $\sqrt{100} - \sqrt{1} = 9$.

Surd Equations

Isolate and square. Get the root alone on one side, square both sides, then solve.

$$\sqrt{3x - 2} = 5 \;\Rightarrow\; 3x - 2 = 25 \;\Rightarrow\; x = 9$$

Always put the answer back. Squaring can create an answer that doesn't really fit. With $\sqrt{x + 5} = x - 1$, squaring gives $x + 5 = x^2 - 2x + 1$, so $x^2 - 3x - 4 = 0$ and $x = 4$ or $x = -1$. Check: $x = 4$ gives $\sqrt{9} = 3$ on the left and 3 on the right, so it works. $x = -1$ gives $\sqrt{4} = 2$ on the left but $-2$ on the right, so it is thrown out.

Two roots whose insides differ by a fixed number. For $\sqrt{x + 24} + \sqrt{x} = 12$, squaring gets messy. Use this instead:

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

Here $A - B = (x + 24) - x = 24$, and the sum of the roots is 12, so their difference is 24 ÷ 12 = 2. Add the sum and the difference: $2\sqrt{x + 24} = 14$, so $\sqrt{x + 24} = 7$ and $x = 25$. Check: $\sqrt{49} + \sqrt{25} = 7 + 5 = 12$.

Componendo and dividendo. If $\frac{m}{n} = \frac{p}{q}$, then

$$\frac{m + n}{m - n} = \frac{p + q}{p - q}$$

This is made for equations of the form $\frac{\sqrt{A} + \sqrt{B}}{\sqrt{A} - \sqrt{B}} = k$. Apply it once and the two roots separate:

$$\frac{2\sqrt{A}}{2\sqrt{B}} = \frac{k + 1}{k - 1} \;\Rightarrow\; \frac{A}{B} = \left(\frac{k + 1}{k - 1}\right)^2$$

Example: $\frac{\sqrt{x + 6} + \sqrt{x - 6}}{\sqrt{x + 6} - \sqrt{x - 6}} = 3$. Then $\frac{\sqrt{x + 6}}{\sqrt{x - 6}} = \frac{4}{2} = 2$, so $x + 6 = 4(x - 6)$, which gives $3x = 30$ and $x = 10$. Check: $\sqrt{16} = 4$ and $\sqrt{4} = 2$, and $\frac{4 + 2}{4 - 2} = 3$.

A ready formula. For the form $\frac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = k$, doing componendo and dividendo a second time gives

$$x = \frac{2ak}{k^2 + 1}$$

With $a = 13$ and $k = 5$: $x = \frac{2 \times 13 \times 5}{26} = 5$. Check: $\sqrt{18} = 3\sqrt{2}$ and $\sqrt{8} = 2\sqrt{2}$, and $\frac{5\sqrt{2}}{\sqrt{2}} = 5$.

[Tier 2] The question turned round. Sometimes $x$ is given as such a fraction and you need an equation in $x$. If $x = \frac{\sqrt{a + 2b} + \sqrt{a - 2b}}{\sqrt{a + 2b} - \sqrt{a - 2b}}$, then

$$bx^2 - ax + b = 0$$

For $x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$, $a + 2b = 5$ and $a - 2b = 3$ give $a = 4$ and $b = \frac{1}{2}$, so $\frac{1}{2}x^2 - 4x + \frac{1}{2} = 0$, which is $x^2 - 8x + 1 = 0$. That fits what we found earlier: this $x$ is $4 + \sqrt{15}$, and $x + \frac{1}{x} = 8$.

With options, putting each one back into a surd equation is often faster than solving it. Start with the option that makes the numbers under the roots perfect squares.

Using Given Root Values, and the Usual Traps

Some questions give a value like $\sqrt{3} = 1.732$ and want a decimal answer. Keep these handy anyway, because they also settle comparisons and rough checks:

$\sqrt{2}$	$\sqrt{3}$	$\sqrt{5}$	$\sqrt{6}$	$\sqrt{7}$	$\sqrt{8}$	$\sqrt{10}$
1.414	1.732	2.236	2.449	2.646	2.828	3.162

Rationalise before you substitute. Dividing by a long decimal is slow. Move the surd to the top first.

• Given $\sqrt{6} = 2.449$: $\frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6} \approx 0.408$, and $\sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{6}}{2} = 1.2245$.
• Given $\sqrt{5} = 2.236$: $\frac{\sqrt{5} + 1}{\sqrt{5} - 1} = \frac{(\sqrt{5} + 1)^2}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2} = 2.618$.

Traps that cost marks.

• $\sqrt{a} + \sqrt{b} \neq \sqrt{a + b}$, and $(\sqrt{a} + \sqrt{b})^2 \neq a + b$. The middle term $2\sqrt{ab}$ is the one people drop.
• A square root is never negative. $\sqrt{(2 - \sqrt{5})^2}$ is $\sqrt{5} - 2$, not $2 - \sqrt{5}$, because $2 - \sqrt{5}$ is below zero. The same idea puts the bigger number first in $\sqrt{a - 2\sqrt{b}}$.
• When you rationalise $\frac{1}{\sqrt{a} - \sqrt{b}}$, the bottom becomes $a - b$, in that order. Writing $b - a$ flips the sign of the answer.
• In comparisons of $\sqrt{a} + \sqrt{b}$, a bigger total under the roots does not always mean a bigger sum. Check the totals first, then the products.
• In telescoping sums the first term is often not $\sqrt{1}$. Read where the chain starts.
• After squaring a surd equation, put the answer back in. False answers creep in.
• $\sqrt{2} \times \sqrt[3]{2}$ is not $\sqrt{4}$ or $\sqrt[5]{4}$. Make the orders the same first: $\sqrt[6]{8} \times \sqrt[6]{4} = \sqrt[6]{32}$.

[SSC Tip] A rough decimal check on any surd answer takes ten seconds. Say you simplified $\frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}$ to $\frac{5 + \sqrt{21}}{2}$. That is about (5 + 4.58) ÷ 2 ≈ 4.8. The original is about 4.38 ÷ 0.91 ≈ 4.8 too, so the working is right. If the two rough values don't match, look for a dropped term or a flipped sign.

Solved Examples

Question 1

Simplify $\sqrt{98} + \sqrt{50} - \sqrt{32}$.

Answer. I break each one into a perfect square times 2: $\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$, $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$ and $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$. Now they are all like surds, so I just work with the numbers in front: 7 + 5 − 4 = 8. The answer is $8\sqrt{2}$.

Question 2

Which of these is not a like surd with the other three? (a) $\sqrt{45}$ (b) $\sqrt{80}$ (c) $\sqrt{90}$ (d) $\sqrt{125}$

Answer. I simplify all four. $\sqrt{45} = 3\sqrt{5}$, $\sqrt{80} = 4\sqrt{5}$ and $\sqrt{125} = 5\sqrt{5}$, but $\sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}$. Three of them are multiples of $\sqrt{5}$ and one is a multiple of $\sqrt{10}$. The answer is (c) $\sqrt{90}$.

Question 3

Write $\sqrt{3} \times \sqrt[3]{2}$ as a single surd.

Answer. The orders are 2 and 3, so I bring both to order 6. $\sqrt{3} = \sqrt[6]{3^3} = \sqrt[6]{27}$ and $\sqrt[3]{2} = \sqrt[6]{2^2} = \sqrt[6]{4}$. Now both are sixth roots, so I multiply under the root: 27 × 4 = 108. The product is $\sqrt[6]{108}$.

Question 4

If $\sqrt{3} = 1.732$, find the value of $\frac{\sqrt{3} - 1}{\sqrt{3} + 1}$.

Answer. Putting 1.732 straight in means dividing 0.732 by 2.732, which is slow. So I rationalise first, multiplying top and bottom by $\sqrt{3} - 1$. The top becomes $(\sqrt{3} - 1)^2 = 3 - 2\sqrt{3} + 1 = 4 - 2\sqrt{3}$ and the bottom becomes 3 − 1 = 2. So the value is $2 - \sqrt{3}$ = 2 − 1.732 = 0.268.

Question 5

Simplify $\frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}$.

Answer. I multiply top and bottom by the conjugate $\sqrt{7} + \sqrt{3}$. The bottom becomes 7 − 3 = 4. The top becomes $(\sqrt{7} + \sqrt{3})^2 = 7 + 3 + 2\sqrt{21} = 10 + 2\sqrt{21}$. So the fraction is $\frac{10 + 2\sqrt{21}}{4}$, and cancelling 2 gives $\frac{5 + \sqrt{21}}{2}$.

Question 6

If $\frac{\sqrt{6} + \sqrt{5}}{\sqrt{6} - \sqrt{5}} = a + b\sqrt{30}$, where $a$ and $b$ are rational, find $a + b$.

Answer. Here 6 − 5 = 1, so after multiplying by the conjugate the bottom is just 1. The top is $(\sqrt{6} + \sqrt{5})^2 = 6 + 5 + 2\sqrt{30} = 11 + 2\sqrt{30}$. So $11 + 2\sqrt{30} = a + b\sqrt{30}$. Matching the plain parts and the $\sqrt{30}$ parts gives $a = 11$ and $b = 2$, so $a + b$ = 13.

Question 7

Rationalise the denominator of $\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{5}}$.

Answer. I group $\sqrt{2} + \sqrt{3}$ together, because 2 + 3 = 5 will cancel the 5 that comes from $(\sqrt{5})^2$. So I multiply top and bottom by $(\sqrt{2} + \sqrt{3}) - \sqrt{5}$. The bottom becomes

$$(\sqrt{2} + \sqrt{3})^2 - 5 = 5 + 2\sqrt{6} - 5 = 2\sqrt{6}$$

That leaves $\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{2\sqrt{6}}$. One more step: multiply top and bottom by $\sqrt{6}$. The top becomes $\sqrt{12} + \sqrt{18} - \sqrt{30} = 2\sqrt{3} + 3\sqrt{2} - \sqrt{30}$ and the bottom becomes 12. The answer is $\frac{3\sqrt{2} + 2\sqrt{3} - \sqrt{30}}{12}$.

Question 8 [Tier 2]

Rationalise $\frac{1}{\sqrt[3]{9} + \sqrt[3]{3} + 1}$.

Answer. Let $t = \sqrt[3]{3}$. Then $\sqrt[3]{9} = t^2$, so the bottom is $t^2 + t + 1$. I know $(t - 1)(t^2 + t + 1) = t^3 - 1$, and $t^3 = 3$. So if I multiply top and bottom by $t - 1$, the bottom becomes 3 − 1 = 2. The answer is $\frac{\sqrt[3]{3} - 1}{2}$.

Question 9

If $x = \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}}$ and $y = \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}}$, find $x^2 + xy + y^2$.

Answer. $x$ and $y$ are reciprocals, so $xy = 1$. For $x$, the bottom after rationalising is 7 − 5 = 2 and the top is $(\sqrt{7} + \sqrt{5})^2 = 12 + 2\sqrt{35}$, so $x = 6 + \sqrt{35}$ and $y = 6 - \sqrt{35}$. That gives $x + y = 12$. Then $x^2 + y^2 = (x + y)^2 - 2xy$ = 144 − 2 = 142, and adding $xy = 1$ gives 143.

Question 10

If $x = 7 - 4\sqrt{3}$, find $\sqrt{x} + \frac{1}{\sqrt{x}}$.

Answer. First I write $4\sqrt{3}$ as $2\sqrt{12}$. Two numbers that add to 7 and multiply to 12 are 4 and 3, so $x = (\sqrt{4} - \sqrt{3})^2 = (2 - \sqrt{3})^2$, and $\sqrt{x} = 2 - \sqrt{3}$ (2 is bigger than $\sqrt{3}$, so this is positive). Since $(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$, I get $\frac{1}{\sqrt{x}} = 2 + \sqrt{3}$. Adding the two, the $\sqrt{3}$ parts cancel, and the answer is 4.

Question 11

If $x = \sqrt{5} + 2$, find the value of $x^3 - 4x^2 - x + 7$.

Answer. Cubing $\sqrt{5} + 2$ is a lot of work, so I find a simple equation for $x$ first. $x - 2 = \sqrt{5}$, and squaring gives $x^2 - 4x + 4 = 5$, so $x^2 - 4x - 1 = 0$. Now I notice that $x^3 - 4x^2 - x = x(x^2 - 4x - 1)$, which is $x \times 0 = 0$. So the whole expression is 0 + 7 = 7.

Question 12

Find $\sqrt{9 + 2\sqrt{14}}$.

Answer. I need two numbers that add to 9 and multiply to 14. That's 7 and 2. So $9 + 2\sqrt{14} = (\sqrt{7} + \sqrt{2})^2$, and the square root is $\sqrt{7} + \sqrt{2}$. Quick check: $(\sqrt{7} + \sqrt{2})^2 = 7 + 2 + 2\sqrt{14}$.

Question 13

Find the square root of $21 - 4\sqrt{5}$.

Answer. The method needs a 2 in front, so I rewrite $4\sqrt{5} = 2 \times 2\sqrt{5} = 2\sqrt{20}$. Now I need two numbers that add to 21 and multiply to 20, which are 20 and 1. With a minus sign the bigger one goes first, so the root is $\sqrt{20} - 1$ = $2\sqrt{5} - 1$. Check: $(2\sqrt{5} - 1)^2 = 20 - 4\sqrt{5} + 1 = 21 - 4\sqrt{5}$.

Question 14 [Tier 2]

Find $\sqrt{5 + \sqrt{21}} + \sqrt{5 - \sqrt{21}}$.

Answer. $\sqrt{21}$ isn't 2 times the root of a whole number, so I multiply and divide inside by 2: $5 + \sqrt{21} = \frac{10 + 2\sqrt{21}}{2}$. Since 7 + 3 = 10 and 7 × 3 = 21, the top is $(\sqrt{7} + \sqrt{3})^2$. So

$$\sqrt{5 + \sqrt{21}} = \frac{\sqrt{7} + \sqrt{3}}{\sqrt{2}} \qquad \sqrt{5 - \sqrt{21}} = \frac{\sqrt{7} - \sqrt{3}}{\sqrt{2}}$$

Adding them, the $\sqrt{3}$ parts cancel and I get $\frac{2\sqrt{7}}{\sqrt{2}} = \sqrt{2} \times \sqrt{7}$ = $\sqrt{14}$.

A faster way is to square the whole thing. The square is $(5 + \sqrt{21}) + (5 - \sqrt{21}) + 2\sqrt{25 - 21} = 10 + 4 = 14$, so the sum is $\sqrt{14}$.

Question 15

Which is greater, $\sqrt{10} + \sqrt{3}$ or $\sqrt{8} + \sqrt{5}$?

Answer. The numbers under the roots add to 13 in both. So I only compare the products: 10 × 3 = 30 and 8 × 5 = 40. The pair 8 and 5 is closer together and has the bigger product, so $\sqrt{8} + \sqrt{5}$ is greater. Squaring shows the same thing: $13 + 2\sqrt{30}$ against $13 + 2\sqrt{40}$.

Question 16

Which of these is the largest? (a) $\sqrt{13} - \sqrt{11}$ (b) $\sqrt{11} - 3$ (c) $\sqrt{7} - \sqrt{5}$ (d) $\sqrt{17} - \sqrt{15}$

Answer. 3 is $\sqrt{9}$, so every option is a difference of two roots whose numbers are 2 apart. Rationalising, each one equals 2 divided by the sum of its two roots. The smallest numbers give the smallest sum at the bottom, and so the biggest value. That's (c) $\sqrt{7} - \sqrt{5}$.

Question 17

Arrange in ascending order: $3\sqrt{5}$, $4\sqrt{3}$, $2\sqrt{11}$, $5\sqrt{2}$.

Answer. I take each outside number back inside the root: $3\sqrt{5} = \sqrt{45}$, $4\sqrt{3} = \sqrt{48}$, $2\sqrt{11} = \sqrt{44}$ and $5\sqrt{2} = \sqrt{50}$. Now it's just 44 < 45 < 48 < 50, so the order is $2\sqrt{11}, 3\sqrt{5}, 4\sqrt{3}, 5\sqrt{2}$.

Question 18

Find $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \frac{1}{\sqrt{80} + \sqrt{81}}$.

Answer. In every term the two numbers under the roots differ by 1, so rationalising gives a bottom of 1: $\frac{1}{\sqrt{n} + \sqrt{n + 1}} = \sqrt{n + 1} - \sqrt{n}$. Written out, the sum is $(\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{81} - \sqrt{80})$, and everything in the middle cancels. What's left is $\sqrt{81} - 1$ = 9 − 1 = 8.

Question 19

Find $\frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \cdots + \frac{1}{\sqrt{47} + \sqrt{49}}$.

Answer. This time the numbers go up by 2, so each term rationalises to $\frac{\sqrt{n + 2} - \sqrt{n}}{2}$. The middle terms still cancel, leaving $\frac{\sqrt{49} - \sqrt{3}}{2}$ = $\frac{7 - \sqrt{3}}{2}$.

Question 20

Find the value of

$$\frac{1}{4 - \sqrt{15}} - \frac{1}{\sqrt{15} - \sqrt{14}} + \frac{1}{\sqrt{14} - \sqrt{13}} - \frac{1}{\sqrt{13} - \sqrt{12}} + \frac{1}{\sqrt{12} - \sqrt{11}} - \frac{1}{\sqrt{11} - \sqrt{10}} + \frac{1}{\sqrt{10} - 3}$$

Answer. Each bottom is a difference of roots of numbers 1 apart (4 is $\sqrt{16}$ and 3 is $\sqrt{9}$), so each term turns into a sum: $\frac{1}{\sqrt{n + 1} - \sqrt{n}} = \sqrt{n + 1} + \sqrt{n}$. The expression becomes $(4 + \sqrt{15}) - (\sqrt{15} + \sqrt{14}) + (\sqrt{14} + \sqrt{13}) - \cdots + (\sqrt{10} + 3)$. Every middle root shows up once with + and once with −, so they all cancel. There are 7 terms, an odd number, so both ends stay positive: 4 + 3 = 7.

Question 21

Solve $\frac{\sqrt{10 + x} + \sqrt{10 - x}}{\sqrt{10 + x} - \sqrt{10 - x}} = 3$.

Answer. I use componendo and dividendo. On the left, (top + bottom) ÷ (top − bottom) gives $\frac{2\sqrt{10 + x}}{2\sqrt{10 - x}}$. On the right it gives $\frac{3 + 1}{3 - 1} = 2$. So $\sqrt{10 + x} = 2\sqrt{10 - x}$. Squaring, 10 + x = 4(10 − x) = 40 − 4x, so 5x = 30 and x = 6. Check: $\sqrt{16} = 4$ and $\sqrt{4} = 2$, and $\frac{4 + 2}{4 - 2} = 3$. The ready formula $x = \frac{2ak}{k^2 + 1} = \frac{2 \times 10 \times 3}{10}$ gives the same 6.

Question 22

Solve $\sqrt{x + 12} + \sqrt{x} = 6$.

Answer. The two insides differ by 12, and $(\sqrt{x + 12} + \sqrt{x})(\sqrt{x + 12} - \sqrt{x}) = 12$. The sum is 6, so the difference must be 12 ÷ 6 = 2. Adding the sum and the difference gives $2\sqrt{x + 12} = 8$, so $\sqrt{x + 12} = 4$ and x = 4. Check: $\sqrt{16} + \sqrt{4} = 4 + 2 = 6$.