Surds: simplifying, rationalising and solving equations

A surd is an irrational root written exactly, such as $\sqrt{2}$ or $3\sqrt{5}$. Surds let us preserve exact values instead of replacing them with rounded decimals. This matters throughout A-level mathematics, especially in coordinate geometry, trigonometry, calculus and proof.

For example,

$$\sqrt{2}=1.4142135\ldots$$

The decimal is approximate, but $\sqrt{2}$ is exact.

Prerequisites

You should be able to:

• find square factors of integers
• factorise expressions and expand brackets
• use the difference of two squares
• solve linear and quadratic equations
• work confidently with fractions

Review the laws of indices if you need to connect roots with fractional powers. That lesson develops expressions such as $x^{3/2}$ and the general laws of powers. This lesson focuses specifically on exact arithmetic with irrational roots.

What counts as a surd?

A root is a surd only when its value is irrational. Therefore,

$$\sqrt{3},\quad \sqrt[3]{7},\quad 2+\sqrt{11}$$

contain surds, whereas

$$\sqrt{9}=3,\qquad \sqrt[3]{64}=4$$

do not.

Unless stated otherwise, $\sqrt{x}$ denotes the principal square root, which is non-negative. Thus,

$$\sqrt{25}=5,$$

not $\pm5$. The equation $x^2=25$ has two solutions, $x=\pm5$, but the symbol $\sqrt{25}$ names only the non-negative root.

The product and quotient rules

For $a\geq0$ and $b\geq0$,

$$\sqrt{ab}=\sqrt{a}\sqrt{b}.$$

For $a\geq0$ and $b>0$,

$$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}.$$

These rules allow a root to be split across multiplication or division. They do not allow a root to be split across addition:

$$\sqrt{a+b}\ne\sqrt{a}+\sqrt{b}$$

in general. For instance, $\sqrt{9}=3$, while $\sqrt{4}+\sqrt{5}=2+\sqrt{5}$.

Simplifying surds

To simplify $\sqrt{n}$, identify the largest square factor of $n$. If $n=k^2m$, then

$$\sqrt{n}=\sqrt{k^2m}=k\sqrt{m}.$$

The final radicand, the number inside the root, should have no square factor greater than $1$.

Example 1: simplify a square root

Simplify $\sqrt{72}$.

The largest square factor of $72$ is $36$:

$$\begin{aligned} \sqrt{72} &=\sqrt{36\times2}\\ &=\sqrt{36}\sqrt{2}\\ &=6\sqrt{2}. \end{aligned}$$

Using $72=9\times8$ would also work, but it would need another step:

$$\sqrt{72}=3\sqrt{8}=3\sqrt{4\times2}=6\sqrt{2}.$$

Example 2: simplify a coefficient and a root

Simplify $5\sqrt{180}$.

$$\begin{aligned} 5\sqrt{180} &=5\sqrt{36\times5}\\ &=5\times6\sqrt{5}\\ &=30\sqrt{5}. \end{aligned}$$

Example 3: simplify an algebraic surd

Suppose $x\geq0$. Simplify $\sqrt{48x^4}$.

$$\begin{aligned} \sqrt{48x^4} &=\sqrt{16\times3}\sqrt{x^4}\\ &=4\sqrt{3}\,x^2\\ &=4x^2\sqrt{3}. \end{aligned}$$

There is no need to assume $x\geq0$ for $\sqrt{x^4}=x^2$, because $x^2$ is already non-negative. In contrast,

$$\sqrt{x^2}=|x|,$$

not always $x$.

Adding and subtracting surds

Surds can be combined only when their simplified irrational parts match. They behave like algebraic terms:

$$a\sqrt{k}+b\sqrt{k}=(a+b)\sqrt{k}.$$

Example 4: collect like surds

Simplify $3\sqrt{5}-8\sqrt{5}+2\sqrt{5}$.

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

Example 5: simplify before collecting

Simplify $\sqrt{50}+3\sqrt{8}-\sqrt{18}$.

First put every surd into simplest form:

$$\sqrt{50}=5\sqrt{2},\qquad 3\sqrt{8}=6\sqrt{2},\qquad \sqrt{18}=3\sqrt{2}.$$

Then collect:

$$5\sqrt{2}+6\sqrt{2}-3\sqrt{2}=8\sqrt{2}.$$

The expression $\sqrt{2}+\sqrt{3}$ cannot be simplified further because the irrational parts differ.

Multiplying surds

Multiply coefficients together and roots together, then simplify.

Example 6: multiply two simple surds

$$\begin{aligned} (3\sqrt{2})(4\sqrt{6}) &=12\sqrt{12}\\ &=12\sqrt{4\times3}\\ &=24\sqrt{3}. \end{aligned}$$

Example 7: expand two brackets

Expand and simplify $(2+\sqrt{3})(5-2\sqrt{3})$.

$$\begin{aligned} (2+\sqrt{3})(5-2\sqrt{3}) &=10-4\sqrt{3}+5\sqrt{3}-2(\sqrt{3})^2\\ &=10+\sqrt{3}-6\\ &=4+\sqrt{3}. \end{aligned}$$

The identity $(\sqrt{a})^2=a$ is used in the second line.

Conjugates

The expressions

$$a+b\sqrt{c}\quad\text{and}\quad a-b\sqrt{c}$$

are conjugates. Multiplying conjugates removes the surd cross-terms:

$$\begin{aligned} (a+b\sqrt{c})(a-b\sqrt{c}) &=a^2-ab\sqrt{c}+ab\sqrt{c}-b^2c\\ &=a^2-b^2c. \end{aligned}$$

This is the difference of two squares. Conjugates are central to rationalising a denominator that contains two terms.

Example 8: evaluate a conjugate product

$$(4+\sqrt{7})(4-\sqrt{7})=4^2-(\sqrt{7})^2=16-7=9.$$

Example 9: simplify a less obvious pair

$$\begin{aligned} (3+2\sqrt{5})(3-2\sqrt{5}) &=3^2-(2\sqrt{5})^2\\ &=9-4\times5\\ &=-11. \end{aligned}$$

The rational product may be negative. Rational does not mean positive.

Rationalising a denominator

To rationalise the denominator is to rewrite a fraction so that its denominator contains no surd. The value of the fraction does not change because numerator and denominator are multiplied by the same non-zero expression.

A denominator with one surd term

If the denominator is $\sqrt{b}$, multiply by $\sqrt{b}/\sqrt{b}$.

Example 10: one square root in the denominator

$$\begin{aligned} \frac{5}{\sqrt{3}} &=\frac{5}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}\\ &=\frac{5\sqrt{3}}{3}. \end{aligned}$$

Example 11: simplify as well as rationalise

Simplify $\dfrac{6}{\sqrt{12}}$.

First, $\sqrt{12}=2\sqrt{3}$:

$$\frac{6}{\sqrt{12}}=\frac{6}{2\sqrt{3}}=\frac{3}{\sqrt{3}}.$$

Now rationalise:

$$\frac{3}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} =\frac{3\sqrt{3}}{3} =\sqrt{3}.$$

You can verify this because $6/\sqrt{12}=\sqrt{36/12}=\sqrt{3}$.

A denominator with two terms

If the denominator is $a+b\sqrt{c}$, multiply by its conjugate $a-b\sqrt{c}$. Multiplying by the surd alone will not remove both terms.

Example 12: rationalise using a conjugate

Simplify $\dfrac{1}{3+\sqrt{2}}$.

$$\begin{aligned} \frac{1}{3+\sqrt{2}} &=\frac{1}{3+\sqrt{2}}\times\frac{3-\sqrt{2}}{3-\sqrt{2}}\\ &=\frac{3-\sqrt{2}}{3^2-(\sqrt{2})^2}\\ &=\frac{3-\sqrt{2}}{9-2}\\ &=\frac{3-\sqrt{2}}{7}. \end{aligned}$$

Example 13: expand the numerator carefully

Simplify $\dfrac{2+\sqrt{3}}{2-\sqrt{3}}$.

The conjugate of the denominator is $2+\sqrt{3}$:

$$\begin{aligned} \frac{2+\sqrt{3}}{2-\sqrt{3}} &=\frac{(2+\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}\\ &=\frac{4+4\sqrt{3}+3}{4-3}\\ &=7+4\sqrt{3}. \end{aligned}$$

Notice that $(2+\sqrt{3})^2$ contains the middle term $4\sqrt{3}$.

Example 14: rationalise with coefficients

Simplify $\dfrac{4}{\sqrt{5}-2}$.

$$\begin{aligned} \frac{4}{\sqrt{5}-2} &=\frac{4(\sqrt{5}+2)}{(\sqrt{5}-2)(\sqrt{5}+2)}\\ &=\frac{4\sqrt{5}+8}{5-4}\\ &=8+4\sqrt{5}. \end{aligned}$$

The denominator becoming $1$ is not suspicious. In fact,

$$(\sqrt{5}-2)(\sqrt{5}+2)=1,$$

so $\sqrt{5}+2$ is the reciprocal of $\sqrt{5}-2$.

Solving equations involving surds

When an equation contains a square root of an expression, isolate the root before squaring. Squaring may introduce solutions that do not satisfy the original equation, so every answer must be checked.

Example 15: a linear expression under a root

Solve

$$\sqrt{2x+3}=x.$$

Since the left side is non-negative, any solution must satisfy $x\geq0$. Squaring gives

$$2x+3=x^2,$$

so

$$x^2-2x-3=0.$$

Factorise:

$$(x-3)(x+1)=0.$$

The candidates are $x=3$ and $x=-1$. Check them in the original equation:

$$x=3:\quad \sqrt{2(3)+3}=\sqrt9=3,$$

so $x=3$ works. For $x=-1$, the left side is $\sqrt1=1$, not $-1$. Therefore,

$$\boxed{x=3}.$$

The rejected value is an extraneous solution, created because squaring loses sign information.

Example 16: isolate before squaring

Solve

$$\sqrt{x+7}=x-1.$$

The right side must be non-negative, so $x\geq1$. Squaring gives

$$x+7=(x-1)^2=x^2-2x+1.$$

Hence

$$x^2-3x-6=0.$$

Using the quadratic formula,

$$x=\frac{3\pm\sqrt{9+24}}{2}=\frac{3\pm\sqrt{33}}{2}.$$

Only the positive candidate is at least $1$. The negative candidate cannot satisfy the original equation. Therefore,

$$\boxed{x=\frac{3+\sqrt{33}}{2}}.$$

Substitution confirms that this value works.

Example 17: an equation in the form $a+b\sqrt{c}$

Solve

$$x+2\sqrt{x}=15,$$

where $x\geq0$.

Let $u=\sqrt{x}$, so $x=u^2$ and $u\geq0$. Then

$$u^2+2u-15=0.$$

Factorise:

$$(u+5)(u-3)=0.$$

Since $u\geq0$, $u=3$. Therefore,

$$x=u^2=9.$$

This substitution turns an equation involving $x$ and $\sqrt{x}$ into a quadratic.

Comparing exact surd values

Avoid decimals when an exact comparison is possible.

Example 18: compare two positive surds

Which is larger, $3\sqrt{5}$ or $2\sqrt{11}$?

Both numbers are positive, so squaring preserves their order:

$$(3\sqrt{5})^2=45, \qquad (2\sqrt{11})^2=44.$$

Therefore,

$$3\sqrt{5}>2\sqrt{11}.$$

This argument is exact. A decimal comparison would be less reliable and would not prove the result unless error bounds were also controlled.

Surds in proof

Surds often appear in proofs of irrationality and exact identities.

Example 19: prove that $3+2\sqrt{5}$ is irrational

Assume, for contradiction, that $3+2\sqrt{5}$ is rational. Subtracting $3$ and dividing by $2$ would then show that

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

is rational. But $\sqrt{5}$ is irrational. This is a contradiction, so $3+2\sqrt{5}$ is irrational.

The proof depends on the rational coefficients. It would fail if the coefficient of $\sqrt{5}$ were zero.

Example 20: prove an exact reciprocal identity

Show that

$$\frac{1}{2+\sqrt{3}}=2-\sqrt{3}.$$

Rationalise the left side:

$$\begin{aligned} \frac{1}{2+\sqrt{3}} &=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}\\ &=\frac{2-\sqrt{3}}{4-3}\\ &=2-\sqrt{3}. \end{aligned}$$

Equivalently, multiply the claimed reciprocal pair:

$$(2+\sqrt{3})(2-\sqrt{3})=1.$$

Example 21: prove a surd expression is an integer

Let $a=\sqrt{3}+\sqrt{2}$. Show that

$$a^2+\frac{1}{a^2}$$

is an integer.

First,

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

Also,

$$(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=1,$$

so

$$\frac1a=\sqrt{3}-\sqrt{2}.$$

Therefore,

$$\frac{1}{a^2}=(\sqrt{3}-\sqrt{2})^2=5-2\sqrt{6}.$$

Adding gives

$$a^2+\frac1{a^2}=(5+2\sqrt6)+(5-2\sqrt6)=10,$$

which is an integer.

Exact arithmetic in geometry

Exact surds should normally remain exact until the final stage of a calculation.

Example 22: distance between two points

Find the exact distance between $A(-1,2)$ and $B(5,7)$.

$$\begin{aligned} AB &=\sqrt{(5-(-1))^2+(7-2)^2}\\ &=\sqrt{6^2+5^2}\\ &=\sqrt{61}. \end{aligned}$$

Since $61$ has no square factor greater than $1$, $\sqrt{61}$ is already simplified. A calculator approximation such as $7.81$ should be given only if the question requests it.

Example 23: an exact trigonometric length

A right-angled triangle has hypotenuse $10$ and an angle of $30^\circ$. The side adjacent to the angle is

$$10\cos30^\circ =10\left(\frac{\sqrt3}{2}\right) =5\sqrt3.$$

The exact answer preserves structure and can be used in later algebra without accumulating rounding error.

Common misconceptions

Splitting a sum inside a root

In general,

$$\sqrt{a+b}\ne\sqrt a+\sqrt b.$$

The product rule works because $(\sqrt a\sqrt b)^2=ab$. There is no corresponding identity for addition.

Combining unlike surds

$$2\sqrt3+5\sqrt2$$

cannot be written as $7\sqrt5$ or $7\sqrt6$. Only like surds combine, just as $2x+5y$ cannot be simplified when $x$ and $y$ are different variables.

Forgetting to simplify first

$\sqrt8$ and $\sqrt2$ initially look unlike, but $\sqrt8=2\sqrt2$. Thus,

$$\sqrt8+\sqrt2=3\sqrt2.$$

Using the wrong conjugate

The conjugate changes the sign between the two terms. The conjugate of $4-3\sqrt2$ is $4+3\sqrt2$, not $-4+3\sqrt2$.

Squaring a binomial incorrectly

$$(a+b\sqrt c)^2=a^2+2ab\sqrt c+b^2c.$$

The middle term is essential.

Accepting every solution after squaring

Squaring is not reversible without a sign check. Always substitute candidates into the original equation.

Replacing exact values with decimals too early

If $\sqrt2$ is rounded before further multiplication or subtraction, the final answer may lose accuracy. Retain exact surds until a decimal is explicitly required.

Check your understanding

Try these without a calculator. Give exact answers in their simplest form.

1. Simplify $\sqrt{108}$.
2. Simplify $2\sqrt{75}-\sqrt{27}+4\sqrt3$.
3. Expand and simplify $(3-\sqrt2)(4+2\sqrt2)$.
4. Rationalise $\dfrac{7}{2\sqrt5}$.
5. Rationalise $\dfrac{3}{4-\sqrt7}$.
6. Simplify $\dfrac{1+\sqrt2}{1-\sqrt2}$.
7. Solve $\sqrt{3x+4}=x$.
8. Solve $x-4\sqrt{x}+3=0$, where $x\geq0$.
9. Decide which is larger: $4\sqrt7$ or $5\sqrt5$.
10. Given $a=\sqrt5+2$, find $1/a$ in the form $p+q\sqrt5$.
11. Prove that $7-3\sqrt2$ is irrational.
12. Find the exact distance between $(2,-3)$ and $(-4,5)$.

Answers

1. $6\sqrt3$.
2. $10\sqrt3-3\sqrt3+4\sqrt3=11\sqrt3$.
3. $12+6\sqrt2-4\sqrt2-4=8+2\sqrt2$.
4. $\dfrac{7\sqrt5}{10}$.
5. $\dfrac{3(4+\sqrt7)}{16-7}=\dfrac{4+\sqrt7}{3}$.
6. $\dfrac{(1+\sqrt2)^2}{1-2}=-3-2\sqrt2$.
7. Squaring gives $x^2-3x-4=0$, so the candidates are $4$ and $-1$. Only $x=4$ satisfies the original equation.
8. Let $u=\sqrt{x}$. Then $u^2-4u+3=0$, so $u=1$ or $u=3$. Hence $x=1$ or $x=9$.
9. Squaring gives $112$ and $125$ respectively, so $5\sqrt5$ is larger.
10. $\dfrac1{\sqrt5+2}=\dfrac{\sqrt5-2}{5-4}=\sqrt5-2$.
11. If $7-3\sqrt2$ were rational, rearrangement would make $\sqrt2$ rational, which is a contradiction.
12. $\sqrt{(-4-2)^2+(5-(-3))^2}=\sqrt{36+64}=10$.

What to learn next

• Study quadratics to see surds arise naturally from the quadratic formula and discriminant.
• Study polynomials to strengthen the expansion and factorisation used in surd manipulation.
• Study coordinate geometry to apply exact roots to distances and intersections.
• Return to the laws of indices for fractional powers and general $n$th roots.