Year 10AC9M10A01

Simplifying and Operating with Surds

Expand, factorise and simplify expressions and solve equations algebraically, applying exponent laws involving products, quotients and powers of variables, and the distributive property.

A surd is an exact value. Leaving it as a root keeps the answer exact, and simplifying makes surds addable.

Method

Split off the largest square factor

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

Use the biggest perfect square you can find.

Only like surds add

3\sqrt{2} + 5\sqrt{2} = 8\sqrt{2}

Treat $\sqrt{2}$ like a letter.

Roots do not add

\sqrt{9} + \sqrt{16} \neq \sqrt{25}

3 + 4 = 7, not 5.

Rationalise by multiplying by the surd over itself

\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}

Clears the root from the bottom.

Worked example

e.g.

\text{Simplify } \sqrt{72} + \sqrt{50} \text{ and rationalise } \frac{6}{\sqrt{3}}.

Find the largest square factor of 72. $72 = 36 \times 2$
Split and simplify. $\sqrt{72} = 6\sqrt{2}$
Do the same for 50. $50 = 25 \times 2 \ \Rightarrow\ 5\sqrt{2}$
Now they are like surds, so add them. $6\sqrt{2} + 5\sqrt{2} = 11\sqrt{2}$
To rationalise, multiply top and bottom by the root. $\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3}$
Simplify. $= 2\sqrt{3}$

Practice

Simplify first. Surds can only be added once they match.

\sqrt{48}

Answer

4\sqrt{3}

\sqrt{200}

Answer

10\sqrt{2}

2\sqrt{5} + 7\sqrt{5}

Answer

9\sqrt{5}

\sqrt{12} + \sqrt{27}

Answer

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

\frac{10}{\sqrt{5}}

Answer

2\sqrt{5}

\text{Why is } \sqrt{2} + \sqrt{3} \text{ not } \sqrt{5}?

Answer

\text{Roots do not add. Check numerically: } 3{,}15 \neq 2{,}24.