Bradley's Maths

Solution

This question is an early-exam non-calculator question, so GCSE Paper 1 around question 5 to 9 and IGCSE Paper 2 in about the same place. Whilst simplifying surds is a relatively low level concept, it nevertheless seems to cause a lot of problems for students and marks are frequently lost in this type of question.

Beware! Some surds don't factorise. And please note that according to the syllabus you are expected to know square numbers up to $15^2=225$ and this question is a way of testing that knowledge.

a) To simplify a surd we need to find the largest square number which is a factor of the number under the square root symbol.

\begin{align*} 75 &= 25\times 3\\ \sqrt{75} &= \sqrt{25\times3}\\ &= \sqrt{25}\sqrt{3}\\ &= 5\sqrt{3} \end{align*}

Final Answer:

$$5\sqrt{3}$$

b) It isn't always obvious that a surd has a square number factor, but in this case $1+8+9=18$ which is a multiple of 9 so 189 is also a multiple of 9.

\begin{align*} 189 &= 9\times 21\\ \sqrt{189} &= \sqrt{9\times21}\\ &= \sqrt{9}\sqrt{21}\\ &= 3\sqrt{21} \end{align*}

Final Answer:

$$3\sqrt21$$

c) In order to add surds they need to be in the form $a\sqrt{b}$ with $b$ being the same for each surd

\begin{align*} \sqrt{72} &= \sqrt{36\times2}\\ &=6\sqrt{2}\\ \sqrt{98} &= \sqrt{49\times2}\\ &=7\sqrt{2}\\ \sqrt{72} + \sqrt{98} &= 6\sqrt{2} + 7\sqrt{2}\\ &= 13\sqrt{2} \end{align*}

Final Answer:

$$13\sqrt{2}$$

d) Subtraction works in the same way as addition.

\begin{align*} \sqrt{150} &= \sqrt{25\times6}\\ &=5\sqrt{6}\\ \sqrt{54} &= \sqrt{9\times6}\\ &=3\sqrt{6}\\ \sqrt{150} - \sqrt{54} &= 5\sqrt{6} - 3\sqrt{6}\\ &= 2\sqrt{6} \end{align*}

Final Answer:

$$2\sqrt{6}$$