Simplification with Surds (SQA National 5 Maths): Revision Note

Surds & exact values

What is a surd?

• A surd is the square root of a non-square integer

• Using surds lets you leave answers in exact form

  • e.g. $5\sqrt{2}$  rather than $7.071067812...$

How do I do calculations with surds?

•  Multiplying surds

  • You can multiply numbers under square roots together

  • $\sqrt{3}\times \sqrt{5}=\sqrt{3\times 5}=\sqrt{15}$

• Dividing surds

  • You can divide numbers under square roots

  • $\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{21}÷\sqrt{7}=\sqrt{21÷7}=\sqrt{3}$

• Factorising surds

  • You can factorise numbers under square roots

  • $\sqrt{35}=\sqrt{5\times 7}=\sqrt{5}\times \sqrt{7}$

• Adding or subtracting surds

  • You can only add or subtract multiples of “like” surds

    • This is similar to collecting like terms when simplifying algebra

  • $3\sqrt{5}+8\sqrt{5}=11\sqrt{5}$

  • $7\sqrt{3}–4\sqrt{3}=3\sqrt{3}$

    • However $2\sqrt{3}+4\sqrt{6}$ cannot be combined in this way

  • You cannot add or subtract numbers under square roots

  • Consider $\sqrt{9}+\sqrt{4}=3+2=5$ 

    • This is not equal to $\sqrt{9+4}=\sqrt{13}=3.60555\dots$

Simplifying surds

How do I simplify surds?

• To simplify a surd, factorise the number using a square number, if possible

  • If multiple square numbers are factors, use the largest

• Use the fact that $\sqrt{ab}=\sqrt{a}\times \sqrt{b}$ and then work out any square roots of square numbers

  • E.g. $\sqrt{48}=\sqrt{16\times 3}=\sqrt{16}\times \sqrt{3}=4\times \sqrt{3}=4\sqrt{3}$

• When simplifying multiple surds, simplify each separately

  • This may produce surds which can then be collected together

    • E.g. $\sqrt{32}+\sqrt{8}$ can be rewritten as $\sqrt{16}\sqrt{2}+\sqrt{4}\sqrt{2}$

    • This simplifies to $4\sqrt{2}+2\sqrt{2}$

    • These surds can then be collected together

    • $6\sqrt{2}$

• You may have to expand brackets containing surds

  • This can be done in the same way as expanding brackets algebraically, and then simplifying where possible

  • The property ${\left(\sqrt{a}\right)}^2=a$ can be used to simplify the expression, once expanded

  • E.g. $\sqrt{7}\left(\sqrt{7}-\sqrt{3}\right)$ expands to ${\left(\sqrt{7}\right)}^2-\sqrt{7}\times \sqrt{3}$

    • This simplifies to $7-\sqrt{21}$

  • Or $\left(\sqrt{6}-2\right)\left(\sqrt{6}+4\right)$ expands to ${\left(\sqrt{6}\right)}^2+4\sqrt{6}-2\sqrt{6}-8$

    • This simplifies to $6+2\sqrt{6}-8$ which gives «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«semantics»«mrow»«mo»-«/mo»«mn»2«/mn»«mo»+«/mo»«mn»2«/mn»«msqrt»«mn»6«/mn»«/msqrt»«/mrow»«annotation encoding=¨application/vnd.wiris.mtweb-params+json¨»{¨fontFamily¨:¨Times New Roman¨,¨fontSize¨:¨18¨,¨autoformat¨:true,¨toolbar¨:¨«toolbar ref=`general`»«tab ref=`general`»«removeItem ref=`setColor`/»«removeItem ref=`bold`/»«removeItem ref=`italic`/»«removeItem ref=`autoItalic`/»«removeItem ref=`setUnicode`/»«removeItem ref=`mtext` /»«removeItem ref=`rtl`/»«removeItem ref=`forceLigature`/»«removeItem ref=`setFontFamily` /»«removeItem ref=`setFontSize`/»«/tab»«/toolbar»¨}«/annotation»«/semantics»«/math»