Simplifying Surds

Surds & Exact Valuesvideo

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 simplified

  • 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

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 a factor, 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 double brackets containing surds

  • This can be done in the same way as multiplying out double brackets algebraically, and then simplifying

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

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

    • This simplifies to $6+2\sqrt{6}-8$ which gives $-2+2\sqrt{6}$