Manipulating Surds (Edexcel A-Level Mathematics): Revision Notes

Definition of Surds:

• A surd is an irrational root of a rational number, e.g., $\sqrt{2}$, $\sqrt{3}$.

Fact: $\sqrt{p}$ where p is prime is a surd.

• Surds are irrational numbers, i.e., cannot be written as a ratio of integers $\frac{a}{b}$ where a, b are integers $(a, b \in \mathbb{Z})$.

Examples:

$3x + 2x = 5x$

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

$5x - 3y$ cannot be simplified

$5\sqrt{2} - 3\sqrt{3}$ cannot be simplified

Examples:

$\sqrt{2} \times \sqrt{3} = \sqrt{6}$

$\sqrt{7} \times \sqrt{3} = \sqrt{21}$

Example: Write $\sqrt{45}$ in simplified surd form

$\sqrt{45} = \sqrt{9 \times 5} = \sqrt{9} \times \sqrt{5}$

$\sqrt{9} \times \sqrt{5} = 3 \times \sqrt{5} = 3\sqrt{5}$

Example: Write $\sqrt{42}$ in simplified surd form

$\sqrt{42} = \sqrt{6 \times 7} = \sqrt{2 \times 3 \times 7}$

$= \sqrt{2} \times \sqrt{3} \times \sqrt{7} = \sqrt{42}$

Example: Write $25^6$ as a power of $5$ :

• Notice that $25$ can be rewritten as $5^2$.

$25^6 = (5^2)^6 = 5^{12}$

• Both represent $25^6$ in two different ways.

Example: Write $64^3$ as a power of $4$ :

$(4^3)^3 = 4^9$

Example: Write $32$ as a power of $4$:

• This is more difficult as $32$ is not an integer power of $4$. Think of numbers that they both have in common and go via that number. $4^{\frac{1}{2}} = 2$

$2^5 = 32$

Using both of these facts:

$32 = 2^5 = (4^{\frac{1}{2}})^5 = 4^{\frac{5}{2}}$

Example: Write $64$ in the form $16^n$ where n is rational:

• Way 1:

$4 = 16^{\frac{1}{2}}$

$64 = 4^3 \Rightarrow 64 = (16^{\frac{1}{2}})^3 = 16^{\frac{3}{2}}$

• Way 2:

$2 = 16^{\frac{1}{4}}$

$64 = 2^6 \Rightarrow 64 = (16^{\frac{1}{4}})^6 = 16^{\frac{6}{4}} = 16^{\frac{3}{2}}$

Challenge (Q5. OCR 4721, Jun 2016, Q5)

Express the following in the form $2^p$:

1. $(2^5 \div 2^7)^3$:

$(2^5 \div 2^7)^3$

$ = (2^{-2})^3 = 2^{-6}

$

2. $5 \times 4^{\frac{2}{3}} + 3 \times 16^{\frac{1}{3}}:$

$5 \times 4^{\frac{2}{3}} + 3 \times 16^{\frac{1}{3}}$

$= 5 \times (2^2)^{\frac{2}{3}}+ 3 \times (2^4)^{\frac{1}{3}}$

$= 5 \times 2^{{\frac{4}{3}}} + 3 \times 2^{\frac{4}{3}} = 8 \times 2^{{\frac{4}{3}}}$

$

= 2^3 \times 2^{{\frac{4}{3}}} =2^{{\frac{9}{3}}} \times 2^{\frac{4}{3}} = 2^{{\frac{13}{3}}} $

Example: Solve $2^{x+3} = 4^{3x-2}$:

$\Rightarrow 2^{x+3} = (2^2)^{3x-2}$

$\Rightarrow 2^{x+3} = 2^{6x-4}$

$\Rightarrow x + 3 = 6x - 4$

$5x = 7 \Rightarrow x = \frac{7}{5}$

Example: Solve $16^{2x+7} = 8^{5-3x}$:

$\Rightarrow(2^4)^{2x+7} = (2^3)^{5-3x}$

$\Rightarrow2^{8x+28} = 2^{15-9x}$

$\Rightarrow8x + 28 = 15 - 9x$

$17x = -13 \Rightarrow x = -\frac{13}{17}$

Key Tips

1. Identify square factors: Always look for the largest square factor to simplify surds.
2. Like terms only: Only combine surds with the same radicand (the number inside the square root).
3. Use conjugates for binomials: To rationalise binomials in the denominator, multiply by the conjugate.
4. Practice: Regularly practice manipulating surds to build fluency and confidence.