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GCSE Edexcel Maths › Rationalising Surds (GCSE Maths)

Rationalising Surds (GCSE Maths)

Skill Check

Q1
Question 1

Rationalise the denominator and simplify $\frac{8}{3\sqrt{2}}$.

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SOLUTION

Step 1  ·  Multiply the numerator and denominator by $\sqrt{2}$

To rationalise the denominator, we multiply both the top and bottom by $\sqrt{2}$:

$$\frac{8}{3\sqrt{2}} = \frac{8 \times \sqrt{2}}{3\sqrt{2} \times \sqrt{2}}$$
$$= \frac{8\sqrt{2}}{3 \times 2}$$
$$= \frac{8\sqrt{2}}{6}$$

Step 2  ·  Simplify the fraction

Divide both the numerator and the denominator by their highest common factor, which is $2$:

$$\frac{8\sqrt{2}}{6} = \frac{4\sqrt{2}}{3}$$

Final Answer

$\dfrac{4\sqrt{2}}{3}$

Q2
Question 2

Rationalise the denominator and simplify $\frac{10}{\sqrt{20}}$.

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SOLUTION

Step 1  ·  Simplify the denominator

First, simplify $\sqrt{20}$ by finding its largest square factor ($\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$):

$$\frac{10}{\sqrt{20}} = \frac{10}{2\sqrt{5}} = \frac{5}{\sqrt{5}}$$

Step 2  ·  Rationalise the denominator

Multiply the top and bottom by $\sqrt{5}$:

$$\frac{5}{\sqrt{5}} = \frac{5 \times \sqrt{5}}{\sqrt{5} \times \sqrt{5}}$$
$$\frac{5\sqrt{5}}{5} = \sqrt{5}$$

Final Answer

$\sqrt{5}$

Q3
Question 3

Rationalise the denominator of $\frac{\sqrt{3}}{\sqrt{7}}$.

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SOLUTION

Step 1  ·  Multiply the numerator and denominator by $\sqrt{7}$

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

Final Answer

$\dfrac{\sqrt{21}}{7}$

Q4
Question 4

Rationalise the denominator of $\frac{15}{\sqrt{5}}$.

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SOLUTION

Step 1  ·  Rationalise the denominator

Multiply both the numerator and denominator by $\sqrt{5}$:

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

Final Answer

$3\sqrt{5}$

Q5
Question 5

Rationalise the denominator of $\frac{6}{\sqrt{3}}$.

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SOLUTION

Step 1  ·  Rationalise the denominator

Multiply both the numerator and denominator by $\sqrt{3}$:

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

Final Answer

$2\sqrt{3}$

Problem Solving

Q6
Question 6

Rationalise the denominator of $\frac{3\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}$.

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SOLUTION

Step 1  ·  Multiply top and bottom by the conjugate

Multiply the numerator and denominator by $(\sqrt{5} - \sqrt{2})$:

$$\frac{3\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} = \frac{(3\sqrt{5} - \sqrt{2})(\sqrt{5} - \sqrt{2})}{(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})}$$

Step 2  ·  Expand the numerator

$$(3\sqrt{5} - \sqrt{2})(\sqrt{5} - \sqrt{2}) = 15 - 3\sqrt{10} - \sqrt{10} + 2 = 17 - 4\sqrt{10}$$

Step 3  ·  Expand the denominator

$$(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = 5 - 2 = 3$$

Step 4  ·  Write the final fraction

Combine the simplified numerator and denominator:

Final Answer

$\frac{17 - 4\sqrt{10}}{3}$

Q7
Question 7

Rationalise the denominator and simplify $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$.

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SOLUTION

Step 1  ·  Rationalise the denominator

Multiply the numerator and denominator by $(\sqrt{3} + 1)$:

$$\frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}$$

Step 2  ·  Expand the numerator

$$(\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}$$

Step 3  ·  Expand the denominator

$$(\sqrt{3} - 1)(\sqrt{3} + 1) = 3 - 1 = 2$$

Step 4  ·  Simplify the fraction

Divide each term in the numerator by $2$:

$$\frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

Final Answer

$2 + \sqrt{3}$

Q8
Question 8

Show that $\frac{3 + \sqrt{2}}{3 - \sqrt{2}}$ can be written in the form $a + b\sqrt{2}$, where a and b are fractions.

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SOLUTION

Step 1  ·  Rationalise the denominator

Multiply the numerator and denominator by $(3 + \sqrt{2})$:

$$\frac{3 + \sqrt{2}}{3 - \sqrt{2}} = \frac{(3 + \sqrt{2})(3 + \sqrt{2})}{(3 - \sqrt{2})(3 + \sqrt{2})}$$

Step 2  ·  Expand the numerator and denominator

Expand the numerator:

$$(3 + \sqrt{2})^2 = 9 + 3\sqrt{2} + 3\sqrt{2} + 2 = 11 + 6\sqrt{2}$$

Expand the denominator:

$$(3 - \sqrt{2})(3 + \sqrt{2}) = 9 - 2 = 7$$

Step 3  ·  Separate into distinct fractions

Divide both terms to form two distinct fractions:

$$\frac{11 + 6\sqrt{2}}{7} = \frac{11}{7} + \frac{6}{7}\sqrt{2}$$

Final Answer

$\frac{11}{7} + \frac{6}{7}\sqrt{2}$

Q9
Question 9

Rationalise the denominator of $\frac{14}{5 - \sqrt{2}}$.

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SOLUTION

Step 1  ·  Multiply by the conjugate

Multiply the numerator and denominator by the conjugate $(5 + \sqrt{2})$:

$$\frac{14}{5 - \sqrt{2}} = \frac{14(5 + \sqrt{2})}{(5 - \sqrt{2})(5 + \sqrt{2})}$$

Step 2  ·  Expand the denominator

Calculate the denominator as a difference of two squares:

$$(5 - \sqrt{2})(5 + \sqrt{2}) = 5^2 - (\sqrt{2})^2 = 25 - 2 = 23$$

Step 3  ·  Write the final fraction

Combine the simplified numerator and denominator:

Final Answer

$\frac{70 + 14\sqrt{2}}{23}$

Q10
Question 10

Rationalise the denominator and simplify $\frac{4}{3 + \sqrt{5}}$.

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SOLUTION

Step 1  ·  Multiply by the conjugate

Multiply the numerator and denominator by the conjugate $(3 - \sqrt{5})$:

$$\frac{4}{3 + \sqrt{5}} = \frac{4(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}$$

Step 2  ·  Expand the denominator

Expand the denominator using the difference of two squares:

$$(3 + \sqrt{5})(3 - \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4$$

Step 3  ·  Simplify the expression

Divide the numerator by the denominator:

$$\frac{4(3 - \sqrt{5})}{4} = 3 - \sqrt{5}$$

Final Answer

$3 - \sqrt{5}$

Exam-Style Questions

Q11
Question 11

Show that $\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}}$ can be written in the form $a - b\sqrt{6}$, where $a$ and $b$ are integers.

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SOLUTION

Step 1  ·  Multiply by the conjugate

Multiply top and bottom by $(3\sqrt{2} - 2\sqrt{3})$:

$$\frac{3\sqrt{2} - 2\sqrt{3}}{3\sqrt{2} + 2\sqrt{3}} = \frac{(3\sqrt{2} - 2\sqrt{3})^2}{(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} - 2\sqrt{3})}$$

Step 2  ·  Expand the numerator

$$(3\sqrt{2} - 2\sqrt{3})^2 = (3\sqrt{2})^2 - 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2$$
$$= 18 - 12\sqrt{6} + 12 = 30 - 12\sqrt{6}$$

Step 3  ·  Expand the denominator

$$(3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6$$

Step 4  ·  Simplify the fraction

Divide each term in the numerator by the denominator:

$$\frac{30 - 12\sqrt{6}}{6} = 5 - 2\sqrt{6}$$

Final Answer

$5 - 2\sqrt{6}$

Q12
Question 12

Given that x = 3 - $\sqrt{8}$, show that:

x + $\frac{1}{x}$ = 6

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SOLUTION

Step 1  ·  Simplify $x$

First, simplify $\sqrt{8}$ to $2\sqrt{2}$, so:

$$x = 3 - 2\sqrt{2}$$

Step 2  ·  Find $\dfrac{1}{x}$ by rationalising the denominator

$$\frac{1}{x} = \frac{1}{3 - 2\sqrt{2}} = \frac{3 + 2\sqrt{2}}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}$$

Expand the denominator:

$$(3 - 2\sqrt{2})(3 + 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1$$
$$\frac{1}{x} = 3 + 2\sqrt{2}$$

Step 3  ·  Calculate $x + \dfrac{1}{x}$

Add $x$ and $\frac{1}{x}$ together:

$$x + \frac{1}{x} = (3 - 2\sqrt{2}) + (3 + 2\sqrt{2}) = 6$$

Final Answer

$6$

Q13
Question 13

Rationalise the denominator of $\frac{1}{\sqrt{x + 4} - \sqrt{x}}$.

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SOLUTION

Step 1  ·  Multiply by the conjugate

Multiply top and bottom by the conjugate $(\sqrt{x + 4} + \sqrt{x})$:

$$\frac{1}{\sqrt{x + 4} - \sqrt{x}} = \frac{1 \times (\sqrt{x + 4} + \sqrt{x})}{(\sqrt{x + 4} - \sqrt{x})(\sqrt{x + 4} + \sqrt{x})}$$

Step 2  ·  Expand the denominator

Expand the denominator using the difference of two squares:

$$(\sqrt{x + 4})^2 - (\sqrt{x})^2 = (x + 4) - x = 4$$

Step 3  ·  Write the final simplified expression

Combine the numerator and the expanded denominator:

Final Answer

$\dfrac{\sqrt{x + 4} + \sqrt{x}}{4}$

Q14
Question 14

Express $\frac{5}{\sqrt{8} - \sqrt{3}} - \frac{6}{\sqrt{18}}$ in the form $a\sqrt{2} + b\sqrt{3}$, where a and b are integers.

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SOLUTION

Step 1  ·  Rationalise and simplify the first fraction

$$\frac{5}{\sqrt{8} - \sqrt{3}} = \frac{5(\sqrt{8} + \sqrt{3})}{8 - 3}$$
$$= \frac{5(2\sqrt{2} + \sqrt{3})}{5} = 2\sqrt{2} + \sqrt{3}$$

Step 2  ·  Rationalise and simplify the second fraction

$$\frac{6}{\sqrt{18}} = \frac{6}{3\sqrt{2}}$$
$$= \frac{2}{\sqrt{2}} = \frac{2\sqrt{2}}{2} = \sqrt{2}$$

Step 3  ·  Subtract the second term from the first

$$(2\sqrt{2} + \sqrt{3}) - \sqrt{2} = \sqrt{2} + \sqrt{3}$$

Final Answer

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

Q15
Question 15

Show that $\frac{1}{2 + \sqrt{3}} + \frac{1}{2 - \sqrt{3}}$ is an integer.

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SOLUTION

Step 1  ·  Combine the fractions over a common denominator

$$\frac{1}{2 + \sqrt{3}} + \frac{1}{2 - \sqrt{3}} = \frac{(2 - \sqrt{3}) + (2 + \sqrt{3})}{(2 + \sqrt{3})(2 - \sqrt{3})}$$

Step 2  ·  Simplify the numerator

$$(2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$$

Step 3  ·  Expand the denominator

$$(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$$

Step 4  ·  Evaluate the final expression

$$\frac{4}{1} = 4$$

Final Answer

$4$

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