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GCSE Edexcel Maths › Surds (GCSE Maths)

Surds (GCSE Maths)

Skill Check

Q1
Question 1

Rationalise the denominator of $\frac{14}{\sqrt{7}}$.

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SOLUTION

Step 1  ·  Rationalise the denominator

Multiply both the numerator and denominator by $\sqrt{7}$:

$$\frac{14}{\sqrt{7}} = \frac{14 \times \sqrt{7}}{\sqrt{7} \times \sqrt{7}}$$
$$\frac{14}{\sqrt{7}} = \frac{14\sqrt{7}}{7}$$

Step 2  ·  Simplify the fraction

$$\frac{14\sqrt{7}}{7} = 2\sqrt{7}$$

Final Answer

$2\sqrt{7}$

Q2
Question 2

Write $(3 - \sqrt{2})^2$ in the form $a + b\sqrt{2}$, where $a$ and $b$ are integers.

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SOLUTION

Step 1  ·  Write out as double brackets and expand

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

Step 2  ·  Simplify by collecting like terms

$$= 11 - 6\sqrt{2}$$

Final Answer

$11 - 6\sqrt{2}$

Q3
Question 3

Expand and simplify $(4 + \sqrt{5})(4 - \sqrt{5})$.

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SOLUTION

Step 1  ·  Apply difference of two squares

Using the identity $(a + b)(a - b) = a^2 - b^2$:

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

Step 2  ·  Simplify the squared terms

Evaluate $4^2$ and $(\sqrt{5})^2$:

$$(4 + \sqrt{5})(4 - \sqrt{5}) = 16 - 5$$

Final Answer

$11$

Q4
Question 4

Write $3\sqrt{32}$ in the form $k\sqrt{2}$, where $k$ is an integer.

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SOLUTION

Step 1  ·  Find the largest square factor

Find the largest square factor of $32$, which is $16$:

$$3\sqrt{32} = 3 \times \sqrt{16 \times 2}$$

Step 2  ·  Simplify the surd

Take the square root of $16$ outside the surd:

$$3\sqrt{32} = 3 \times 4\sqrt{2}$$
$$3\sqrt{32} = 12\sqrt{2}$$

Final Answer

$12\sqrt{2}$

Q5
Question 5

Write $\sqrt{75}$ in the form $k\sqrt{3}$, where $k$ is an integer.

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SOLUTION

Step 1  ·  Find the largest square factor

Find the largest square factor of $75$, which is $25$:

$$\sqrt{75} = \sqrt{25 \times 3}$$

Step 2  ·  Simplify the surd

$$\sqrt{75} = \sqrt{25} \times \sqrt{3}$$
$$\sqrt{75} = 5\sqrt{3}$$

Final Answer

$5\sqrt{3}$

Problem Solving

Q6
Question 6

Show that $\frac{4\sqrt{3} + 3}{4 + \sqrt{3}}$ can be written as $\sqrt{3}$.

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SOLUTION

Step 1  ·  Multiply by the conjugate

Multiply the numerator and denominator by the conjugate $(4 - \sqrt{3})$:

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

Step 2  ·  Expand the numerator and denominator

Expand the numerator:

$$(4\sqrt{3} + 3)(4 - \sqrt{3}) = 16\sqrt{3} - 12 + 12 - 3\sqrt{3} = 13\sqrt{3}$$

Expand the denominator:

$$(4 + \sqrt{3})(4 - \sqrt{3}) = 16 - 3 = 13$$

Step 3  ·  Divide and simplify

Divide the numerator by the denominator:

$$\frac{13\sqrt{3}}{13} = \sqrt{3}$$

Final Answer

$\sqrt{3}$

Q7
Question 7

Simplify fully $\frac{(5 + 2\sqrt{3})(5 - 2\sqrt{3})}{\sqrt{13}}$.

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SOLUTION

Step 1  ·  Expand the numerator

Using the difference of two squares:

$$(5 + 2\sqrt{3})(5 - 2\sqrt{3}) = 25 - (2\sqrt{3})^2$$
$$25 - 12 = 13$$

Step 2  ·  Rationalise the denominator

Substitute this back into the fraction and multiply numerator and denominator by $\sqrt{13}$:

$$\frac{13}{\sqrt{13}} = \frac{13 \times \sqrt{13}}{\sqrt{13} \times \sqrt{13}}$$
$$\frac{13\sqrt{13}}{13} = \sqrt{13}$$

Final Answer

$\sqrt{13}$

Q8
Question 8

Expand and simplify $(3 + \sqrt{5})^2 - (3 - \sqrt{5})^2$.

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SOLUTION

Step 1  ·  Expand the first squared bracket

Expand $(3 + \sqrt{5})^2$:

$$(3 + \sqrt{5})^2 = 9 + 6\sqrt{5} + 5 = 14 + 6\sqrt{5}$$

Step 2  ·  Expand the second squared bracket

Expand $(3 - \sqrt{5})^2$:

$$(3 - \sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5}$$

Step 3  ·  Subtract the second expansion from the first

Substitute the expanded expressions back into the original problem and simplify:

$$(14 + 6\sqrt{5}) - (14 - 6\sqrt{5}) = 14 + 6\sqrt{5} - 14 + 6\sqrt{5}$$
$$= 12\sqrt{5}$$

Final Answer

$12\sqrt{5}$

Q9
Question 9

Expand and simplify $(4 + \sqrt{3})(2 - \sqrt{3})$.

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SOLUTION

Step 1  ·  Expand the brackets using FOIL

Expand using FOIL (First, Outside, Inside, Last):

$$(4 + \sqrt{3})(2 - \sqrt{3}) = 8 - 4\sqrt{3} + 2\sqrt{3} - 3$$

Step 2  ·  Collect like terms and simplify

Collect the integer terms and the surd terms:

$$(4 + \sqrt{3})(2 - \sqrt{3}) = (8 - 3) + (-4\sqrt{3} + 2\sqrt{3})$$
$$(4 + \sqrt{3})(2 - \sqrt{3}) = 5 - 2\sqrt{3}$$

Final Answer

$5 - 2\sqrt{3}$

Q10
Question 10

Simplify $\frac{3 + \sqrt{6}}{\sqrt{3}}$.

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SOLUTION

Step 1  ·  Multiply top and bottom by $\sqrt{3}$

$$\frac{3 + \sqrt{6}}{\sqrt{3}} = \frac{(3 + \sqrt{6}) \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$$
$$\frac{3 + \sqrt{6}}{\sqrt{3}} = \frac{3\sqrt{3} + \sqrt{18}}{3}$$

Step 2  ·  Simplify the numerator

Simplify $\sqrt{18}$ to $3\sqrt{2}$:

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

Final Answer

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

Exam-Style Questions

Q11
Question 11

Show that $\frac{\sqrt{98} + \sqrt{32}}{\sqrt{18}}$ can be written as the mixed number $3\frac{2}{3}$.

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SOLUTION

Step 1  ·  Simplify each surd

Simplify each surd by finding their largest square factors:

$$\sqrt{98} = \sqrt{49 \times 2} = 7\sqrt{2}$$
$$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$$
$$\sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$$

Step 2  ·  Substitute into the fraction

Substitute these simplified surds back into the fraction:

$$\frac{7\sqrt{2} + 4\sqrt{2}}{3\sqrt{2}} = \frac{11\sqrt{2}}{3\sqrt{2}}$$

Step 3  ·  Cancel and convert to a mixed number

Cancel the $\sqrt{2}$ from top and bottom and convert to a mixed number:

$$\frac{11}{3} = 3\frac{2}{3}$$

Final Answer

$3\frac{2}{3}$

Q12
Question 12

Simplify fully $(\sqrt{3a} + \sqrt{2b})^2 - (\sqrt{3a} - \sqrt{2b})^2$.

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SOLUTION

Step 1  ·  Expand the first squared bracket

$$(\sqrt{3a} + \sqrt{2b})^2 = 3a + 2\sqrt{6ab} + 2b$$

Step 2  ·  Expand the second squared bracket

$$(\sqrt{3a} - \sqrt{2b})^2 = 3a - 2\sqrt{6ab} + 2b$$

Step 3  ·  Subtract the second expression from the first

$$(3a + 2\sqrt{6ab} + 2b) - (3a - 2\sqrt{6ab} + 2b)$$
$$= 3a + 2\sqrt{6ab} + 2b - 3a + 2\sqrt{6ab} - 2b$$
$$= 4\sqrt{6ab}$$

Final Answer

$4\sqrt{6ab}$

Q13
Question 13

Show that $\frac{2}{\frac{1}{\sqrt{3}} - 1}$ can be written as $-3 - \sqrt{3}$.

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SOLUTION

Step 1  ·  Combine the denominator into a single fraction

$$\frac{1}{\sqrt{3}} - 1 = \frac{1 - \sqrt{3}}{\sqrt{3}}$$

Step 2  ·  Rewrite the main fraction

Divide by the new denominator:

$$\frac{2}{\frac{1 - \sqrt{3}}{\sqrt{3}}} = \frac{2\sqrt{3}}{1 - \sqrt{3}}$$

Step 3  ·  Rationalise the denominator

Multiply the top and bottom by $(1 + \sqrt{3})$:

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

Final Answer

$-3 - \sqrt{3}$

Q14
Question 14

Show that $\frac{1}{\frac{1}{\sqrt{5}} + \sqrt{5}}$ can be written as $\frac{\sqrt{5}}{6}$.

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SOLUTION

Step 1  ·  Combine the terms in the denominator

First, combine the terms in the denominator into a single fraction:

$$\frac{1}{\sqrt{5}} + \sqrt{5} = \frac{1}{\sqrt{5}} + \frac{5}{\sqrt{5}} = \frac{6}{\sqrt{5}}$$

Step 2  ·  Substitute and simplify

Now substitute this back into the original fraction:

$$\frac{1}{\frac{6}{\sqrt{5}}}$$

Taking the reciprocal of $\dfrac{6}{\sqrt{5}}$ gives:

$$\frac{\sqrt{5}}{6}$$

Final Answer

$\dfrac{\sqrt{5}}{6}$

Q15
Question 15

Show that $\frac{4 + \sqrt{3}}{2 - \sqrt{3}}$ can be written in the form $a + b\sqrt{3}$, where $a$ and $b$ are integers.

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SOLUTION

Step 1  ·  Multiply by the conjugate

Multiply the top and bottom by the conjugate of the denominator, $(2 + \sqrt{3})$:

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

Step 2  ·  Expand the numerator

Expand the brackets in the numerator:

$$(4 + \sqrt{3})(2 + \sqrt{3}) = 8 + 4\sqrt{3} + 2\sqrt{3} + 3 = 11 + 6\sqrt{3}$$

Step 3  ·  Expand the denominator

Expand the brackets in the denominator using the difference of two squares:

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

Step 4  ·  Write the final expression

Combine the simplified numerator and denominator:

$$\frac{11 + 6\sqrt{3}}{1} = 11 + 6\sqrt{3}$$

Final Answer

$11 + 6\sqrt{3}$

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