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Show that \[ \frac{\sqrt{18} + \sqrt{2}}{\sqrt{8} - 2} \] can be written in the form \( \alpha(\beta + \sqrt{2}) \) where \( \alpha \) and \( \beta \) are integers. - Edexcel - GCSE Maths - Question 21 - 2018 - Paper 1

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Question 21

Show-that--\[-\frac{\sqrt{18}-+-\sqrt{2}}{\sqrt{8}---2}-\]--can-be-written-in-the-form-\(-\alpha(\beta-+-\sqrt{2})-\)-where-\(-\alpha-\)-and-\(-\beta-\)-are-integers.-Edexcel-GCSE Maths-Question 21-2018-Paper 1.png

Show that \[ \frac{\sqrt{18} + \sqrt{2}}{\sqrt{8} - 2} \] can be written in the form \( \alpha(\beta + \sqrt{2}) \) where \( \alpha \) and \( \beta \) are integers... show full transcript

Worked Solution & Example Answer:Show that \[ \frac{\sqrt{18} + \sqrt{2}}{\sqrt{8} - 2} \] can be written in the form \( \alpha(\beta + \sqrt{2}) \) where \( \alpha \) and \( \beta \) are integers. - Edexcel - GCSE Maths - Question 21 - 2018 - Paper 1

Step 1

Step 1: Simplify the Numerator

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Answer

First, simplify the numerator: 18+2=9⋅2+2=32+2=42.\sqrt{18} + \sqrt{2} = \sqrt{9 \cdot 2} + \sqrt{2} = 3\sqrt{2} + \sqrt{2} = 4\sqrt{2}.

Step 2

Step 2: Simplify the Denominator

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Answer

Next, simplify the denominator: 8−2=22−2.\sqrt{8} - 2 = 2\sqrt{2} - 2.

Step 3

Step 3: Rationalize the Denominator

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Answer

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator: 42(22+2)(22−2)(22+2).\frac{4\sqrt{2}(2\sqrt{2} + 2)}{(2\sqrt{2} - 2)(2\sqrt{2} + 2)}.

This becomes: 8+828−4=8+824=2+22.\frac{8 + 8\sqrt{2}}{8 - 4} = \frac{8 + 8\sqrt{2}}{4} = 2 + 2\sqrt{2}.

Thus, we can write it as: 2(1+2),2(1 + \sqrt{2}), where ( \alpha = 2 ) and ( \beta = 1 ).

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