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Show that \( \frac{8 + \sqrt{12}}{5 + \sqrt{3}} \) can be written in the form \( \frac{a + \sqrt{3}}{b} \), where \( a \) and \( b \) are integers. - Edexcel - GCSE Maths - Question 20 - 2021 - Paper 1
Question 20
Show that \( \frac{8 + \sqrt{12}}{5 + \sqrt{3}} \) can be written in the form \( \frac{a + \sqrt{3}}{b} \), where \( a \) and \( b \) are integers.
Worked Solution & Example Answer:Show that \( \frac{8 + \sqrt{12}}{5 + \sqrt{3}} \) can be written in the form \( \frac{a + \sqrt{3}}{b} \), where \( a \) and \( b \) are integers. - Edexcel - GCSE Maths - Question 20 - 2021 - Paper 1
Step 1
Rationalizing the denominator
Answer
To rationalize the denominator ( 5 + \sqrt{3} ), we multiply the numerator and denominator by the conjugate of the denominator, which is ( 5 - \sqrt{3} ):
[ \frac{(8 + \sqrt{12})(5 - \sqrt{3})}{(5 + \sqrt{3})(5 - \sqrt{3})} ]
Calculating the denominator:
[ (5 + \sqrt{3})(5 - \sqrt{3}) = 5^2 - (\sqrt{3})^2 = 25 - 3 = 22 ]
Calculating the numerator:
[ (8 + \sqrt{12})(5 - \sqrt{3}) = 8 \cdot 5 - 8 \cdot \sqrt{3} + \sqrt{12} \cdot 5 - \sqrt{12} \cdot \sqrt{3} ]
Simplifying further: [ = 40 - 8\sqrt{3} + 5\sqrt{12} - \sqrt{36} = 40 - 8\sqrt{3} + 10\sqrt{3} - 6 ]
Combining like terms: [ = 34 + 2\sqrt{3} ]
Thus, we have: [ \frac{34 + 2\sqrt{3}}{22} ]
Step 2
Final simplification
Answer
Now we can express this as:
[ \frac{34}{22} + \frac{2\sqrt{3}}{22} = \frac{17}{11} + \frac{\sqrt{3}}{11} ]
Therefore, combining these gives: [ \frac{17 + \sqrt{3}}{11} ]
This is in the form ( \frac{a + \sqrt{3}}{b} ) where ( a = 17 ) and ( b = 11 ).
Thus, we have shown that ( \frac{8 + \sqrt{12}}{5 + \sqrt{3}} ) can be expressed as required.