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Show that \( rac{2}{\sqrt{12}-\sqrt{8}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\), where a and b are integers. - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 2

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Show-that--\(-rac{2}{\sqrt{12}-\sqrt{8}}\)-can-be-written-in-the-form-\(\sqrt{a}+\sqrt{b}\),-where-a-and-b-are-integers.-Edexcel-A-Level Maths Pure-Question 5-2012-Paper 2.png

Show that \( rac{2}{\sqrt{12}-\sqrt{8}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\), where a and b are integers.

Worked Solution & Example Answer:Show that \( rac{2}{\sqrt{12}-\sqrt{8}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\), where a and b are integers. - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 2

Step 1

Correct Method to Rationalise the Denominator

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Answer

To simplify (\frac{2}{\sqrt{12}-\sqrt{8}}), we multiply the numerator and the denominator by the conjugate of the denominator, which is (\sqrt{12} + \sqrt{8}):

2(12+8)(12−8)(12+8)\frac{2(\sqrt{12} + \sqrt{8})}{(\sqrt{12} - \sqrt{8})(\sqrt{12} + \sqrt{8})}

The denominator simplifies to:

(12)2−(8)2=12−8=4.(\sqrt{12})^2 - (\sqrt{8})^2 = 12 - 8 = 4.

Thus, we get:

2(12+8)4=12+82.\frac{2(\sqrt{12} + \sqrt{8})}{4} = \frac{\sqrt{12} + \sqrt{8}}{2}.

Step 2

Simplifying Further

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Answer

Now, simplify (\frac{\sqrt{12} + \sqrt{8}}{2}):

=4⋅3+4⋅22=23+222=3+2.= \frac{\sqrt{4 \cdot 3} + \sqrt{4 \cdot 2}}{2} = \frac{2\sqrt{3} + 2\sqrt{2}}{2} = \sqrt{3} + \sqrt{2}.

This shows that:

212−8=3+2,\frac{2}{\sqrt{12}-\sqrt{8}} = \sqrt{3} + \sqrt{2},

where (a = 3) and (b = 2).

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