Find the value of \(a+2b\) from the equation
\(\displaystyle\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=a+b\sqrt{6}\)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 9
A fraction with \(\sqrt{3}-\sqrt{2}\) in the denominator is simplified by rationalising: multiplying the top and bottom by \(\sqrt{3}+\sqrt{2}\), because \((\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=3-2=1\). Then \(a\) and \(b\) are found by comparing the rational parts and the \(\sqrt{6}\) parts of the two sides.
Rationalised, the denominator becomes 1: \(\displaystyle \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}+\sqrt{2})^2}{3-2}=(\sqrt{3}+\sqrt{2})^2\)
Squaring the numerator: \((\sqrt{3}+\sqrt{2})^2=3+2+2\sqrt{6}=5+2\sqrt{6}\)
Comparing \(5+2\sqrt{6}\) with \(a+b\sqrt{6}\): \(a=5\) and \(b=2\)
Value asked for: \(a+2b=5+4=9\)