Question
Simplify by rationalising the denominator:
$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$
$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$
✓
Answer
$\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}$
$=\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\times\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}$
$=\frac{2\sqrt{6}\times3\sqrt{5}+\big(2\sqrt{6}\big)^2-\sqrt{5}\times3\sqrt{5}-\sqrt{5}\times2\sqrt{6}}{\big(3\sqrt{5}\big)^2-\big(2\sqrt{6}\big)^2}$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{45-24}$
$=\frac{4\sqrt{30}+9}{21}$
$=\frac{2\sqrt{6}-\sqrt{5}}{3\sqrt{5}-2\sqrt{6}}\times\frac{3\sqrt{5}+2\sqrt{6}}{3\sqrt{5}+2\sqrt{6}}$
$=\frac{2\sqrt{6}\times3\sqrt{5}+\big(2\sqrt{6}\big)^2-\sqrt{5}\times3\sqrt{5}-\sqrt{5}\times2\sqrt{6}}{\big(3\sqrt{5}\big)^2-\big(2\sqrt{6}\big)^2}$
$=\frac{6\sqrt{30}+24-15-2\sqrt{30}}{45-24}$
$=\frac{4\sqrt{30}+9}{21}$
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