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Question

# $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$ = a + b$\sqrt{3}$ then find the value of a and b.

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Solution

## We have : $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$ = a + b$\sqrt{3}$ Now, rationalising the denominator of the left side of the given equation by multiplying both the numerator and denominator with $7-4\sqrt{3}$, we get: $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=\frac{5+2\sqrt{3}}{7+4\sqrt{3}}×\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\phantom{\rule{0ex}{0ex}}=\frac{\left(5+2\sqrt{3}\right)\left(7-4\sqrt{3}\right)}{{\left(7\right)}^{2}-{\left(4\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{5\left(7-4\sqrt{3}\right)+2\sqrt{3}\left(7-4\sqrt{3}\right)}{49-48}\phantom{\rule{0ex}{0ex}}=\frac{35-20\sqrt{3}+14\sqrt{3}-24}{1}\phantom{\rule{0ex}{0ex}}=11-6\sqrt{3}$ ∴ $11-6\sqrt{3}$ = a + b$\sqrt{3}$ So, on comparing the LHS and RHS, we get: a = 11 and b = $-6$

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