Question

Solve this:
Q. If $\frac{5+\sqrt{3}}{7-4\sqrt{3}}=47a+\sqrt{3}b$, find the value of a and b.

Answer 1

Dear Student,
$$\begin{gathered} \frac{5+\sqrt{3}}{7-4\sqrt{3}}=47\mathrm{a}+\mathrm{b}\sqrt{3} \\ \Rightarrow \frac{5+\sqrt{3}}{7-4\sqrt{3}}\times \frac{7+4\sqrt{3}}{7+4\sqrt{3}}=47\mathrm{a}+\mathrm{b}\sqrt{3} \\ \Rightarrow \frac{35+20\sqrt{3}+7\sqrt{3}+12}{49-48}=47\mathrm{a}+\mathrm{b}\sqrt{3} \\ \Rightarrow \frac{47+27\sqrt{3}}{1}=47\mathrm{a}+\mathrm{b}\sqrt{3} \\ \Rightarrow 47\mathrm{a}=47\Rightarrow \mathrm{a}=1\mathrm{and}\mathrm{b}=27 \end{gathered}$$

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