Question

If both $A$ and $B$ are rational number, find the value of $A$ and $B$ of the equality $\frac{5+\sqrt{3}}{7-4\sqrt{3}}=47A+B\sqrt{3}$

Answer 1

Step 1: Solve the L.H.S

$L.H.S=\frac{5+\sqrt{3}}{7-4\sqrt{3}}$

Now, by rationalizing the L.H.S we have

$$\begin{gathered} L.H.S=\frac{(5+\sqrt{3})\left(7+4\sqrt{3}\right)}{(7-4\sqrt{3})(7+4\sqrt{3})} \\ =\frac{35+20\sqrt{3}+7\sqrt{3}+12}{7^2-{\left(4\sqrt{3}\right)}^2}\left[\because \left(\mathrm{a}+\mathrm{b}\right)(\mathrm{a}-\mathrm{b})={\mathrm{a}}^2-{\mathrm{b}}^2\right] \\ =\frac{47+27\sqrt{3}}{49-48} \\ =\frac{47+27\sqrt{3}}{1} \\ =47+27\sqrt{3} \\ \therefore L.H.S=47+27\sqrt{3} \end{gathered}$$

Step 2: Compare the L.H.S with R.H.S

$R.H.S=47A+B\sqrt{3}$

Equate L.H.S and R.H.S we have

$47+27\sqrt{3}=47A+B\sqrt{3}$

Now, by comparing both the sides, we get,

$A=1$ and $B=27$

Hence, the value of $A$ is $1$ and the value of $B$ is $27$.