Question

If $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a+b\sqrt{3}$. Then, the value of a = and b =

• A. 9
• B. 11
• C. 6
• D. -6

Answer 1

Answer: (B), (D)

-6

$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$can be simplified by rationalizing the denominator $(7+4\sqrt{3})$. We can rationalize the given denominator by multiplying the given number $(7+4\sqrt{3})$with its rationalizing factor $(7-4\sqrt{3})$.

Multiplying the numerator and denominator with $(7-4\sqrt{3})$, we get

$\frac{(5+2\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}$=$\frac{35+14\sqrt{3}-20\sqrt{3}-8\left(\sqrt{3}\sqrt{3}\right)}{\left(7\right)\left(7\right)-(4\times 4)(\sqrt{3}\times \sqrt{3})}$

=$\frac{35-6\sqrt{3}-24}{49-48}$

=$11-6\sqrt{3}$

On comparing, with $a+b\sqrt{3}$, we get $a=11,b=-6$