Question

$x=\frac{1}{7+4\sqrt{3}}$ and $y=\frac{1}{7-4\sqrt{3}}$. Find the value of $5x^2-7xy-5y^2$

• A. $-7(1+80\sqrt{3})$
• B. $8(1+80\sqrt{3})$
• C. $0$
• D. None of these

Answer 1

Answer: (A)

$-7(1+80\sqrt{3})$

$x=\frac{1}{7+4\sqrt{3}}$ and $y=\frac{1}{7-4\sqrt{3}}$

Consider, $5x^2-7xy-5y^2$

$\Rightarrow \frac{5.1}{(7+4\sqrt{3}{)}^2}-\frac{7.1}{(7+4\sqrt{3})(7-4\sqrt{3})}-\frac{5.1}{(7-4\sqrt{3}{)}^2}$

$\Rightarrow \frac{5}{49+48+56\sqrt{3}}-\frac{7}{49-48}-\frac{5}{49+48-56\sqrt{3}}$

$\Rightarrow 5\left[\frac{97-56\sqrt{3}-97-56\sqrt{3}}{(97{)}^2-(56\sqrt{3}{)}^2}\right]-7$

$\Rightarrow 5\left[\frac{-112\sqrt{3}}{1}\right]-7$

$\Rightarrow -7[1+80\sqrt{3}]$.