Question

If $x=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$ and $y=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$, then find the value of $x^2+y^2-5xy$.

• A. 0
• B. 1
• C. 57
• D. $16\sqrt{5}+57$

Answer 1

The correct option is D $16\sqrt{5}+57$

$x=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$x=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\times \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\frac{5+3-2\sqrt{15}}{2}=4-\sqrt{15}$

$y=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

$y=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{5+3+2\sqrt{15}}{2}=4+\sqrt{15}$

Now,

$x^2+y^2-5xy$

$=(4-\sqrt{15}{)}^2+(4+\sqrt{15}{)}^2-5(4-\sqrt{15})(4+\sqrt{15})$

$=(16+15-8\sqrt{15})+(16+15+8\sqrt{15})-5(4-\sqrt{15})(4+\sqrt{15})$

$=62-5(16-15)=62-5=57$