Question

If $a=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $b=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of
$\frac{a^2+ab+b^2}{a^2-ab+b^2}$ is

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{3}{5}$
• D. $\frac{5}{3}$

Answer 1

Answer: (B)

$\frac{4}{3}$

$a=\frac{(\sqrt{5}+1)}{(\sqrt{5}-1)}\times \frac{(\sqrt{5}+1)}{(\sqrt{5}+1)}=\frac{(\sqrt{5}+1{)}^2}{5-1}$
$=\frac{5+1+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}$
$b=\frac{\sqrt{5}-1}{\sqrt{5}+1}\times \frac{\sqrt{5}-1}{\sqrt{5}-1}=\frac{(\sqrt{5}-1{)}^2}{5-1}$
$=\frac{5+1-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$
$\therefore a^2+b^2={\left(\frac{3+\sqrt{5}}{2}\right)}^2+{\left(\frac{3-\sqrt{5}}{2}\right)}^2$
$=\frac{9+5+6\sqrt{5}+9+5-6\sqrt{5}}{4}$
$=\frac{28}{4}=7$
$ab=\left(\frac{3+\sqrt{5}}{2}\right)\times \left(\frac{3-\sqrt{5}}{2}\right)=\frac{9-5}{4}=\frac{4}{4}=1$
$\therefore \frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{7+1}{7-1}=\frac{8}{6}=\frac{4}{3}$