Question

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

• A. $\frac{4}{3}$
• B. $\frac{5}{3}$
• C. $2$
• D. $\frac{7}{3}$

Answer 1

The correct option is A $\frac{4}{3}$

Here $ab=1$

& $(a+b)=\frac{\sqrt{5}+1}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{\sqrt{5}+1}$

$\Rightarrow (a+b)=\frac{(\sqrt{5}+1{)}^2+(\sqrt{5}-1{)}^2}{4}$

$=\frac{5+1+2\sqrt{5}+5+1-25}{4}$

$\Rightarrow (a+b)=3$

$(a^2+b^2)=(a+b{)}^2-2ab$

$=9-2=7$

$\therefore \frac{a^2+b^2+ab}{a^2+b^2-ab}=\frac{7+1}{7-1}=\frac{4}{3}$