Question 86
If $$a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$$ and $$b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1}$$, then the value of $$a^2 + ab + b^2$$ is
Solution
Multiplying both numerator and denominator of a by $$\sqrt{\ 5}$$, we get a = $$\frac{\left(\sqrt{\ 5}+1\right)^2}{4}$$
Multiplying both numerator and denominator of b by $$\sqrt{\ 5}-1$$, we get b = $$\frac{\left(\sqrt{\ 5}-1\right)^2}{4}$$
Now, $$a^2=\frac{\left(14+6\sqrt{\ 5}\right)}{4}$$
$$b^2=\frac{\left(14-6\sqrt{\ 5}\right)}{4}$$
$$a\cdot b=1$$
Adding them up, we get 8 as the answer
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