If a = (√5+1)/(√5-1) and b = (√5-1)(√5-1), the value of (a^2 + ab + b^2)/(a^2-ab-b^2)

Question

If a = \(\frac{\sqrt5+1}{\sqrt5-1}\) and b = \(\frac{\sqrt5-1}{\sqrt5-1}\), the value of \(\frac{a^2+ab+b^2}{a^2-ab+b^2}\) is

(a) \(\frac34\)

(b) \(\frac43\)

(c) \(\frac35\)

(d) \(\frac53\)

Answer 1

(b) \(\frac43\)

a = \(\frac{\sqrt5+1}{\sqrt5-1}\) x \(\frac{\sqrt5+1}{\sqrt5+1}\) = \(\frac{(\sqrt5+1)^2}{5-1}\)

= \(\frac{5+1+2\sqrt5}{4}= \frac{3+\sqrt5}{2}\)

b = \(\frac{\sqrt5-1}{\sqrt5+1}\) x \(\frac{\sqrt5-1}{\sqrt5-1}\) = \(\frac{(\sqrt5-1)^2}{5-1}\)

= \(\frac{5+1-2\sqrt5}{4}= \frac{3-\sqrt5}{2}\)

∴ a² + b² = \(\bigg(\frac{3+\sqrt5}{2}\bigg)^2 +\)\(\bigg(\frac{3-\sqrt5}{2}\bigg)^2 \)

= \(\frac{9+5+6\sqrt5+9+5-6\sqrt5}{4} = \frac{28}4 = 7\)

ab = \(\bigg(\frac{3+\sqrt5}{2}\bigg)\) x \(\bigg(\frac{3-\sqrt5}{2}\bigg)\) = \(\frac{9-5}{4} = \frac44= 1\)

\(\frac{a^2+ab+b^2}{a^2-ab+b^2}\) = \(\frac{7+1}{7-1} = \frac8 6= \frac43.\)