Question

If $x = (9 + 4\sqrt{5})^{48} = [x] + f$, where $[x]$ is defined as integral part of $x$ and $f$ is a fraction, then $x(1 - f)$ equals

Accepted Answer

A. Let $y = (9 - 4\sqrt{5})^{48}$. Since $0 < 9 - 4\sqrt{5} < 1$, $0 < y < 1$. Thus $y = 1 - f$. Then $x(1 - f) = x y = ((9 + 4\sqrt{5})(9 - 4\sqrt{5}))^{48} = (81 - 80)^{48} = 1$.

Algebra: Binomial Expansion Fraction

Explanation

Practise this under exam conditions