Answer to Question #228837 in Algebra for Benjamin

Question

Simplify (√5+1)÷(√5-1)

Answer 1

Answer:

To solve this problem we should multiply both numerator and denominator by  $\frac{(\sqrt{\smash[b]{5}}+1)}{(\sqrt{\smash[b]{5}}+1)}$ =1 (because multiplying any number by 1 doesn't change it) by itself to get a 'normal' number on the bottom. So in this case we do:

$\frac{(\sqrt{\smash[b]{5}}+1)}{(\sqrt{\smash[b]{5}}-1)}*\frac{(\sqrt{\smash[b]{5}}+1)}{(\sqrt{\smash[b]{5}}+1)}=\frac{(\sqrt{\smash[b]{5}}*\sqrt{\smash[b]{5}}+\sqrt{\smash[b]{5}}+\sqrt{\smash[b]{5}}+1)}{(\sqrt{\smash[b]{5}}*\sqrt{\smash[b]{5}}+\sqrt{\smash[b]{5}}-\sqrt{\smash[b]{5}}-1)}=\frac{(5+2\sqrt{\smash[b]{5}}+1)}{(5-1)}$

$\implies \frac{(5+2\sqrt{\smash[b]{5}}+1)}{(5-1)}=\frac{(6+2\sqrt{\smash[b]{5}})}{(4)}=\frac{2(3+\sqrt{\smash[b]{5}})}{4}=\frac{3+\sqrt{\smash[b]{5}}}{2}$

Answer:

$\frac{3+\sqrt{\smash[b]{5}}}{2}$