Question

Find the value of $sqrt5$ correct to 2 decimal places; then use it to find the square root of $\frac{3-\sqrt{5}}{3+\sqrt{5}}$ correct to 2 significant digits.

Answer 1

$sqrt5=2.236=2.24$

$\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{(3-\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}}=\sqrt{\frac{(3-\sqrt{5}{)}^2}{(3{)}^2-(\sqrt{5}{)}^2}}=\sqrt{\frac{(3-\sqrt{5}{)}^2}{9-5}}=\sqrt{\frac{(3-\sqrt{5}{)}^2}{4}}=\frac{(3-2.24)}{2}=\frac{\left(0.76\right)}{2}=0.38$