Question

Solve:
(i) $\sqrt{98}÷\sqrt{2}$

(ii) $\sqrt[8]{28}÷\sqrt[2]{7}$

(iii) $\sqrt[4]{27}÷\sqrt[4]{3}$

(iv) $\sqrt[3]{5}÷\sqrt[4]{3}$

(v) $5\sqrt[3]{4}÷\sqrt[3]{2}$

Answer 1

$\left(\mathrm{i}\right)\sqrt{98}÷\sqrt{2}=\frac{\sqrt{98}}{\sqrt{2}}=\frac{\sqrt{49}\times \sqrt{2}}{\sqrt{2}}=\sqrt{49}=7$

$\left(\mathrm{ii}\right)8\sqrt{28}÷2\sqrt{7}=\frac{8\sqrt{28}}{2\sqrt{7}}=\frac{8\times \sqrt{4}\times \sqrt{7}}{2\times \sqrt{7}}=\frac{16}{2}=8$

$\left(\mathrm{iii}\right)\sqrt[4]{27}÷\sqrt[4]{3}=\frac{\sqrt[4]{27}}{\sqrt[4]{3}}=\sqrt[4]{\frac{27}{3}}=\sqrt[4]{9}$

$\left(\mathrm{iv}\right)\sqrt[3]{5}÷\sqrt[4]{3}=\frac{\sqrt[3]{5}}{\sqrt[4]{3}}=\frac{\sqrt[3\times 4]{5^4}}{\sqrt[4\times 3]{3^3}}=\frac{\sqrt[12]{625}}{\sqrt[12]{27}}=\sqrt[12]{\frac{625}{27}}$

$\left(\mathrm{v}\right)5\sqrt[3]{4}÷\sqrt[3]{2}=\frac{5\times \sqrt[3]{4}}{\sqrt[3]{2}}=5\sqrt[3]{\frac{4}{2}}=5\sqrt[3]{2}$