Question

Pls urgent answer Q7 viii) and Q10 iv) and vi).

Answer 1

$\underline{\text{7. (viii)}}$

$\text{Now,}\:\frac{\sqrt[5]{4}}{\sqrt[5]{50}}$

$=\frac{4^{1/5}}{50^{1/5}}$

$=\frac{(2^{2})^{1/5}}{(2\times 5^{2})^{1/5}}$

$=\frac{2^{2/5}}{2^{1/5}\times 5^{2/5}}$

$=\frac{2^{2/5}\times 2^{-1/5}}{5^{2/5}}$

$=\frac{2^{1/5}\times 5^{3/5}}{5^{2/5}\times 5^{3/5}}$

$=\frac{(2\times 5^{3})^{1/3}}{5^{5/5}}$

$=\frac{250^{1/5}}{5^{1}}$

$=\frac{\sqrt[5]{250}}{5}$

$\to \boxed{\frac{\sqrt[5]{4}}{\sqrt[5]{50}}=\frac{\sqrt[5]{250}}{5}}$ ,


$\text{which is the required rationalisation.}$

$\underline{\text{10. (iv)}}$

$\text{Now,}\:\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$

$=\frac{(\sqrt{5}+\sqrt{3})^{2}+(\sqrt{5}-\sqrt{3})^{2}}{(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})}$

$=\frac{5+2\sqrt{5}\sqrt{3}+3+5-2\sqrt{5}\sqrt{3}+3}{5-3}$

$=\frac{16}{2}=8$

$\to \boxed{\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=8}$

$\underline{\text{10. (vi)}}$

$\text{Given,}\:x=5+\sqrt{5}$

$\text{Then,}\:x^{2}=(5+\sqrt{5})^{2}$

$=5^{2}+2.5.\sqrt{5}+(\sqrt{5})^{2}$

$=25+10\sqrt{5}+5$

$=30+10\sqrt{5}$

$\implies x^{2}=30+10\sqrt{5}$

$\text{and}\:x^{3}=(5+\sqrt{5})^{3}$

$=5^{3}+3.5^{2}.\sqrt{5}+3.5.(\sqrt{5})^{2}+(\sqrt{5})^{3}$

$=125+75\sqrt{5}+75+5\sqrt{5}$

$=200+80\sqrt{5}$

$\implies x^{3}=200+80\sqrt{5}$

$\therefore 10x+7x^{2}-x^{3}$

$=10(5+\sqrt{5})+7(30+10\sqrt{5})-(200+80\sqrt{5})$

$=50+10\sqrt{5}+210+70\sqrt{5}-200-80\sqrt{5}$

$=60$

$\implies \boxed{10x+7x^{2}-x^{3}=60}$