Question

1. Simplify ${\left(\sqrt[3]{x^2}\right)}^{\frac{3}{4}}$.
2. Find the value of a, for which the polynomial $2x^2+ax+\sqrt{2}$ has 1 as its zero.

Answer 1

Dear Student,

$$\begin{gathered} 1.{\left(\sqrt[3]{x^2}\right)}^{\frac{3}{4}} \\ ={\left\{{\left(x^2\right)}^{\frac{3}{4}}\right\}}^{\frac{1}{3}}[\because \sqrt[m]{a}=a^{\frac{1}{m}}] \\ =x^{2\times \frac{\cancel{3}}{4}\times \frac{1}{\cancel{3}}}[\because {\left(a^m\right)}^n=a^{mn}] \\ =x^{\frac{1}{2}}=\sqrt{x} \\ 2.Thepolynomialp\left(x\right)=2x^2+ax+\sqrt{2}has1asitszero. \\ Thus,forx=1,p\left(x\right)=0. \\ \therefore 2{\left(1\right)}^2+a\left(1\right)+\sqrt{2}=0 \\ \Rightarrow 2+a+\sqrt{2}=0 \\ \Rightarrow a=-2-\sqrt{2} \end{gathered}$$