Question

Verify whether the following are zeroes of the polynomial, indicated against them: $\mathrm{p}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-1,\mathrm{x}=\frac{-1}{\sqrt{3}},\frac{2}{\sqrt{3}}$

Answer 1

Step 1: Verifying whether $\mathrm{x}=\frac{-1}{\sqrt{3}}$ is a zero of the polynomial $\mathrm{p}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-1$

We have,

$\mathrm{p}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-1,\mathrm{x}=\frac{-1}{\sqrt{3}},\frac{2}{\sqrt{3}}$

Put the value of $\mathrm{x}=\frac{-1}{\sqrt{3}}$ in $\mathrm{p}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-1$

$$\begin{gathered} \mathrm{p}\left(\frac{-1}{\sqrt{3}}\right)=3{\left(\frac{-1}{\sqrt{3}}\right)}^2-1 \\ \Rightarrow \mathrm{p}\left(\frac{-1}{\sqrt{3}}\right)=3\left(\frac{1}{3}\right)-1 \\ \Rightarrow \mathrm{p}\left(\frac{-1}{\sqrt{3}}\right)=1-1 \\ \Rightarrow \mathrm{p}\left(\frac{-1}{\sqrt{3}}\right)=0 \end{gathered}$$

Hence, $\mathrm{x}=\frac{-1}{\sqrt{3}}$ is a zero of the polynomial $\mathrm{p}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-1$.

Step 2: Verifying whether $\mathrm{x}=\frac{2}{\sqrt{3}}$ is a zero of the polynomial $\mathrm{p}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-1$

Now,

Put the value of $\mathrm{x}=\frac{2}{\sqrt{3}}$ in $\mathrm{p}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-1$

$$\begin{gathered} \mathrm{p}\left(\frac{2}{\sqrt{3}}\right)=3{\left(\frac{2}{\sqrt{3}}\right)}^2-1 \\ \Rightarrow \mathrm{p}\left(\frac{2}{\sqrt{3}}\right)=3\left(\frac{4}{3}\right)-1 \\ \Rightarrow \mathrm{p}\left(\frac{2}{\sqrt{3}}\right)=4-1 \\ \Rightarrow \mathrm{p}\left(\frac{2}{\sqrt{3}}\right)=3\ne 0 \end{gathered}$$

Hence, $\mathrm{x}=\frac{2}{\sqrt{3}}$ is not a zero of the polynomial $\mathrm{p}\left(\mathrm{x}\right)=3{\mathrm{x}}^2-1$.