Question

Verify whether the following are zeros of the polynomial indicated against them:

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

• A. $x=-\frac{1}{\sqrt{3}}$ is a zero
• B. $x=\frac{2}{\sqrt{3}}$ is a zero
• C. $x=-\frac{1}{\sqrt{3}}$ is not a zero
• D. $x=\frac{2}{\sqrt{3}}$ is not a zero

Answer 1

Answer: (C), (D)

The correct options are
C $x=-\frac{1}{\sqrt{3}}$ is not a zero
D $x=\frac{2}{\sqrt{3}}$ is not a zero

Given, $p\left(x\right)=-3x^2-1$

Now, $p(-\frac{1}{\sqrt{3}})=-3{(-\frac{1}{\sqrt{3}})}^2-1=-1-1=-2$ and

$p\left(\frac{2}{\sqrt{3}}\right)=-3{\left(\frac{2}{\sqrt{3}}\right)}^2-1=-12-1=-13$

Clearly, $p(-\frac{1}{\sqrt{3}})\ne 0$ and $p\left(\frac{2}{\sqrt{3}}\right)\ne 0$

$\therefore x=-\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}}$ are not zeroes of given polynomial.

That is, option $C$ and $D$ are correct.