Question:
The zeroes of the polynomial $p(x)=3 x^{2}-1$ are
(a) $\frac{1}{3}$ and 3
(b) $\frac{1}{\sqrt{3}}$ and $\sqrt{3}$
(c) $\frac{-1}{\sqrt{3}}$ and $\sqrt{3}$
(d) $\frac{1}{\sqrt{3}}$ and $\frac{-1}{\sqrt{3}}$
Solution:
(d) $\frac{1}{\sqrt{3}}$ and $\frac{-1}{\sqrt{3}}$
Let:
$p(x)=3 x^{2}-1$
To find the zeroes of $p(x)$, we have :
$p(x)=0 \Rightarrow 3 x^{2}-1=0$
$\Rightarrow 3 x^{2}=1$
$\Rightarrow x^{2}=\frac{1}{3}$
$\Rightarrow x=\pm \frac{1}{\sqrt{3}}$
$\Rightarrow x=\frac{1}{\sqrt{3}}$ and $x=-\frac{1}{\sqrt{3}}$
Hence, the correct answer is option D.
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