Question

Write the zeros of the quadratic polynomial $f\left(x\right)=6x^2-3$.

Answer 1

The given polynomial is,

$f\left(x\right)=6x^2-3$

$=3(2x^2-1)$

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

$=3(\sqrt{2}x+1)(\sqrt{2}x-1)$

[by the identity $a^2-b^2=(a+b)(a-b)$]

The value of $6x^2-3$ is zero when

$3(\sqrt{2}x+1)(\sqrt{2}x-1)=0$

$\Rightarrow \sqrt{2}x+1=0$ or $\sqrt{2}x-1=0$

$\Rightarrow x=\frac{-1}{\sqrt{2}}$ or $x=\frac{1}{\sqrt{2}}$

The zeroes are $\frac{-1}{\sqrt{2}}and\frac{1}{\sqrt{2}}$