Question

Find the zeros of the quadratic polynomial $f\left(x\right)=6x^2-3$, and verify the relation-ship between the zeros and its coefficients:

Answer 1

We have,

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

$\to$ $f\left(X\right)=(\sqrt{6}x{)}^2-(\sqrt{3}{)}^2$

$\to$ $f\left(x\right)=(\sqrt{6}x-\sqrt{3})(\sqrt{6}x+\sqrt{3})$

The zeros of f(x) are given by f(x) =0

ie., $(\sqrt{6}x-\sqrt{3})(\sqrt{6}x+\sqrt{3})=0$

$\Rightarrow$ $\sqrt{6}x-\sqrt{3}=0$ or, $\sqrt{6}x+\sqrt{3}=0$

$\Rightarrow$ $x=\frac{\sqrt{3}}{\sqrt{6}}$ or, $x=\frac{-\sqrt{3}}{\sqrt{6}}$

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

Hence, the zeros of $f\left(x\right)=6x^2-3$ are: $\alpha =\frac{1}{\sqrt{2}}$= and $\beta =-\frac{1}{\sqrt{2}}$

Now,

Sum of the zeros = $\alpha +\beta =\frac{1}{\sqrt{2}}+(-\frac{1}{\sqrt{2}})=0$

Sum of the zeros = -$\frac{Coefficientofx}{Coefficientof{x}^2}$

$\therefore$ -$\frac{Coefficientofx}{Coefficientof{x}^2}=-\frac{0}{6}=0$

Also,

Product of the zeros = $\alpha \beta$ = $\frac{1}{\sqrt{2}}\times \frac{-1}{\sqrt{2}}=\frac{-1}{2}$

and,

$\therefore$ $Productofthezeros=$ $\frac{Constantterm}{Coefficientof{x}^2}=\frac{-3}{6}=\frac{-1}{2}$

Hence verified.