Question

Find the zeros of the following polynomial by factorization method and verify the relations between the zeroes and the coefficient of the polynomials :
$2s^2-(1+2\sqrt{2})s+\sqrt{2}$

Answer 1

Let $f\left(s\right)=2s^2-(1+2\sqrt{2})s+\sqrt{2}$

For zeroes of $f\left(s\right),f\left(s\right)=0$

$\Rightarrow 2s^2-(1+2\sqrt{2})s+\sqrt{2}=0$ $[\because a\times c=(2\sqrt{2}\left)\right]$

$\Rightarrow 2s^2-1s-2\sqrt{2}s+\sqrt{2}=0$ [Open the brackets]

$\Rightarrow s(2s-1)-\sqrt{2}(2s-1)=0$

$\Rightarrow (2s-1)(s-\sqrt{2})=0$

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

$\Rightarrow s=\frac{1}{2}$ or $s=\sqrt{2}$

verification:

$\alpha =\frac{1}{2},\beta =\sqrt{2}$,

$a=2,b=-(1+2\sqrt{2}),c=\sqrt{2}$

$\alpha +\beta =\frac{-b}{a}$

$\Rightarrow \frac{1}{2}+\sqrt{2}=\frac{+(1+2\sqrt{2})}{2}$

$\Rightarrow \frac{1}{2}+\sqrt{2}=\frac{1}{2}+\frac{2\sqrt{2}}{2}$

$\Rightarrow \frac{1}{2}+\sqrt{2}=\frac{1}{2}+\sqrt{2}$

$\Rightarrow LHS=RHS$

Hence verified.

$\alpha .\beta =\frac{c}{a}$

$\Rightarrow \left(\frac{1}{2}\right)\left(\sqrt{2}\right)=\frac{\sqrt{2}}{2}$

$\Rightarrow \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$

$\Rightarrow LHS=RHS$

Hence, verified.