Question

Find the zeros of the following polynomial by factorization method and verify the relations between the zeroes and the coefficient of the polynomials :
$y^2+\frac{3}{2}\sqrt{5}y-5$

Answer 1

Let $f\left(y\right)=y^2+\frac{3}{2}\sqrt{5}y-5$

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

$\Rightarrow y^2+\frac{3}{2}\sqrt{5}y-5=0$

$\Rightarrow 2y^2+3.\sqrt{5}y-10=0$

$\Rightarrow 2y^2+4\sqrt{5}y-1\sqrt{5}y-10=0$

$\Rightarrow 2y(y+2\sqrt{5})-\sqrt{5}[y+2\sqrt{5}]=0$

$\Rightarrow (y+2\sqrt{5})(2y-\sqrt{5})=0$

$\Rightarrow y+2\sqrt{5}=0$ or $2y-\sqrt{5}=0$

$\Rightarrow y=-2\sqrt{5}$ or $y=\frac{\sqrt{5}}{2}$

verification:

$\alpha =-2\sqrt{5},\beta =\frac{\sqrt{5}}{2}$,

$a=1,b=\frac{3}{2}\sqrt{5}$ and $c=-5$

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

$\Rightarrow -2\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{\frac{-3}{2}\sqrt{5}}{1}$

$\Rightarrow \frac{-4\sqrt{5}+\sqrt{5}}{2}=\frac{-3}{2}\sqrt{5}$

$\Rightarrow \frac{-3\sqrt{5}}{2}=\frac{-3}{2}\sqrt{5}$

$\Rightarrow LHS=RHS$

Hence, verified.

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

$\Rightarrow (-2\sqrt{5})\left(\frac{\sqrt{5}}{2}\right)=\frac{-5}{1}$

$\Rightarrow -5=-5$

$\Rightarrow LHS=RHS$

Hence, verified.