Question

Question 1 (ix)
Find the zeroes of the following polynomials by factorization method and verify the relations between the zeroes and the coefficient of the polynomials

(ix) $y^2+\frac{3}{2}\sqrt{5}y-5$

Answer 1

ix) Let $f\left(y\right)=y^2+\frac{3}{2}\sqrt{5}y-5$
$=2y^2+3\sqrt{5}y–10$
$=2y^2+4\sqrt{5}y–\sqrt{5}y–10$ [by splitting the middle term]
$=2y(y+2\sqrt{5})-\sqrt{5}(y+2\sqrt{5})$
$=(y+2\sqrt{5})(2y-\sqrt{5})$
So, the value of $y^2+\frac{3}{2}\sqrt{5}y-5$ is zero when $(y+2\sqrt{5})=0or(2y-\sqrt{5})=0$
i.e., when $y=-2\sqrt{5}ory=\frac{\sqrt{5}}{2}$
so, the zeroes of $2y^2+3\sqrt{5}y-10are–2\sqrt{5}and\frac{\sqrt{5}}{2}$
$\therefore$ Sum of zeroes = $-2\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{-3\sqrt{5}}{2}$
$=-\frac{\left(Coefficientofy\right)}{\left(Coefficientof{y}^2\right)}$
And product of zeroes =$-2\sqrt{5}\times \frac{\sqrt{5}}{2}=-5$
$=\frac{Constantterm}{Coefficientof{y}^2}$
Hence, verified the relations between the zeroes and the coefficients of the polynomial