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Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:

y^{2}+\frac{3}{2}\sqrt{5}y-5

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Answer.    \left [-2\sqrt{5}, \frac{\sqrt{5}}{2} \right ]

Solution. Zeroes : zeroes of the polynomial are the value(s) that makes it equal to 0.

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

Multiply by 2

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

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

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

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

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

y=-2\sqrt{5}                                         y=\frac{\sqrt{5}}{2}

Hence, \left [-2\sqrt{5}, \frac{\sqrt{5}}{2} \right ]  are the zeroes of the polynomial

Here, a = 2, b = 3\sqrt{5} , c = –10

Sum of zeroes =\frac{-b}{a}=\frac{-3\sqrt{5}}{2}

Here, -2\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{-4\sqrt{5}+\sqrt{5}}{2}=\frac{-3\sqrt{5}}{2}=\frac{-b}{a}

Product of zeroes =\frac{c}{a}=\frac{-10}{2}=-5

Here, -2\sqrt{5}\times \frac{\sqrt{5}}{2}=-5=\frac{c}{a}

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