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Polynomials — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsPolynomials5 Marks
Question
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
$\text{y}^2+\frac{3}{5}\sqrt{5}\text{y}-5$.
✓

Answer

Let $\text{f}\text{(x)}=\text{y}^2+\frac{3}{5}\sqrt{5}\text{y}-5$
$=2\text{y}^2+3\sqrt{5}\text{y}-10$
$=2\text{y}^2+4\sqrt{5}\text{ y}-\sqrt{5}\text{ y}-10$
$=2\text{y}(\text{y}+2\sqrt{5})-\sqrt{5}(\text{y}+2\sqrt{5})$
$=(\text{y}+2\sqrt{5})(2\text{y}-\sqrt{5})$
So, the value of $\text{y}^2+\frac{3}{2}\sqrt{5}\text{y}-5$ is zero when $(\text{y}+2\sqrt{5})=0$ or $(2\text{y}-\sqrt{5})=0,$
i.e., when $\text{y}=-2\sqrt{5}$ or $\text{y}=\frac{\sqrt{5}}{2}.$
So, the zeroes of $2\text{y}^2+3\sqrt{5}\text{y}-10$ are $-2\sqrt{5}$ and $\frac{\sqrt{5}}{2}$
$\therefore\ \text{Sum of zeroes}=-2\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{-3\sqrt{5}}{2}=-\frac{(\text{Coefficient of y})}{(\text{Coefficient of y}^2)}$
and $\text{product of zeroes}=-2\sqrt{5}\times\frac{\sqrt{5}}{2}=-5=\frac{\text{Constant term}}{\text{Coefficient of y}^2}$
Hence, verified the relations between the zeroes and the coefficients of the polynomial.

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