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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:
$2\text{s}^2-(1+2\sqrt{2})\text{s}+\sqrt{2}$.
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Answer

Let $\text{f}(\text{s})=2\text{s}^2-(1+2\sqrt{2})\text{s}+\sqrt{2}$
$=2\text{s}^2-\text{s}+2\sqrt{2}\text{ s}+\sqrt{2}$
$=\text{s}(2\text{s}-1)-\sqrt{2}(2\text{s}-1)$
$=(2\text{s}-1)(\text{s}-\sqrt{2})$
So, the value of $2\text{s}^2-(1+2\sqrt{2})\text{s}+\sqrt{2}$ is zero when $2\text{s}-1=0$ or $\text{s}-\sqrt{2} = 0,$
i.e., when $\text{s}=\frac{1}{2}$ or $\text{s}=\sqrt{2}$
So, the zeroes of $2\text{s}^2-(1+2\sqrt{2})\text{s}+\sqrt{2}$ are $\frac{1}{2}$ and $\sqrt{2}$
$\therefore\ \text{Sum of zeroes}=\frac{1}{2}+\sqrt{2}=\frac{1+2\sqrt{2}}{2}=\frac{-[-(1+2\sqrt{2})]}{2}$
$=\frac{\text{(Coefficient of s)}}{\text{(Coefficient of s)}^2}$
and $\text{product of zeroes}=\frac{1}{2}.\sqrt{2}=\frac{1}{\sqrt{2}}=\frac{\text{Constant term}}{\text{Coefficient of s}^2}$
Hence, verified the relations between the zeroes and the coefficients of the polynomial.

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