Question

4s^2-4s+1 relationship between zeros and the coefficient
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Answer 1

Given:

$f(x) = 4s^2-4s+1$

To derive:

The relationship between zeroes and coefficients of f(x)

Solution:

$\sf f(x) = 4s^2-4s+1 - - - - [i] \\\\ \tt Let ~\alpha~ and ~\beta~ be ~the~ zeroes~ of~ f(x) \\\\ \sf Now,~ (s- \alpha)~and ~(s- \beta)~ are ~the~ factors~ of~ f(x) \\\\ \sf f(x) = 4s^2-4s+1~ can~ be~ written~ as~ product~ of ~its~ factors~ \\\\ \sf i.e~f(x) =k(s- \alpha)(s- \beta) ~where~ k~ is~ constant. \\\\ \sf f(x) = k(s^2-s \beta - s \alpha + \alpha \beta) \\\\ \sf f(x)= k \{s^2-s(\alpha+ \beta) + \alpha \beta \} \\\\ \sf f(x) = ks^2-k(\alpha+\beta) S +k \alpha \beta - - - -[ii]$

Comparing [i] and [ii], we have

$\sf k=4 - - - - [iv] \\\\ \sf k(\alpha + \beta)=4 - - - - [v] \\\\ \sf k \alpha \beta = 1- - - - [vi]\\\\ \sf Now,~ putting~ k=4 ~in~ [v] ~and ~[vi],~ we ~get~ \\\\ \sf 4(\alpha + \beta)=4 \\\\ \sf \boxed{(\alpha + \beta)=4/4 =>1} \\\\ \sf and~ 4 \alpha \beta = 1 \\\\ \sf \boxed{ \alpha \beta = \frac{1}{4}}$