Question

Answer :

The zeroes of the given polynomial are 1/2 and 1/2

Given :

The quadratic polynomial is :

4s² - 4s + 1

Task :

To find out the zeroes of the given polynomial

Also to verify the relationship between the zeroes and the coefficients .

Solution :

Splitting the middle term for factorization :

\begin{lgathered}\sf4 {s}^{2} - 4s + 1 \\ \\ \longrightarrow \sf4 {s}^{2} - 2s - 2s + 1 \\ \\ \longrightarrow \sf2s(2s - 1) - 1(2s - 1) \\ \\ \longrightarrow \sf(2s - 1)(2s - 1)\end{lgathered}4s2−4s+1⟶4s2−2s−2s+1⟶2s(2s−1)−1(2s−1)⟶(2s−1)(2s−1)​

The zeroes of the polynomial are :

\begin{lgathered}\sf2s - 1 = 0\: \: and \: \: 2s - 1 = 0 \\ \\ \implies \sf2s = 1 \: \: and \implies2s = 1 \\ \\ \implies \sf s = \dfrac{1}{2} \: \: and \implies s = \dfrac{1}{2}\end{lgathered}2s−1=0and2s−1=0⟹2s=1and⟹2s=1⟹s=21​and⟹s=21​​

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Verification of the relationship between the zeroes and the coefficients :

\sf{Sum \: \: of \: \: the \: \: roots = \dfrac{ -coefficient \: of \: x}{coefficient \: \: of \: \: {x}^{2} } }Sumoftheroots=coefficientofx2−coefficientofx​

\begin{lgathered}\implies \sf \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{ - ( - 4)}{4} \\ \\ \sf\implies \dfrac{1 + 1}{2} = \dfrac{4}{4} \\ \\ \implies \sf \dfrac{2}{2} = \dfrac{4}{4} \\ \\ \bf\implies1 = 1\end{lgathered}⟹21​+21​=4−(−4)​⟹21+1​=44​⟹22​=44​⟹1=1​

Again :

\sf{Product \: \: of \: \: the \: roots \: = \dfrac{constant \: term}{coefficient \: \: of \: \: {x}^{2} } }Productoftheroots=coefficientofx2constantterm​

\begin{lgathered}\sf \implies \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4} \\ \\ \bf \implies \dfrac{1}{4} = \dfrac{1}{4}\end{lgathered}⟹21​×21​=41​⟹41​=41​​

\bold{Hence \: \: Verified}HenceVerified

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Answer 1

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Answer 2

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