Question

if a and bita are the zeros of the polynomial 4s²+4s+1 then the value of alpha × bita is​

Answer 1

Answer:

your answer

thanks my answer

Answer 2

Given : The polynomial $\rm 4s^2+4s+1$ has two zeroes $\alpha$ and $\rm\beta.$

To find : Here, we shall find the value of $\alpha\beta.$

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When we compare $\rm 4s^2+4s+1$ to $\rm ax^2+bx+c,$ we get the values :

• $\rm a = 4$
• $\rm b=4$
• $\rm c = 1$

Also, we know that :

$~~~\rm \dashrightarrow ~~\alpha\beta~\lgroup Product~ of~ zeroes\rgroup ~=~\dfrac{c}{a}$

$~~~\rm \dashrightarrow ~~\alpha\beta~=~\dfrac{1}{4}$

$\bf {{\therefore~ The~ value~ of ~\alpha\beta~ is ~\bf{\dfrac{1}{4}.}}}$

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Verification : We'll verify by splitting the middle term and knowing the zeroes.

$~~~\rm \dashrightarrow~~4s^2+4s+1$

$~~~\rm \dashrightarrow ~~ 4s^2+2s+2s+1$

$~~~\rm \dashrightarrow~~ 2s(2s+1)+1(2s+1)$

$~~~\rm \dashrightarrow ~~(2s+1)(2s+1)$

$~~~\rm \dashrightarrow~~ s~ = ~\dfrac{-1}{2}$

$\bf {{\therefore~ The~ zeroes~ of~ 4s^2+4s+1~are~ \bf{\dfrac{-1}{2},~ \dfrac{-1}{2}.}}}$

Multiplying their zeroes :

$~~~\rm \dashrightarrow ~~\alpha\beta~=~\dfrac{1}{2}$

$~~~\rm \dashrightarrow ~~ \bigg\lgroup\dfrac{-1}{2}\bigg\rgroup \times\bigg\lgroup\dfrac{-1}{2}\bigg\rgroup~=~\dfrac{1}{4}$

$~~~\rm \dashrightarrow~~\dfrac{1}{4}~=~\dfrac{1}{4}$

LHS = RHS

${\bf Henceforth,~ verified!}$