Question

find the zeroes of the following polynomial and verify the relationship between the zeroes and the coefficient 4s2-4s+1​

Answer 1

$\large\underline{\sf{Solution-}}$

Given quadratic polynomial is

$\rm :\longmapsto\:f(s) = {4s}^{2} - 4s + 1$

On splitting the middle terms, we get

$\rm :\longmapsto\:f(s) = {4s}^{2} - 2s - 2s + 1$

$\rm :\longmapsto\:f(s) = 2s(2s - 1) - 1(2s - 1)$

$\rm :\longmapsto\:f(s) = (2s - 1)(2s - 1)$

So,

$\rm \implies\:zeroes \: of \: f(s) = \dfrac{1}{2} , \: \dfrac{1}{2}$

Let assume that,

$\rm \implies\: \alpha = \dfrac{1}{2} \:and \: \beta = \dfrac{1}{2}$

So,

$\red{\boxed{ \tt{ \: \rm :\longmapsto\: \alpha + \beta = \dfrac{1}{2} + \dfrac{1}{2} = 1}}}$

and

$\boxed{ \tt{ \: \red{\rm :\longmapsto\: \alpha \beta = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}}}}$

Now, Verification

Given polynomial is

$\rm :\longmapsto\:f(s) = {4s}^{2} - 4s + 1$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}} = \frac{4}{4} = 1 }}$

and

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}} = \frac{1}{4} }}$

Hence, We concluded that,

$\boxed{\red{\sf \alpha \beta =\frac{Constant}{coefficient\ of\ x^{2}}}}$

and

$\boxed{\red{\sf \alpha + \beta =\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

Know to More :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given quadratic polynomial is

$\rm :\longmapsto\:f(s) = {4s}^{2} - 4s + 1$

On splitting the middle terms, we get

$\rm :\longmapsto\:f(s) = {4s}^{2} - 2s - 2s + 1$

$\rm :\longmapsto\:f(s) = 2s(2s - 1) - 1(2s - 1)$

$\rm :\longmapsto\:f(s) = (2s - 1)(2s - 1)$

So,

$\rm \implies\:zeroes \: of \: f(s) = \dfrac{1}{2} , \: \dfrac{1}{2}$

Let assume that,

$\rm \implies\: \alpha = \dfrac{1}{2} \:and \: \beta = \dfrac{1}{2}$

So,

$\red{\boxed{ \tt{ \: \rm :\longmapsto\: \alpha + \beta = \dfrac{1}{2} + \dfrac{1}{2} = 1}}}$

and

$\boxed{ \tt{ \: \red{\rm :\longmapsto\: \alpha \beta = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}}}}$

Now, Verification

Given polynomial is

$\rm :\longmapsto\:f(s) = {4s}^{2} - 4s + 1$

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}} = \frac{4}{4} = 1 }}$

and

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}} = \frac{1}{4} }}$

Hence, We concluded that,

$\boxed{\red{\sf \alpha \beta =\frac{Constant}{coefficient\ of\ x^{2}}}}$

and

$\boxed{\red{\sf \alpha + \beta =\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

Know to More :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$