Math, asked by sukhwaltanushree3007, 1 month ago

findthe zeroes of the quadratic polynomial 7y^2-11/3y-2/3and verify the relationship between the cofficeient​

Answers

Answered by mathdude500
2

\large\underline{\bf{Solution-}}

\bf :\longmapsto\:Let \: f(y) =  {7y}^{2}  - \dfrac{11}{3} y - \dfrac{2}{3}

 \rm \:=  \: \:\dfrac{ {21y}^{2} - 11y - 2 }{3}

 \rm \:=  \: \:\dfrac{1}{3}\bigg( {21y}^{2} - 14y + 3y - 2 \bigg)

 \rm \:=  \: \:\dfrac{1}{3}\bigg(7y(3y - 2)  + 1(3y - 2)\bigg)

 \rm \:=  \: \:\dfrac{1}{3}(7y + 1)(3y - 2)

Hence,

\rm :\longmapsto\:Zeroes \: of \: f(y) \:  =  - \dfrac{1}{7}  \:  \: or \:  \: \dfrac{2}{3}

\rm :\longmapsto\:Let \:  \alpha  =  - \dfrac{1}{7}  \:  \: and \:  \: \beta  =  \dfrac{2}{3}

Consider,

\rm :\longmapsto\: \alpha   + \beta

 \rm \:=  \: \: - \dfrac{1}{7}  + \dfrac{2}{3}

 \rm \:=  \: \: \dfrac{ - 3 + 14}{21}

 \rm \:=  \: \: \dfrac{ 11}{21}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \boxed{ \bf\implies \:\bf \:  \alpha  +  \beta  = \dfrac{11}{21}}

Consider,

\rm :\longmapsto\: \alpha  \beta

 \rm \:=  \: \: - \dfrac{1}{7}  \times \dfrac{2}{3}

 \rm \:=  \: \: - \dfrac{2}{21}

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \boxed{ \bf\implies \:\bf \:  \alpha \beta  = -  \:  \dfrac{2}{21}}

Now, Verification

On comparing the given polynomial with

\rm :\longmapsto\: {ay}^{2} + by + c

We get

 \red{\rm :\longmapsto\:a =  \: 7 \:  \:  \:  \:  \:  \: } \\  \red{\rm :\longmapsto\: b = \:  - \dfrac{11}{3} } \\  \red{\rm :\longmapsto\: c = \: - \dfrac{2}{3}   \:  \: }

We know that

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ y}{coefficient\ of\ y^{2}}}}

\rm :\implies\:Sum \:  of\ the\ zeroes = -  \dfrac{ - \dfrac{11}{3} }{ \:  \: 7 \:  \: }  = \dfrac{11}{21}

And

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ y^{2}}}}

\rm :\implies\:{{\sf Product\ of\ the\ zeroes=\dfrac{\dfrac{ - 2}{3} }{7}}} = -  \:  \dfrac{2}{21}

Hence,

We conclude that

\boxed{\red{\rm :\longmapsto\:\sf Sum\ of\ the\ zeroes= \alpha  +  \beta }}

and

\boxed{\red{\rm :\longmapsto\:\sf Product\ of\ the\ zeroes= \alpha  \beta }}

Hence, Verified

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