Question

If α and β are the zeros of the polynomial x
2
+8x+6, form a quadratic polynomial whose zeros are α+β/α, α+β/ β​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

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

Given that

$\rm :\longmapsto\: \alpha \: and \: \beta \: are \: zeroes \: of \: {x}^{2} + 8x + 6$

We know,

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

$\rm :\implies\: \alpha + \beta = - \dfrac{8}{1} = - \: 8$

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

$\rm :\implies\: \alpha \beta = \dfrac{6}{1} = 6$

Now, we have to form a quadratic polynomial whom zeroes are

$\rm :\longmapsto\:\dfrac{ \alpha + \beta }{ \alpha } , \: \dfrac{ \alpha + \beta }{ \beta }$

Consider,

Sum of zeroes,

$\rm :\longmapsto\:S = \rm \:\dfrac{ \alpha + \beta }{ \alpha } \: + \: \dfrac{ \alpha + \beta }{ \beta }$

$\rm  \:  =  \: \:( \alpha + \beta)\bigg(\dfrac{1}{ \alpha } + \dfrac{1}{ \beta } \bigg)$

$\rm  \:  =  \: \:( \alpha + \beta)\bigg(\dfrac{ \beta + \alpha }{ \alpha \beta}\bigg)$

$\rm  \:  =  \: \:( - 8) \times \dfrac{( - 8)}{6}$

$\rm  \:  =  \: \:\dfrac{32}{3}$

$\bf\implies \:S \: = \: \dfrac{32}{3}$

Product of zeroes,

$\rm :\longmapsto\:P = \:\dfrac{ \alpha + \beta }{ \alpha } \times \: \dfrac{ \alpha + \beta }{ \beta }$

$\rm  \:  =  \: \:\dfrac{ { (\alpha + \beta ) }^{2} }{ \alpha \beta }$

$\rm  \:  =  \: \:\dfrac{ {( - 8)}^{2} }{6}$

$\rm  \:  =  \: \:\dfrac{64}{6}$

$\rm  \:  =  \: \:\dfrac{32}{3}$

$\bf\implies \:P\: = \: \dfrac{32}{3}$

Hence,

The required Quadratic polynomial is

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - Sx + P\bigg), \: where \: k \: \ne \: 0$

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - \dfrac{32}{3} x + \dfrac{32}{3} \bigg), \: where \: k \: \ne \: 0$

Additional Information :-

$\rm :\longmapsto\:\sf \: \alpha , \beta , \gamma \: are \: zeroes \: of \: {ax}^{3} + {bx}^{2} + cx + d, \: then \:$

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

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

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