Question

If α and β are the zeroes of the polynomial x²-4x+1 then the value of α+ 1/α is​

Answer 1

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

Given quadratic polynomial is

$\red{\rm :\longmapsto\: {x}^{2} - 4x + 1}$

Further, given that

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: {x}^{2} - 4x + 1$

We know,

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

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

$\rm \implies\: \alpha \beta = 1$

$\rm \implies\: \beta = \dfrac{1}{ \alpha } - - - (1)$

Further, we know that

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

$\rm \implies\: \alpha + \beta = - \: \dfrac{( - 4)}{1}$

$\rm \implies\: \alpha + \dfrac{1}{ \alpha } = 4$

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More to Know

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

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

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

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