Question

if alpha and beta are the roots of the equation ax2+bx+c=0 then find √alpha / beta + √beta/ alpha = √b/a​

Answer 1

Step-by-step explanation:

We have,

$\tt{ \green{ax^{2} - bx + c = 0}}$

Roots are α and β

so,

$\sf{ \purple{ Sum \: \: of \: \: roots} , \alpha + \beta = - \dfrac{( - b)}{a} = \dfrac{b}{a} }$

$\sf{ \purple{ Product \: \: of \: \: roots} , \alpha \beta = \dfrac{c}{a} }$

Now,

$\sf{ \sqrt{ \dfrac{ \alpha }{ \beta }} + \sqrt{ \dfrac{ \beta }{ \alpha } } }$

$\sf{ = \dfrac{ \alpha + \beta }{ \sqrt{ \alpha \beta }} }$

$\sf{ = \dfrac{ \dfrac{b}{a} }{ \sqrt{ \dfrac{c}{a} }} }$

$\sf{ = \dfrac{ b }{ \sqrt{ ac }} }$

Answer 2

Appropriate Question :-

$\sf \: \alpha, \beta \: are \: the \: roots \: of \: {ax}^{2} - bx + b = 0, \: prove \: that$

$\sf \: \sqrt{\dfrac{ \alpha }{ \beta } } + \sqrt{\dfrac{ \beta }{ \alpha } } = \sqrt{\dfrac{b}{a} }$

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

Given that,

$\rm :\longmapsto\:\alpha, \beta \: are \: the \: roots \: of \: {ax}^{2} - bx + b = 0, \:$

We know

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

$\rm \implies\: \alpha + \beta = - \: \dfrac{( - b)}{a}$

$\rm \implies\: \alpha + \beta = \: \dfrac{b}{a}$

And

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

$\rm \implies\: \alpha \beta = \dfrac{b}{a}$

Now, Consider

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

$\rm \:  =  \: \dfrac{ \alpha + \beta }{ \sqrt{ \alpha \beta } }$

$\rm \:  =  \: \dfrac{\dfrac{b}{a} }{ \sqrt{\dfrac{b}{a} } }$

$\rm \:  =  \: \sqrt{\dfrac{b}{a} }$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \sqrt{\dfrac{ \alpha }{ \beta } } + \sqrt{\dfrac{ \beta }{ \alpha } } = \sqrt{\dfrac{b}{a} } \: }}$

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More to know :-

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

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

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

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