Question

If alpha and beta be the roots of ax2 -bx+b=0 the value of rootalpha/beta+rootbeta/alpha is

Answer 1

roots are α and β
$\alpha + \beta = - \frac{b}{a} = - \frac{ - b}{a} \\ \alpha \beta = \frac{c}{a} = \frac{b}{a} \\ \sqrt{ \frac{ \alpha }{ \beta } } + \sqrt{ \frac{ \beta }{ \alpha } } \\ \frac{ \sqrt{ \alpha } }{ \sqrt{ \beta } } + \frac{ \sqrt{ \beta } }{ \sqrt{ \alpha } } \\ \frac{ \alpha + \beta }{ \sqrt{ \alpha \beta } } \\ \frac{ \frac{ - b}{a} }{ \sqrt{ \frac{b}{a} } } \\ - \frac{b \sqrt{a} }{a \sqrt{b} } \\ - \sqrt{ \frac{b}{a} }$
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Answer 2

Step-by-step explanation:

roots are α and β

\alpha + \beta = - \frac{b}{a} = - \frac{ - b}{a} \\ \alpha \beta = \frac{c}{a} = \frac{b}{a} \\ \sqrt{ \frac{ \alpha }{ \beta } } + \sqrt{ \frac{ \beta }{ \alpha } } \\ \frac{ \sqrt{ \alpha } }{ \sqrt{ \beta } } + \frac{ \sqrt{ \beta } }{ \sqrt{ \alpha } } \\ \frac{ \alpha + \beta }{ \sqrt{ \alpha \beta } } \\ \frac{ \frac{ - b}{a} }{ \sqrt{ \frac{b}{a} } } \\ - \frac{b \sqrt{a} }{a \sqrt{b} } \\ - \sqrt{ \frac{b}{a} }

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