Question

$x^{2} -3x+2=0,\frac{\alpha }{\beta +\frac{\beta }{\alpha }$

Answer 1

Step-by-step explanation:

Correct question -

$x^{2} -3x+2=0 \: , \: \: \: \: find \: → \large\frac{ \alpha }{ \beta + \frac{ \beta }{ \alpha } }$

Solution -

${x}^{2} - 3x + 2 = 0$

$→ {x}^{2} - x - 2x + 2 = 0$

$→x( x - 1) - 2(x - 1) = 0$

$→(x - 2)(x - 1) = 0$

$→x = 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: , \: \: \: \: \: \: \: \: \: \: \: \: \: x = 2$

Case 1 -

$let \: \: \alpha = 1 \: \: and \: \: \beta = 2$

$Now \: , \: \large\frac{ \alpha }{ \beta + \frac{ \beta }{ \alpha } } = \large\frac{ 1 }{ 2 + \frac{ 2 }{ 1 } }$

$→\large\frac{ \alpha }{ \beta + \frac{ \beta }{ \alpha } } = \large\frac{ 1 }{ 2 + { 2 }}$

$→\large \orange{\frac{ \alpha }{ \beta + \frac{ \beta }{ \alpha } } = \large\frac{ 1 }{ 4}}$

Case 2 -

$let \: \: \alpha = 2 \: \: and \: \: \beta = 1$

$Now \: , \: \large\frac{ \alpha }{ \beta + \frac{ \beta }{ \alpha } } = \Large\frac{ 2 }{ 1 + \frac{ 1 }{ 2 } }$

$→\large\frac{ \alpha }{ \beta + \frac{ \beta }{ \alpha } } = \Large\frac{ 2 }{ \frac{2 + 1 }{ 2 } }$

$→ \large\frac{ \alpha }{ \beta + \frac{ \beta }{ \alpha } } = \Large\frac{ 2 }{ \frac{3 }{ 2 } }$

$→\large\frac{ \alpha }{ \beta + \frac{ \beta }{ \alpha } } = \Large\frac{ 2 × 2 }{{3 }}$

$→\large \green{\frac{ \alpha }{ \beta + \frac{ \beta }{ \alpha } } = \Large\frac{ 4 }{{3 }}}$