Question

Solve:

${x}^{2} + ( \frac{a}{a + b} + \frac{a + b}{a} )x + 1 = 0$

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Answer 1

Hey friend

See attachment ☝


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Answer 2

$\mathsf{ \huge{ \green{Dear \: \: \: User!}}}$



$\cancel{\cancel{\implies{ \huge{ \bold{ \blue{Question \ratio}}}}}}$




$\implies{ \mathsf{ \red{{x}^{2} + ( \frac{a}{a + b} + \frac{a + b}{a} )x + 1 = 0}}} \\ \\ \\ \\ \implies{\mathsf{ \red{{x}^{2} + \frac{a}{a + b}x + \frac{a + b}{a} x + \frac{a}{a + b} \times \frac{a + b}{a} = 0}}} \\ \\ \\ \\ \implies{ \mathsf{ \green{x(x + \frac{a}{a + b}) + \frac{a + b}{a} (x + \frac{a}{a + b} )= 0 }}} \\ \\ \\ \implies{ \mathsf{ \green{(x + \frac{a}{a + b} )(x + \frac{a + b}{a} ) = 0}}} \\ \\ \\ \implies{ \mathsf{ \green{(x + \frac{a + b}{a}) = 0 \: \: or \: \: (x + \frac{a}{a + b} ) = 0}}} \\ \\ \\ \\ \implies{ \mathsf{ \blue{x = \frac{ - a}{a + b} \: \: \: or \: \: \: x = \frac{ - (a + b)}{a}}}}$