Question

find the roots(class X) :-
$\frac{m}{n} {x}^{2} + \frac{n}{m} = 1 - 2x$
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Answer 1

Step-by-step explanation:

$\dfrac{m}{n} {x}^{2} + \dfrac{n}{m} = 1 - 2x \\ multiplying \: whole \: equation \: by \: \dfrac{n}{m} \\ {x}^{2} + \dfrac{ {n}^{2} }{ {m}^{2} } = \dfrac{n}{m} - 2 \dfrac{n}{m} x \\ {x}^{2} + 2 \dfrac{n}{m} x + \dfrac{ {n}^{2} }{ {m}^{2} } = \dfrac{n}{m} \\ {(x + \dfrac{n}{m} )}^{2} = {(\sqrt{ \frac{n}{m} } \: )}^{2} \\ if \\ x + \dfrac{n}{m} = \sqrt{ \frac{n}{m} } \\ x = \sqrt{ \dfrac{n}{m} } - \dfrac{n}{m} \\ if \\ x + \dfrac{n}{m} = \sqrt{ \dfrac{n}{m} } \\ x = \sqrt{ \dfrac{n}{m} } - \dfrac{n}{m}$