Question

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CORRECT ANSWER WILL BE BRAINLIEST
solve the following equation by factorization..... no spam plz... onky factorization
$\frac{m}{n} {x}^{2} + \frac{n}{m} = 1 - 2x$
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Answer 1

Answer:

Hey mate.

Here is your answer -

$- 2\frac{m}{n}$

Step-by-step explanation:

$\frac{m}{n} {x}^{2} + \frac{n}{m} = 1 - 2x \\ \frac{m}{n} x \: + \frac{n}{m} = 1 - 2 \: (x \: is \: cancelled \: out) \\ \frac{m}{n \: } \: x \: + \frac{n}{m} = \: - 1 \\ x + \frac{n}{m } = \: - 1 \times \frac{n}{m} \\ x \: = \: - \frac{n}{m} - \frac{n}{m} \\ therefore \: x \: = - 2 \frac{m}{n}$

Hope it helps!!!

Answer 2

Step-by-step explanation:

$\frac{m}{n} {x}^{2} + \frac{n}{m} = 1 - 2x = > \frac{m}{n} {x}^{2} + 2x$

$+ \frac{n}{m} = 1 = >$

Multiply by mn on both sides we get

$\frac{m}{n} {x}^{2} + x + x + \frac{n}{m} = 1 = >$

$\frac{m}{n}x(x + \frac{n}{m}) + 1(x + \frac{n}{m}) = 1 = >$

$( \frac{m}{n}x + 1)(x + \frac{n}{m}) = 1 = >$

Take m/n as common from first term then we get

$\frac{m}{n} {(x + \frac{n}{m})}^{2} - 1 = 0 = >$

let's take (x + n/m) be y so

$\frac{m}{n} {y}^{2} - 1 = 0 = >$

$( \sqrt{ \frac{m}{n} }y - 1)( \sqrt{ \frac{m}{n}}y + 1) = 0 = >$

Substitute y by (x+ n/m)

Then we get

$( \sqrt{ \frac{m}{n} }(x + \frac{n}{m}) - 1)( \sqrt{ \frac{m}{n} }(x + \frac{n}{m} ) + 1)$

$= 0 = >$

$( \sqrt{ \frac{m}{n} }x + \sqrt{ \frac{n}{m} } + 1)( \sqrt{ \frac{m}{n}}x + \sqrt{ \frac{n}{m}} - 1)$

= 0. Hope it helps you.