Question

Find the roots of equation, if they exists, by applying the quadratic formula :

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

Explanation needed

Answer 1

hey mate
here's the solution

Answer 2

Given Equation is (m/n)x^2 + (n/m) = 1 - 2x.

⇒ m²x² + n² = mn(1 - 2x)

⇒ m²x² + n² = mn - 2mnx

⇒ m²x² + 2mnx + n² - mn = 0.

Here, a = m^2, b = 2mn, c = (n² - mn).

The solutions are:

(i)

$=>\frac{-b+\sqrt{b^2 - 4ac}}{2a}$

$=>\frac{-2mn+\sqrt{(2mn)^2-4(m^2)(n^2-mn)}}{2m^2}$

$=>\frac{-2mn+\sqrt{4m^2n^2-4m^2n^2+4m^3n}}{2m^2}$

$=>\frac{-2mn+ \sqrt{4m^3n}}{2m^2}$

$=>\frac{-2mn+ \sqrt{4m^2*mn} }{2m^2}$

$=>\frac{-2mn+2m\sqrt{mn}}{2m^2}$

$=>\frac{2m(-n+\sqrt{mn})}{2m^2}$

$=> \frac{-n+\sqrt{mn}}{m}$

(ii)

$=> x=\frac{-b-\sqrt{b^2 - 4ac}}{2a}$

$=>\frac{-2mn-\sqrt{(2mn)^2-4(m^2)(n^2-mn)}}{2m^2}$

$=>\frac{-2mn-\sqrt{4m^2n^2-4m^2n^2+4mn}}{2m^2}$

$=>\frac{-2mn-\sqrt{4mn}}{2m^2}$

$=>\frac{-2mn-2m\sqrt{mn}}{2m^2}$

$=>\frac{2m(-n-\sqrt{mn})}{2m^2}$

$=> \frac{-n- \sqrt{mn}}{m}$

Therefore, the roots of the equation are:

$=>x=\boxed{\frac{-n+ \sqrt{mn}}{m},\frac{-n- \sqrt{mn}}{m}}$

Hope it helps!