mnx^2 + nm = 1 2x ⇒m^2x^2 + n^2 = mn 2mn x ⇒m^2x^2 + 2mnx + n^2 mn = 0 Here, a = m^2, b = 2mn, c = n^2 mn By using the quadratic ...

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Quadratic Formula

Solve mnx 2+...

Question

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

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Solution

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

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

$\Rightarrow m^{2} x^{2} + 2 m n x + n^{2} - m n = 0$

Here, $a = m^{2}, b = 2 m n, c = n^{2} - m n$

By using the quadratic formula, we have

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$\Rightarrow x = \frac{- 2 m n \pm \sqrt{{(2 m n)}^{2} - 4 m^{2} (n^{2} - m n)}}{2 m^{2}}$

$\Rightarrow x = \frac{- 2 m n \pm \sqrt{4 m^{2} n^{2} - 4 m^{2} n^{2} + 4 m^{3} n}}{2 m^{2}}$

$\Rightarrow x = \frac{- 2 m n \pm 2 m \sqrt{m n}}{2 m^{2}}$

$\Rightarrow x = \frac{- n \pm \sqrt{m n}}{m}$

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

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Find the roots of each of the following equations, if they exist, by applying the quadratic formula:
$\frac{m}{n} x^{2} + \frac{n}{m} = 1 - 2 x$

x - 2

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