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 ...

196

You visited us 196 times! Enjoying our articles? Unlock Full Access!

Quadratic Formula

Solve mnx 2+...

Question

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

Open in App

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

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

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

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

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

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

Suggest Corrections

Similar questions

x

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

(m - n) x^{2} + (n - 1) x + 1 - m = 0

1

m

n

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

x - \frac{1}{2}

n x^{2} - 5 x + m

m = n = 2

x - 2

x - \frac{1}{2}

n x^{2} - 5 x + m

m = n = 2

Join BYJU'S Learning Program