What are the solutions to the equation mnx^2=mn-(m^2-n^2)x ?

The solutions to the equation mnx^2=mn-(m^2-n^2)x are x= n/m ,x=-m/n ;\quad mn\ne 0

Find the zeros of mnx^2=mn-(m^2-n^2)x

The zeros of mnx^2=mn-(m^2-n^2)x are x= n/m ,x=-m/n ;\quad mn\ne 0

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mn x^{2} = mn - (m^{2} - n^{2}) x

x = \frac{n}{m}, x = - \frac{m}{n}; mn \neq = 0

Solution steps

mn x^{2} = mn - (m^{2} - n^{2}) x

Expand

mn - (m^{2} - n^{2}) x : mn - m^{2} x + n^{2} x

mn x^{2} = mn - m^{2} x + n^{2} x

Move

n^{2} x

to the left side

mn x^{2} - n^{2} x = mn - m^{2} x

Move

m^{2} x

to the left side

mn x^{2} - n^{2} x + m^{2} x = mn

Move

mn

to the left side

mn x^{2} - n^{2} x + m^{2} x - mn = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - n ^{2} + m ^{2} ) \pm ( - n ^{2} + m ^{2} ) ^{2} - 4 mn ( - mn )}{2 mn}

Simplify

(- n^{2} + m^{2})^{2} - 4 mn (- mn) : m^{2} + n^{2}

x_{1, 2} = \frac{- ( - n ^{2} + m ^{2} ) \pm ( m ^{2} + n ^{2} )}{2 mn}; mn \neq = 0

Separate the solutions

x_{1} = \frac{- ( - n ^{2} + m ^{2} ) + m ^{2} + n ^{2}}{2 mn}, x_{2} = \frac{- ( - n ^{2} + m ^{2} ) - ( m ^{2} + n ^{2} )}{2 mn}

x = \frac{- ( - n ^{2} + m ^{2} ) + m ^{2} + n ^{2}}{2 mn} : \frac{n}{m}

x = \frac{- ( - n ^{2} + m ^{2} ) - ( m ^{2} + n ^{2} )}{2 mn} : - \frac{m}{n}

The solutions to the quadratic equation are:

x = \frac{n}{m}, x = - \frac{m}{n}; mn \neq = 0

x^{2} + 3 x - 0.9 = 0

4.905 x^{2} - 90 x - 150 = 0

x^{2} - 10 x + 25 = 0

a, 2 - j^{4} = (a + 3 jb)^{2}

7 x^{2} + 18 x = 10 x^{2} + 12 x

What are the solutions to the equation mnx^2=mn-(m^2-n^2)x ?
The solutions to the equation mnx^2=mn-(m^2-n^2)x are x= n/m ,x=-m/n ;\quad mn\ne 0
Find the zeros of mnx^2=mn-(m^2-n^2)x
The zeros of mnx^2=mn-(m^2-n^2)x are x= n/m ,x=-m/n ;\quad mn\ne 0