What are the solutions to the equation 2mnx^2+nx-2mx-1=0 ?

The solutions to the equation 2mnx^2+nx-2mx-1=0 are x= 1/n ,x=-1/(2m);\quad 2mn\ne 0

Find the zeros of 2mnx^2+nx-2mx-1=0

The zeros of 2mnx^2+nx-2mx-1=0 are x= 1/n ,x=-1/(2m);\quad 2mn\ne 0

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

x = \frac{1}{n}, x = - \frac{1}{2 m}; 2 mn \neq = 0

Solution steps

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(n - 2 m)^{2} - 4 \cdot 2 mn (- 1) : 2 m + n

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

Separate the solutions

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

x = \frac{- ( n - 2 m ) + 2 m + n}{2 \cdot 2 mn} : \frac{1}{n}

x = \frac{- ( n - 2 m ) - ( 2 m + n )}{2 \cdot 2 mn} : - \frac{1}{2 m}

The solutions to the quadratic equation are:

x = \frac{1}{n}, x = - \frac{1}{2 m}; 2 mn \neq = 0

x^{2} + (\frac{4}{5})^{2} = 1

0 = y^{2} - 4 y

(\frac{3}{2} x - \frac{1}{2})^{2} - 0.8 (1 - x) = 0

(200 + 20 x) (40 - 2 x) = 9000

t = \frac{1}{2} \cdot (- 9.81) \cdot t^{2} + 20 \cdot t + 5

What are the solutions to the equation 2mnx^2+nx-2mx-1=0 ?
The solutions to the equation 2mnx^2+nx-2mx-1=0 are x= 1/n ,x=-1/(2m);\quad 2mn\ne 0
Find the zeros of 2mnx^2+nx-2mx-1=0
The zeros of 2mnx^2+nx-2mx-1=0 are x= 1/n ,x=-1/(2m);\quad 2mn\ne 0