What are the solutions to the equation 2nx^2=x+1/n ?

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

Find the zeros of 2nx^2=x+1/n

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

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

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

Solution steps

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

Multiply both sides by

n

2 n^{2} x^{2} = n x + 1; n \neq = 0

Move

1

to the left side

2 n^{2} x^{2} - 1 = n x; n \neq = 0

Move

n x

to the left side

2 n^{2} x^{2} - 1 - n x = 0; n \neq = 0

Write in the standard form

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

2 n^{2} x^{2} - n x - 1 = 0; n \neq = 0

Solve with the quadratic formula

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

Simplify

(- n)^{2} - 4 \cdot 2 n^{2} (- 1) : 3 n

x_{1, 2} = \frac{- ( - n ) \pm 3 n}{2 \cdot 2 n ^{2}}; n \neq = 0

Separate the solutions

x_{1} = \frac{- ( - n ) + 3 n}{2 \cdot 2 n ^{2}}, x_{2} = \frac{- ( - n ) - 3 n}{2 \cdot 2 n ^{2}}

x = \frac{- ( - n ) + 3 n}{2 \cdot 2 n ^{2}} : \frac{1}{n}

x = \frac{- ( - n ) - 3 n}{2 \cdot 2 n ^{2}} : - \frac{1}{2 n}

The solutions to the quadratic equation are:

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

- x^{2} - 5 x - 4 = 0

x^{2} - 10 x = 10

x^{2} + 8 x + 12 = 0

4 (x + 1)^{2} = 36

x^{2} + 20 = 7 x

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