What are the solutions to the equation (2n-1)v^2-2nv+1=0 ?

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

Find the zeros of (2n-1)v^2-2nv+1=0

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

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

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

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

v_{1, 2} = \frac{- ( - 2 n ) \pm 2 ( n - 1 )}{2 ( 2 n - 1 )}; n \neq = \frac{1}{2}

Separate the solutions

v_{1} = \frac{- ( - 2 n ) + 2 ( n - 1 )}{2 ( 2 n - 1 )}, v_{2} = \frac{- ( - 2 n ) - 2 ( n - 1 )}{2 ( 2 n - 1 )}

v = \frac{- ( - 2 n ) + 2 ( n - 1 )}{2 ( 2 n - 1 )} : 1

v = \frac{- ( - 2 n ) - 2 ( n - 1 )}{2 ( 2 n - 1 )} : \frac{1}{2 n - 1}

The solutions to the quadratic equation are:

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

2 x (x - 8) = 0

2 y^{2} = - 40

(x - 2) (2 x - 3) = 0

z^{2} + 0

5 x^{2} - 10 x = 6

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