What are the solutions to the equation solvefor x,x^2-4y^2-2x+1=0 ?

The solutions to the equation solvefor x,x^2-4y^2-2x+1=0 are x=1+2y,x=1-2y

Find the zeros of solvefor x,x^2-4y^2-2x+1=0

The zeros of solvefor x,x^2-4y^2-2x+1=0 are x=1+2y,x=1-2y

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solve for

x, x^{2} - 4 y^{2} - 2 x + 1 = 0

x = 1 + 2 y, x = 1 - 2 y

Solution steps

x^{2} - 4 y^{2} - 2 x + 1 = 0

Write in the standard form

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

x^{2} - 2 x - 4 y^{2} + 1 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 y}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 4 y}{2 \cdot 1}

x = \frac{- ( - 2 ) + 4 y}{2 \cdot 1} : 1 + 2 y

x = \frac{- ( - 2 ) - 4 y}{2 \cdot 1} : 1 - 2 y

The solutions to the quadratic equation are:

x = 1 + 2 y, x = 1 - 2 y

36 x^{2} - 60 x + 25 = 0

3 x^{2} - 8 x + 8 = 0

m^{2} - m = 0

3 x - x^{2} = 0

4 x^{2} + 89 = - 36 x

What are the solutions to the equation solvefor x,x^2-4y^2-2x+1=0 ?
The solutions to the equation solvefor x,x^2-4y^2-2x+1=0 are x=1+2y,x=1-2y
Find the zeros of solvefor x,x^2-4y^2-2x+1=0
The zeros of solvefor x,x^2-4y^2-2x+1=0 are x=1+2y,x=1-2y