What are the solutions to the equation x(x-2α+1)+α^2-1=0 ?

The solutions to the equation x(x-2α+1)+α^2-1=0 are x=(2α-1+sqrt(-4α+5))/2 ,x=(2α-1-sqrt(-4α+5))/2

Find the zeros of x(x-2α+1)+α^2-1=0

The zeros of x(x-2α+1)+α^2-1=0 are x=(2α-1+sqrt(-4α+5))/2 ,x=(2α-1-sqrt(-4α+5))/2

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x (x - 2 α + 1) + α^{2} - 1 = 0

x = \frac{2 α - 1 + - 4 α + 5}{2}, x = \frac{2 α - 1 - - 4 α + 5}{2}

Solution steps

x (x - 2 α + 1) + α^{2} - 1 = 0

Expand

x (x - 2 α + 1) + α^{2} - 1 : x^{2} - 2 αx + x + α^{2} - 1

x^{2} - 2 αx + x + α^{2} - 1 = 0

Write in the standard form

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

x^{2} + (- 2 α + 1) x + α^{2} - 1 = 0

Solve with the quadratic formula

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

Simplify

(- 2 α + 1)^{2} - 4 \cdot 1 \cdot (α^{2} - 1) : - 4 α + 5

x_{1, 2} = \frac{- ( - 2 α + 1 ) \pm - 4 α + 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 α + 1 ) + - 4 α + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 2 α + 1 ) - - 4 α + 5}{2 \cdot 1}

x = \frac{- ( - 2 α + 1 ) + - 4 α + 5}{2 \cdot 1} : \frac{2 α - 1 + - 4 α + 5}{2}

x = \frac{- ( - 2 α + 1 ) - - 4 α + 5}{2 \cdot 1} : \frac{2 α - 1 - - 4 α + 5}{2}

The solutions to the quadratic equation are:

x = \frac{2 α - 1 + - 4 α + 5}{2}, x = \frac{2 α - 1 - - 4 α + 5}{2}

4 x^{2} = 1 - 2 x

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

x^{2} - 64 = 0

4 x^{2} + 12 x

6 x^{2} = - 19 x - 15

What are the solutions to the equation x(x-2α+1)+α^2-1=0 ?
The solutions to the equation x(x-2α+1)+α^2-1=0 are x=(2α-1+sqrt(-4α+5))/2 ,x=(2α-1-sqrt(-4α+5))/2
Find the zeros of x(x-2α+1)+α^2-1=0
The zeros of x(x-2α+1)+α^2-1=0 are x=(2α-1+sqrt(-4α+5))/2 ,x=(2α-1-sqrt(-4α+5))/2