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

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

Find the zeros of x(x-2a+1)+a^2-1=0

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

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x (x - 2 a + 1) + a^{2} - 1 = 0

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

Solution steps

x (x - 2 a + 1) + a^{2} - 1 = 0

Expand

x (x - 2 a + 1) + a^{2} - 1 : x^{2} - 2 a x + x + a^{2} - 1

x^{2} - 2 a x + x + a^{2} - 1 = 0

Write in the standard form

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

x^{2} + (- 2 a + 1) x + a^{2} - 1 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

y^{2} + 9 y - 22 = 0

x^{2} - 2021 x + 7 (x - 2021) = 0

p^{2} - 2 p - 24 = 0

1450 \frac{V ^{2}}{24} = 1450 \cdot 9.8

5 x^{2} + 13 x - 28 = 0

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