What are the solutions to the equation 2x-1=x^2 ?

The solutions to the equation 2x-1=x^2 are x=1

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2x-1=x^2

2 x - 1 = x^{2}

x = 1

Solution steps

2 x - 1 = x^{2}

Switch sides

x^{2} = 2 x - 1

Move

1

to the left side

x^{2} + 1 = 2 x

Move

2 x

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 1 \cdot 1 = 0

x_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 \cdot 1}

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

\frac{- ( - 2 )}{2 \cdot 1} = 1

x = 1

The solution to the quadratic equation is:

x = 1

3 (2 x + 2) > 24 - 4 x

v^{2} - 49

2^{x + 2} - 2^{x + 1} + 2^{x - 1} - 2^{x - 2} \leq 9

c^{2} - 12 c + 35

- 5

What are the solutions to the equation 2x-1=x^2 ?
The solutions to the equation 2x-1=x^2 are x=1
Find the zeros of 2x-1=x^2
The zeros of 2x-1=x^2 are x=1