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

The solutions to the equation (x^2)/2-x+1=0 are x=1+i,x=1-i

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

The zeros of (x^2)/2-x+1=0 are x=1+i,x=1-i

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

x = 1 + i, x = 1 - i

Solution steps

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

Multiply both sides by

2

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1 + i, x = 1 - i

(x^{6})^{\frac{2}{3}}

\frac{3}{5} \times \frac{2}{7}

8 m^{2} - 2 m - 7 = 0

(x^{\frac{2}{3}})^{4}

\frac{- x + 2}{x - 4} \geq 0

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