What are the solutions to the equation (x-2)^2+(x+1)^2=3 ?

The solutions to the equation (x-2)^2+(x+1)^2=3 are x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2

Find the zeros of (x-2)^2+(x+1)^2=3

The zeros of (x-2)^2+(x+1)^2=3 are x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2

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(x - 2)^{2} + (x + 1)^{2} = 3

x = \frac{1}{2} + i \frac{3}{2}, x = \frac{1}{2} - i \frac{3}{2}

Solution steps

(x - 2)^{2} + (x + 1)^{2} = 3

Expand

(x - 2)^{2} + (x + 1)^{2} : 2 x^{2} - 2 x + 5

2 x^{2} - 2 x + 5 = 3

Move

3

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 2 \cdot 2 : 23 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 3 i}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 3 i}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 3 i}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 3 i}{2 \cdot 2} : \frac{1}{2} + i \frac{3}{2}

x = \frac{- ( - 2 ) - 2 3 i}{2 \cdot 2} : \frac{1}{2} - i \frac{3}{2}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{3}{2}, x = \frac{1}{2} - i \frac{3}{2}

\frac{- 2 - 4 i}{7 i}

16 x^{2} + 7 = 12

∣ x + 9 ∣ = ∣ 9 - x ∣

x^{2} - x > 0

What are the solutions to the equation (x-2)^2+(x+1)^2=3 ?
The solutions to the equation (x-2)^2+(x+1)^2=3 are x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2
Find the zeros of (x-2)^2+(x+1)^2=3
The zeros of (x-2)^2+(x+1)^2=3 are x= 1/2+i(sqrt(3))/2 ,x= 1/2-i(sqrt(3))/2