What are the solutions to the equation solvefor x,(x+y_{i})^2=3-4i ?

The solutions to the equation solvefor x,(x+y_{i})^2=3-4i are x=-y_{i}+2-i,x=-y_{i}-2+i

Find the zeros of solvefor x,(x+y_{i})^2=3-4i

The zeros of solvefor x,(x+y_{i})^2=3-4i are x=-y_{i}+2-i,x=-y_{i}-2+i

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solve for

x, (x + y_{i})^{2} = 3 - 4 i

x = - y_{i} + 2 - i, x = - y_{i} - 2 + i

Solution steps

(x + y_{i})^{2} = 3 - 4 i

Expand

(x + y_{i})^{2} : x^{2} + 2 x y_{i} + (y_{i})^{2}

x^{2} + 2 x y_{i} + (y_{i})^{2} = 3 - 4 i

Subtract

3 - 4 i

from both sides

x^{2} + 2 x y_{i} + (y_{i})^{2} - (3 - 4 i) = 3 - 4 i - (3 - 4 i)

Simplify

x^{2} + 2 y_{i} x + (y_{i})^{2} - 3 + 4 i = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 y _{i} \pm ( 2 y _{i} ) ^{2} - 4 \cdot 1 \cdot ( ( y _{i} ) ^{2} - 3 + 4 i )}{2 \cdot 1}

Simplify

(2 y_{i})^{2} - 4 \cdot 1 \cdot ((y_{i})^{2} - 3 + 4 i) : 4 - 2 i

x_{1, 2} = \frac{- 2 y _{i} \pm ( 4 - 2 i )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 y _{i} + 4 - 2 i}{2 \cdot 1}, x_{2} = \frac{- 2 y _{i} - ( 4 - 2 i )}{2 \cdot 1}

x = \frac{- 2 y _{i} + 4 - 2 i}{2 \cdot 1} : - y_{i} + 2 - i

x = \frac{- 2 y _{i} - ( 4 - 2 i )}{2 \cdot 1} : - y_{i} - 2 + i

The solutions to the quadratic equation are:

x = - y_{i} + 2 - i, x = - y_{i} - 2 + i

x, 4 x^{2} + 9 x + 3 = 0

2 x^{2} - 3 x + \frac{5}{8} = 0

5 x^{2} - 3 x + 1 = - 3 x^{2} - 2 x

a^{2} + 24 a + 144 = 0

5 x^{2} - 50 x + 30 = 0

What are the solutions to the equation solvefor x,(x+y_{i})^2=3-4i ?
The solutions to the equation solvefor x,(x+y_{i})^2=3-4i are x=-y_{i}+2-i,x=-y_{i}-2+i
Find the zeros of solvefor x,(x+y_{i})^2=3-4i
The zeros of solvefor x,(x+y_{i})^2=3-4i are x=-y_{i}+2-i,x=-y_{i}-2+i