What are the solutions to the equation (1-x)(1-x)+4=0 ?

The solutions to the equation (1-x)(1-x)+4=0 are x=-2i+1,x=2i+1

Find the zeros of (1-x)(1-x)+4=0

The zeros of (1-x)(1-x)+4=0 are x=-2i+1,x=2i+1

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(1 - x) (1 - x) + 4 = 0

x = - 2 i + 1, x = 2 i + 1

Solution steps

(1 - x) (1 - x) + 4 = 0

Move

4

to the right side

(1 - x) (1 - x) = - 4

Simplify

(1 - x)^{2} = - 4

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

1 - x = - 4 : x = - 2 i + 1

Solve

1 - x = - - 4 : x = 2 i + 1

The solutions to the quadratic equation are:

x = - 2 i + 1, x = 2 i + 1

m^{2} + 11 m + 16 = 0

x^{2} - 10 x + c

2 (x^{2} - 7) + 3 = - 3

x^{2} + 20 x + 86 = 0

\frac{x ^{2}}{8} = \frac{24 - x ^{2}}{16}

What are the solutions to the equation (1-x)(1-x)+4=0 ?
The solutions to the equation (1-x)(1-x)+4=0 are x=-2i+1,x=2i+1
Find the zeros of (1-x)(1-x)+4=0
The zeros of (1-x)(1-x)+4=0 are x=-2i+1,x=2i+1