What are the solutions to the equation (f-2)^2=-4f^2 ?

The solutions to the equation (f-2)^2=-4f^2 are f= 2/5+i 4/5 ,f= 2/5-i 4/5

Find the zeros of (f-2)^2=-4f^2

The zeros of (f-2)^2=-4f^2 are f= 2/5+i 4/5 ,f= 2/5-i 4/5

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(f - 2)^{2} = - 4 f^{2}

f = \frac{2}{5} + i \frac{4}{5}, f = \frac{2}{5} - i \frac{4}{5}

Solution steps

(f - 2)^{2} = - 4 f^{2}

Expand

(f - 2)^{2} : f^{2} - 4 f + 4

f^{2} - 4 f + 4 = - 4 f^{2}

Move

4 f^{2}

to the left side

5 f^{2} - 4 f + 4 = 0

Solve with the quadratic formula

f_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 5 \cdot 4}{2 \cdot 5}

Simplify

(- 4)^{2} - 4 \cdot 5 \cdot 4 : 8 i

f_{1, 2} = \frac{- ( - 4 ) \pm 8 i}{2 \cdot 5}

Separate the solutions

f_{1} = \frac{- ( - 4 ) + 8 i}{2 \cdot 5}, f_{2} = \frac{- ( - 4 ) - 8 i}{2 \cdot 5}

f = \frac{- ( - 4 ) + 8 i}{2 \cdot 5} : \frac{2}{5} + i \frac{4}{5}

f = \frac{- ( - 4 ) - 8 i}{2 \cdot 5} : \frac{2}{5} - i \frac{4}{5}

The solutions to the quadratic equation are:

f = \frac{2}{5} + i \frac{4}{5}, f = \frac{2}{5} - i \frac{4}{5}

x (x - 1) - 2 x + 2 = 0

9 a^{2} + 46 a = - 5

- 6 + 3 x^{2} = - 39

2 x^{2} - x - 28 = 0

100 x^{2} = 0

What are the solutions to the equation (f-2)^2=-4f^2 ?
The solutions to the equation (f-2)^2=-4f^2 are f= 2/5+i 4/5 ,f= 2/5-i 4/5
Find the zeros of (f-2)^2=-4f^2
The zeros of (f-2)^2=-4f^2 are f= 2/5+i 4/5 ,f= 2/5-i 4/5