What are the solutions to the equation solvefor a,2-j^4=(a+3jb)^2 ?

The solutions to the equation solvefor a,2-j^4=(a+3jb)^2 are a=sqrt(2-j^4)-3jb,a=-sqrt(2-j^4)-3jb

Find the zeros of solvefor a,2-j^4=(a+3jb)^2

The zeros of solvefor a,2-j^4=(a+3jb)^2 are a=sqrt(2-j^4)-3jb,a=-sqrt(2-j^4)-3jb

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solve for

a, 2 - j^{4} = (a + 3 jb)^{2}

a = 2 - j^{4} - 3 jb, a = - 2 - j^{4} - 3 jb

Solution steps

2 - j^{4} = (a + 3 jb)^{2}

Switch sides

(a + 3 jb)^{2} = 2 - j^{4}

For

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

the solutions are

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

Solve

a + 3 jb = 2 - j^{4} : a = 2 - j^{4} - 3 jb

Solve

a + 3 jb = - 2 - j^{4} : a = - 2 - j^{4} - 3 jb

The solutions to the quadratic equation are:

a = 2 - j^{4} - 3 jb, a = - 2 - j^{4} - 3 jb

7 x^{2} + 18 x = 10 x^{2} + 12 x

- 25 x^{2} - 30 x - 9 = 0

\frac{q ( q + 7 )}{7} - q = 4 - q^{2}

3 y^{2} + 9 y + 15

a x^{2} - (a + 2) x - 1 = 0

What are the solutions to the equation solvefor a,2-j^4=(a+3jb)^2 ?
The solutions to the equation solvefor a,2-j^4=(a+3jb)^2 are a=sqrt(2-j^4)-3jb,a=-sqrt(2-j^4)-3jb
Find the zeros of solvefor a,2-j^4=(a+3jb)^2
The zeros of solvefor a,2-j^4=(a+3jb)^2 are a=sqrt(2-j^4)-3jb,a=-sqrt(2-j^4)-3jb