Solving a Quadratic Equation: Finding the Four Roots of Z⁴ + 4 = 0

[becka12]

the book asks to fnd the four roots of z to the fourth power + 4 = 0 and then use to demonstrate that z to the fourth power + 4 can be factored into two quadratics with real coefficinets. I am clueless on where to start. Please help.

 

[dmoravec]

start by setting $$z^4=-4$$ and solve for z taking the square root function twice (saving both roots)

 

[becka12]

Ok so I did what you said and then did the quadratic formula so i have 4 roots but how do I form quadratic formulas?

 

[dmoravec]

so let a,b,c,d be your roots. the general form of your equation can be written as (z-a)(z-b)(z-c)(z-d)=0. From that you can pick any two, multiply out and you'll have a quadratic ie: $$(z^2-bz-az+ab)(z-c)(z-d)=0$$. Since your roots are complex and you need real coefficients you might have to try different combinations

 

[HallsofIvy]

Ok so I did what you said and then did the quadratic formula so i have 4 roots but how do I form quadratic formulas?

Let y= z² so that you have a quadratic equation
y²+ 4= 0. What are the two solutions to that? Then, for each of the two values of y, set z²= y and take square roots.

 

[FrostScYthe]

Ok, I noticed something while playing with your equation, you can factor out z^4 + 4 = 0 into
(z^2 + 2i)(z^2 - 2i)
then
(z^2 + 2i) = (z + 1 - i)(z - 1 + i)
(z^2 - 2i) = (z + 1 + i)(z - 1 - i)

multiply
(z + 1 - i)(z + 1 + i) = z^2 + 2z + 2
(z - 1 + i)(z - 1 - i) = z^2 - 2z + 2

and there you have your two real number quadratics