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Lampa20
We can use t=x^2 to solve this
Once we do that we will have simple square equation which we know how to solve.

t^2 + 3t + 2 = 0

t1 = -1
t2 = -2

x1 = √-1 = i
x2 = -i

x3 = √-2 = i√2
x4 = -i√2

Make sure you know that i^2 = -1 and (-i)^2 = -1 which gives us solutions we got...
abidemiokin

The solution to the system of equations are ±2i and ±i

Quartic equations

Given the quartic equation

x^4 + 3x^2 + 2 = 0

Let u = x^2 to have:

(x^2)^2 + 3x^2 + 2 = 0

u^2 + 3u + 2 = 0

Factorize

u^2 + u +2u + 2 = 0

u(u + 1) + 2(u + 1) = 0

(u + 2)(u + 1) = 0

u = -2 or -1

If u = x^2

-2 = x^2

x = √-2

x = 2i

Similarly

If u = x^2

-1 = x^2

x = √-1

x = i

Hence the solution to the system of equations are ±2i and ±i

Learn more on equation here: https://brainly.com/question/13763238

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