Asked by Nick
                Solve x^4 - 12x - 5 = 0 given that -1 + 2i is a root.
I’ve used synthetic division to get the answers x = 1 +/- 2i , 1 +/- sqrt 2
I’ve having trouble checking the answers though because of that imaginary number. How can I check these kind of problems?
            
        I’ve used synthetic division to get the answers x = 1 +/- 2i , 1 +/- sqrt 2
I’ve having trouble checking the answers though because of that imaginary number. How can I check these kind of problems?
Answers
                    Answered by
            Damon
            
    If (-1+2i) is a solution, its complex conjugate (-1-2i) must be a solution.
In other words
x = -1+2i
x = -1-2i
are solutions
so
x+1-2iand x+1+2i are both factors
so their product should be a factor
I get x^2+2x+5
now do your division
I get
x^2-2x-1
well
that is
(x-1)(x-1)
so my four solutions are those two complex conjugates and x=1 and x =1 again
    
In other words
x = -1+2i
x = -1-2i
are solutions
so
x+1-2iand x+1+2i are both factors
so their product should be a factor
I get x^2+2x+5
now do your division
I get
x^2-2x-1
well
that is
(x-1)(x-1)
so my four solutions are those two complex conjugates and x=1 and x =1 again
                    Answered by
            Damon
            
    Whoa - a bit hasty there
x^2-2x-1
can not factor that so easily!
I will have to solve it
x^2-2x-1 = 0
x = (1/2)(2 +/-sqrt(4+4))
=(1/2)(2 +/-2sqrt(2))
x = 1+sqrt(2)
or
x = 1-sqrt(2)
are the last two solutions
    
x^2-2x-1
can not factor that so easily!
I will have to solve it
x^2-2x-1 = 0
x = (1/2)(2 +/-sqrt(4+4))
=(1/2)(2 +/-2sqrt(2))
x = 1+sqrt(2)
or
x = 1-sqrt(2)
are the last two solutions
                    Answered by
            SivaArumugam
            
    Solve x^4 - 12x - 5 = 0 given that -1 + 2i is a root.
    
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