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quadratic equation with roots -1+4i and -1-4i?
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(-1+4i)(-1-4i)
Can you help me with three more??
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find the zeroes of the equation -3x^4+27x^2+1200=0
On the first one i meant: (x-(-1+4i))(x-(-1-4i)) =(x+1-4i)(x+1+4i) because if like a is a root then, x-a is a factor
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-3(x^4-9x^2-400)=0 -3((x^2)^2-9x^2-400)=0 -3(x^2-9/2)^2 - (9/2)^2-400=0 -3(x^2-9/2)^2 - (1681/4)=0 -3(x^2-9/2)^2- (41/2)^2=0 -3(x^2+16)(x^2-25)=0 x=5, x=-5 and x=4i, x=-4i
I used complete square method, to solve it
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