How do you find all the solutions of the equation #a^4-a^3+a^2+2=0#?

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Answer 1 · 2018-06-21T13:54:40

#(a^2+a+1)(a^2-2a+2)#

We have

#(a^2+a+1)(a^2-2a+2)=a^4+a^3+a^2-2a^3-2a^2-2a+2a^2+2a+2#

Answer 2 · 2018-06-22T16:05:53

#a^4-a^3+a^2+2#

#=>a^4+a^2+1-a^3+1=0#

#=>(a^2)^2+2a^2+1-a^2-(a^3-1)=0#

#=>(a^2+1)^2-a^2-(a-1)(a^2+a+1)=0#

#=>(a^2+a+1)(a^2-a+1)-(a-1)(a^2+a+1)=0#

#=>(a^2+a+1)(a^2-a+1-a+1)=0#

#=>(a^2+a+1)(a^2-2a+2)=0#

When

#a^2+a+1=0#

#=>a=(-1pmsqrt(1^1-4*1))/2#

#=>a=(-1pmsqrt(-3))/2#

When

#a^2-2a+2=0#

#=>a=(2pmsqrt(4-4*1*2))/2#

#=>a=(2pmsqrt(4-8))/2#

#=>a=(2pmsqrt(-4))/2#

#=>a=(2pm2sqrt(-1))/2#

#=>a=1pmsqrt(-1)#