There is a general solution to the 4th power equation, but it is very complex. Fortunately this equation can be solved quite easily.
x4−2x3+4x2+6x−21=(x2+a)(x2+bx+c)=0x^4-2x^3+4x^2+6x-21=(x^2+a)(x^2+bx+c)=0x4−2x3+4x2+6x−21=(x2+a)(x2+bx+c)=0
x4+bx3+(a+c)x2+abx+ac=0x^4+bx^3+(a+c)x^2+abx+ac=0x4+bx3+(a+c)x2+abx+ac=0
b=−2,a+c=4,ab=6,ac=−21b=-2, a+c=4,ab=6,ac=-21b=−2,a+c=4,ab=6,ac=−21
b=−2,ab=6,a=−3,c=4−a=4+3=7b=-2, ab=6,a=-3,c=4-a=4+3=7b=−2,ab=6,a=−3,c=4−a=4+3=7
Therefore the equation is:
(x2−3)(x2−2x+7)=0(x^2-3)(x^2-2x+7)=0(x2−3)(x2−2x+7)=0
So either
x=±3x=\pm\sqrt{3}x=±3 or
x2−2x+7=0x^2-2x+7=0x2−2x+7=0
D=b2−4ac=4−28=−24,D=26iD=b^2-4ac=4-28=-24,\sqrt{D}=2\sqrt{6}iD=b2−4ac=4−28=−24,D=26i
x=2±26i2=1±6ix=\frac{2\pm2\sqrt{6}i}{2}=1\pm\sqrt{6}ix=22±26i=1±6i
Out of 4 roots that we found (±3,1±6i)(\pm\sqrt{3},1\pm\sqrt{6}i)(±3,1±6i) only the first two are equal in magnitude and opposite in sign, so n=1 (one pair).
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