挑戰19

M.N.M.

證明下列方程式沒有實數根:
(1)x^8-x^5+x^2+x+1=0

(2)x^4+2x^2-x+2=0

(3)x^12-x^9+x^4-x+1=0

scotten

3.
設h(x)=x^12-x^9+x^4-x+1
則
(1)x>=1
顯然h(x)>x^12-x^9+x^4-x=x(x^3-1)(x^8+1)>=0
(2)0<=x<1
有h(x)>=1-x+x^4-x^9=(1-x)+x^4(1-x)(1+x+x^2+x^3+x^4)>0
(3)x<0
顯然h(x)>0
由上述(1)(2)(3)知
x屬於R,h(x)恆大於0
=>h(x)=0無實數解#

scotten

1.
設g(x)=x^8-x^5+x^2+x+1
則
(1)x>=1
顯然g(x)>x^8-x^5=x^5(x^3-1)>0
(2)0<=x<1
有g(x)>1-x^5=(1-x)(1+x+x^2+x^3+x^4)>0
(3)-1<x<0
有g(x)>1+x>0
(4)x<=-1
顯然g(x)>-x^5+x=-x(x^4-1)>=0
由上述(1)(2)(3)(4)知
x屬於R時,g(x)恆大於0
=>g(x)=0無實數解#

scotten

2.
prove that x^4+2x^2-x+2=0 has no real root

if y^4+2y^2-y+2=0 has a real root:x
(1)
x^4+2x^2-x+2=0
=>(x^2+1)^2=x-1
because x^2>=0
=>x^2+1>=1
=>(x^2+1)^2>=1
=>x-1>=1
=>x>=2
(2)
x^4+2x^2-x+2=0
=>x=x^4+2x^2+2
if x>=2
x^4+2x^2+2>=x+2x+2
=>x>=3x+2
=>x<=1 ><

so x^4+2x^2-x+2=0 has no real root #