roots again!
Level
pending
If the equation x^4+ax^3+bx^2+ax+1=0 has a root,then find the minimum value of a^2+b^2.
The answer is 0.8.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
You are easy to see that x=0 is not this equation's root.Suppose that z is this equation's root(z><0),divide both sides of the equation to z^2,you get: z^2+az+b+a/z+1/z^2=0 <=>(z^2+1/z^2)+a(z+1/z)+b=0 Set y=z+1/z=>Abs y=abs z+1/z >=2 y^2-2=z^2+1/z^2=>y^2-2=-ay-b You have:(a^2+b^2)(y^2+1)>=(ay^2+b)^2=(y^2-2)^2 =>a^2+b^2>=(y^4-4y^2+4)/(y^2+1)=A (1) A-0.8=(5y^4-24y^2+16)/(5y^2+5)=((5y^2-4)(y^2-4))/(5y^2+5)>=0(because abs y>=2=>y^2>=4)=>A>=0.8 (2) (1)(2)=>a^2+b^2>=0.8