|
|
|
Let a and b be real numbers for which the equation
x4 + ax3 + bx2 + ax + 1 = 0
has at least one real solution. For all such pairs, (a,b), find the minimum value of a2+b2.
Source: 1973 BMO
Since this polynomial is a "reciprocal equation", we can "center" it by dividing through by x2,
x2 + ax + b + a/x + 1/x2 = 0
x2 + 1/x2 + a(x+1/x) + b = 0
(x+1/x)2 + a(x+1/x) + b - 2 = 0
So now it's a quadratic in x+1/x, so we get
x+1/x = -a/2 ± sqrt(a2-4b+8)/2
Since the left hand side, x+1/x, must be at least 2, then one of the solutions (right hand side) must be at least 2. We can double both sides to get the inequality
|a| + sqrt(a2-4b+8) ≥ 4
sqrt(a2-4b+8) ≥ 4 - |a|
a2-4b+8 ≥ 16 - 8|a| + a2
-4b ≥ 8-8|a|
2|a|-b ≥ 2
-b ≥ 2-2|a|
Since we're looking for a minimum a2+b2, let's assume that |a|<1 (we'll revisit this assumption later), and so both sides of the inequality, above, are positive. Then we can square both sides to get
b2 ≥ (2|a|-2)2
a2+b2 ≥ a2+(2|a|-2)2 = 5a2-8|a|+4 = 5(a2-8/5|a|+20/25) = 5(|a|-4/5)2+4/5
So the least possible value of a2+b2 is 4/5, which is achieved when |a|=4/5, and b=-2/5. So the original equation is one of
x4 + (4/5)x3 - (2/5)x2 + (4/5)x + 1 = 0
x4 - (4/5)x3 - (2/5)x2 - (4/5)x + 1 = 0
Sure enough, these two equations each have a solution, as they factorize as follows:
(x-1)2(x2+(6/5)x+1)=0, and
(x+1)2(x2-(6/5)x+1)=0, respectively.
We're done. Now let's revisit the assumption that |a|<1. If we assume the opposite, then a2+b2 would be at least 1, so a better minimum of a2+b2 would not be found. Now we're really done.
gifted.hkedcity.net/Gifted/Download/notes/phase3/basic/Forth%20Lecture_2007-02-10.pdf (mirror), which contains this puzzle and a number of other useful hints for solving fourth-degree polynomials.
(none)
The webmaster and author of this Math Help site is Graeme McRae.