A Fun Equation from Kettering University Math Olympiads

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This problem is from Kettering University Math Olympiads:
#ChallengingMathProblems #PolynomialEquations

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[SyberMath]

Hello everyone, one fact that's worth mentioning:
(1+x^2)(1+x^4)-4x^3 can be factored as
(x-1)^2(x^4+2x^3+4x^2+2x+1)=0
which cannot be zero

[typha]

Descartes' rule of signs says that we have at most 2 positive solutions, and no negative ones.
Rational root test says that +-1 are the only possible rational roots. (but we already ruled out negatives)
factor out (x-1), for the quotient the only possible rational root is 1 (by previous argument), see that 1 is again a root.
Since 1 is a double root, this accounts for both of the possible roots that Descartes' rule of signs prescribed, and thus there cannot be any other roots.

[satyapalsingh]

When I watch your videos, I get unending and overwhelming joy .You are a great mathematician . Thank you, dear professor .

[paultoutounji]

Love the originality of your problems ! As a math teacher, you don't how much it motivates me ! Thank you as always !

[richardryan]

Excellent solutions are presented here. Another method: It is easy to see that 0 is not a solution. If x is negative, then the left-hand side is positive, but the right-hand side is negative; thus, the given equation has no negative solutions. As SyberMath points out, the given equation can be rewritten as x^6 + x^4 - 4x^3 + x^2 + 1 = 0. Using synthetic division (twice), we see that x = 1 is a double-root, and the given equation can be restated as (x - 1)^2*(x^4 + 2x^3 + 4x^2 + 2x + 1) = 0. When x is positive, but not 1, the left-hand side of the last equation is clearly positive. Thus, x = 1 is the only real solution to the given equation.

[fracaralho]

In the second method, one can recall that it is a well known result* that x + 1/x >= 2 for x > 0 (x cannot be negative in this equation, since the LHS is bigger than 0).
So, if (x + 1/x) * (x² + 1/x²) = 4, this means that both x + 1/x and x² + 1/x² equal 2, from which we immediately arrive to the result x = 1.

*One can prove this result using the AM-GM inequality:
(x + 1 / x) / 2 >= sqrt(x * 1/x) = 1.

[macelomendes]

👏👏👏👏👏 Congratulations! A Hug from Brazil!!

[ezzatabdo]

Elegant explanation, thanks Professor.

[XJWill]

Trying this comment again since it did not show up before
Bringing all terms onto the same side, this factors to
(x-1)^2 * (x^2 + (1-i)*x + 1) * (x^2 + (1+i)*x + 1) = 0
So the 5 possible values for x are:
x = 1
x = i/2 - 1/2 - sqrt(-i/2 - 1)
x = i/2 - 1/2 + sqrt(-i/2 - 1)
x = -i/2 - 1/2 - sqrt( i/2 - 1)
x = -i/2 - 1/2 + sqrt( i/2 - 1)

[ahmadmazbouh]

That's my definition of a fun equation + i love your clever 2nd method (as always)😀

[carloshuertas]

Another great explanation, SyberMath! It looks a bit hard to solve but I actually solved the math problem by using my head. Thanks a lot!

[advaithkumar]

a slightly similar approach to method 2:
expand first to get x^6 + x^4 + x^2 + 1 - 4x^3 = 0
then dividing by x^3 and rearranging, x^3 + 1/x^3 + x + 1/x - 4 = 0,
then let y = x+1/x and proceed to solve the cubic to get y = 2 => x =1

[diogenissiganos]

This one was a very interesting problem. Wasn't very easy imo but not too hard either, and it gave me the opportunity to recall past substitutions that I hadn't used for a while.

[darshanmourya]

Beautiful explanation... This way we are increasing our knowledge further

[alnitaka]

I did it by noting that 1 is a solution, so I factor x-1 out and get x^5+x^4+2x^3-2x^2-x-1=0. The cofficients add to zero, so that tells me that x-1 is a factor of this polynomial too, so x=1 is a double root of the original equation. Dividing out yet another x-1 gives me x^4+2x^3+4x^2+2x+1 = 0 which is palindromic. So I assume it is (x^2+Ax+1)(x^2+Bx+1). Expanding and setting coefficients equal to those of the palindrome gives me A+B=2 and 2+AB=4, so that A^2-2A+2 = 0, with roots 1 +/- i, so that I have factored the palindrome into (x^2+(1+i)x+1)(x^2+(1-i)x+1), giving me complex roots (-1 -/+ i)/2 =/- sqrt(+/-i-2)/sqrt(2). So the original sextic has these four complex numbers as solutions as well as 1 with multiplicity 2. Ignore the strikeout. I don't know why Youtube did that.

[davidgillies]

You could also note that the sum of the coefficients of x^6 + x^4 - 4 x^2 + x^2 + 1 = 0 to give you a root of 1 immediately. Do the polynomial division to get x^5 + x^4 + 2 x^3 - 2 x^2 - x - 1 and again note coefficients sum to 0 so the factor of (x - 1) has multiplicity 2. You're left with x^4 + 2 x^3 + 4 x^2 + 2 x + 1 = 0 which is irreducible in reals (although it factors quite nicely in Gaussian integers).

[mdperpe]

One has that |x+1/x| ≥ 2 for all x≠0 (it's not defined for x=0) and therefore also x²+1/x² ≥ 2 for all x≠0, with equality when x=±1.
Therefore |(x+1/x) (x²+1/x²)| ≥ 4 for all x≠0 with equality when x=±1. One can then see that for x=-1 one gets (x+1/x) (x²+1/x²) = -4, and for x=+1 one gets (x+1/x) (x²+1/x²) = 4.
Thus the solution is x=+1.

[scottleung]

My complex solutions are (4^(1/3)+sqrt(4^(2/3)-4))/2 and (4^(1/3)-sqrt(4^(2/3)-4))/2. As for the real solution, I just plugged in 1 for x and it worked. I didn't do any rigorous analysis on that end.

[brinzanalexandru]

Another method is to take x²=a and x⁴=b then x³ is the geometric mean of a and b so we have (1+a)(1+b)=4√(ab) which we can rewrite simply as (√a-√b)²+(√ab-1)²=0 which has the solution (1, 1) so x=1 the only sol.

[snejpu]

This one looked harder than it really was. I love playing with xs'. : )