2

I'm looking for $r$ such that the equation $$x^4+x^2(1-2r)-2x+1=0$$ has only one real solution. I've made some attempts to this problem, but it seems that I even didn't get close to the solution. The approximation for $r$ is 0.3347498
Is it possible to find analytic solution for $r$ and if yes then how?
Thanks for all the help.

Novice
  • 1,127
Jakub Pawlak
  • 781
  • 3
    Because this is a quartic and the complex roots come in conjugate pairs, the only chance is a double root. A double root of $f(x)$ is a root of the greatest common factor of $f(x)$ and its derivative $f'(x)$. What did checking that give you? – Jyrki Lahtonen Aug 25 '20 at 16:43
  • 2
    You could say that since $r = \frac{x^4 + x^2 - 2x + 1}{2x^2}$, if you graph the right hand side as a function of $x$ the problem is asking what values in the range are achieved exactly once. – Daniel Schepler Aug 25 '20 at 17:02
  • Mathematica gave $$r=\frac{3}{8}+\frac{1}{8 \sqrt{\frac{3}{67+\sqrt[3]{791776-27168 \sqrt{849}}+2\ 2^{2/3} \sqrt[3]{24743+849 \sqrt{849}}}}}-\frac{1}{2} \sqrt{\frac{67}{24}-\frac{1}{48} \sqrt[3]{791776-27168 \sqrt{849}}-\frac{1}{12} \sqrt[3]{\frac{1}{2} \left(24743+849 \sqrt{849}\right)}+\frac{255}{8} \sqrt{\frac{3}{67+\sqrt[3]{791776-27168 \sqrt{849}}+2
    2^{2/3} \sqrt[3]{24743+849 \sqrt{849}}}}}$$ for the key root of the resulting quartic. Don't know if simplification/denesting is possible :-)     – Jyrki Lahtonen Aug 25 '20 at 17:15

3 Answers3

1

We can write the function in following form where $c>0$.

$(x-a)^2((x-b)^2+c)=0$

$(x^2-2ax+a^2)(x^2-2bx+b^2+c)=0$

$x^4-(2a+2b)x^3+(a^2+b^2+4ab+c)x^2-(2ab^2+2ac+2a^2b)x+a^2b^2+a^2c=0$

Therefore,

$b=-a, c={1-a^4\over a^2} > 0$

$2ab^2+2ac+2a^2b=2a^3+{2-2a^4\over a}-2a^3=-2,a^4-a-1=0$

$1-2r=a^2+b^2+4ab+c=2a^2-4a^2+{1-a^4\over a^2}={1-3a^4\over a^2}=-{3a+2\over a^2}$

$r={a^2+3a+2\over 2a^2}$

We take the real root where $a^4<1$ in $a^4-a-1=0$ which is around $-0.72449$ and gives $r$ around $0.33476$. And yes, there is a closed formula for quartic equation using radicals so $r$ also has an exact formula as well, but it should be pretty messy.

cr001
  • 12,598
0

Has noted, the polynomial should have a real double root and two conjugate complex roots, so its general form is $$ (x-a)^2[(x+p)^2+q^2] $$ Expanding this, and comparing with our initial polynomial we have the relations \begin{alignat}{2} &-&&a^2 p^2-a^2 q^2+1=0,\\ &-&&2 a^2 p+2 a p^2+2 a q^2-2=0,\\ &-&&a^2+4 a p-p^2-q^2-2 r+1=0,\\ &&&2 a-2 p=0 \end{alignat} From forth equation we have $p=a$, so the system reduce to \begin{alignat}{2} &-&&a^4-a^2 q^2+1=0,\\ &&&2 a q^2-2=0,\\ &&&2 a^2-q^2-2 r+1=0 \end{alignat} From the second we have $q^2=1/a>0$ so the system reduces to \begin{alignat}{2} &-&&a^4-a+1=0,\\ &&&2 a^2-\frac{1}{a}-2 r+1=0 \end{alignat} So the problem is reduced to look for a positive solution to $$ -a^4-a+1=0 $$ and once found, we have, from the second $$ r=\frac{2 a^3+a-1}{2 a} $$ The only positive solution for $a$ is $$ a=0.7244919590005157 $$ so that $$ r=0.3347498141075979. $$

Vincenzo Tibullo
  • 10,955
0

Equation $$x^4 + x^2 (1 - 2 r) - 2 x + 1 = 0$$ has only one real solution if the curve $y=x^4 + x^2 (1 - 2 r) - 2 x + 1$ is tangent to $x$-axis, that is: the curve has a absolute minimum, provided that $y''=2 (1-2 r)+12 x^2$ is positive, which means $1-2r>0\to r<\frac{1}{2}$

Derivative is $y'=4 x^3 +2 (1-2 r) x-2$. We have $y'=0$ for $r=\frac{2 x^3+x-1}{2 x}$.

Substitute in the given equation $$x^4 + x^2 \left(1 - 2\cdot \frac{2 x^3+x-1}{2 x}\right) - 2 x + 1 = 0$$ which gives $$x^4+x-1=0$$ Real solutions are $$x_1=-1.22074;\;x_2= 0.724492$$ Correspondent $r$ values are $$r_1=2.3998;\; r_2=0.33475$$ $r_1$ must be discarded because it is larger than $1/2$ so the unique solution is $$r = 0.33475$$

Raffaele
  • 26,371