1

Three real roots of the equation $x^3-2p(p+1)x^2+(p^4+4p^3-1)x-3p^3=0$ is the length of three sides of a right triangle, what is the value of $p$?

Ycoronene
  • 25
  • 1
    Ok, ignore my previous comment -- it works out. If you do it correctly (as I didn't at first), you'll get $p = \sqrt{2}$. – quasi Apr 29 '17 at 01:16
  • The condition $p = \sqrt{2}$ is a necessary condition. But then I still get no solutions for $a,b,c$ (since they have to be positive). Perhaps I made a mistake? – quasi Apr 29 '17 at 01:23
  • @quasi $p=\sqrt{2}$ gives the roots $1,,2\sqrt{2},,3,$ which do indeed form a right triangle. – dxiv Apr 29 '17 at 01:26
  • Yes, I simplified incorrectly -- $1,2\sqrt{2},3$ works, and is the only such triangle.The algebra is messy. – quasi Apr 29 '17 at 01:30
  • @quasi The other real values $p=0,,-\sqrt{2},$ lead to ineligible roots, however, so your point about having to actually verify the solutions is a good one. – dxiv Apr 29 '17 at 01:33

1 Answers1

2

Hint: let $a,b,c$ be the roots, then by Vieta's relations:

$$\require{cancel} \begin{align} a^2+b^2+c^2 &= (a+b+c)^2 - 2(ab+bc+ca) \\ &= \left(2p(p+1)\right)^2 - 2(p^4+4p^3-1) \\ &= 4p^2(p^2+\bcancel{2p}+1) - 2p^4-\bcancel{8p^3}+2 \\ &= 2(p^4 + 2p^2+1) \\ &= 2(p^2+1)^2 \end{align} $$

For the roots to be the sides of a right triangle, one of them, say $a\,$, must satisfy $a^2=b^2+c^2$.

dxiv
  • 76,497
  • Then $a=p^2+1$, and substitute it to the equation, simplify, got $2p^5-3p^3-2p=p(2p^2+1)(p^2-2)=0$, then $p=\sqrt{2}$. Thanks so much! – Ycoronene Apr 29 '17 at 01:42
  • @Ycoronene Right, and glad it helped. As noted in a comment, you need to also verify (and reject) the other potential real solutions $p=0$ ($a,b,c$ then form a degenerate triangle) and $p=-\sqrt{2}$ (one root is negative), leaving $p=\sqrt{2},$ which gives the eligible triplet $a,b,c,$. – dxiv Apr 29 '17 at 01:51
  • 1
    Yes, From the Vieta's formula, $abc=3p^3>0$, so $p>0$. Thanks again! – Ycoronene Apr 29 '17 at 01:55