MHB Find the real roots of an equation.

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Roots
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Hello to all members of the forum,

Problem:
Find the real roots of the equation
$\displaystyle x^3+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\;\;\; (0<a<\frac{1}{4}) $

I really have no idea on how to even start to work with this problem, could anyone please help me?

Thanks.
 
Mathematics news on Phys.org
anemone said:
Hello to all members of the forum,

Problem:
Find the real roots of the equation
$\displaystyle x^3+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\;\;\; (0<a<\frac{1}{4}) $

I really have no idea on how to even start to work with this problem, could anyone please help me?

Thanks.


You should probably start with

$\displaystyle \begin{align*} x^3 + 2ax + \frac{1}{16} &= -a + \sqrt{a^2 + x - \frac{1}{16} } \\ x^3 + 2ax + \frac{1}{16} + a &= \sqrt{a^2 + x - \frac{1}{16}} \\ \left( x^3 + 2ax + \frac{1}{16} + a \right)^2 &= a^2 + x - \frac{1}{16} \end{align*}$

Now expand everything, and try to solve the resulting 6th order polynomial equation if possible...
 
anemone said:
Problem:
Find the real roots of the equation
$\displaystyle x^3+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\;\;\; (0<a<\frac{1}{4}) $


Hello anemone,

Could you provide the source and/or context? Is it necessary a closed form solution?
 
anemone said:

A little story: yesterday I spend two hours tryng to solve the problem without success, so I had a suspection. Browsing on the net I saw the solution in a pdf. There is a typo, in the left side the first term is $x^2$ instead of $x^3$. Have a look at the following link

http://www.fernandorevilla.es/wp-content/uploads/2013/02/olympiad.pdf

There are at least two ways of solving the problem. The first way (I tried in the same way with $x^3$) using a graphical method, the second one using an horrible method. Ask your doubts.
 
anemone said:
Hello to all members of the forum,

Problem:
Find the real roots of the equation
$\displaystyle x^2+2ax+\frac{1}{16}=-a+\sqrt{a^2+x-\frac{1}{16}}\;\;\; (0<a<\frac{1}{4}) $

I really have no idea on how to even start to work with this problem, could anyone please help me?

Thanks.

Using Fernando Revilla's correction to the problem, and hoping my algebra is working properly today:

... forget it, it does not work ..
 
Last edited:
zzephod said:
... forget it it does not work ..

At any rate, thanks for your contribution and welcome to MHB.
 

Similar threads

Replies
12
Views
4K
Replies
2
Views
2K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
4
Views
2K
Replies
10
Views
3K
Back
Top