Difficult equation

1. Feb 16, 2009

dirk_mec1

1. The problem statement, all variables and given/known data

Find all real x satisfying:

$$x = \sqrt{1- \frac{1}{x} } + \sqrt{x-\frac{1}{x}}$$

3. The attempt at a solution
I've tried numerous things (substitution, squaring, completing a square) but I can't figure it out. Can someone give me a hint?

2. Feb 16, 2009

MathematicalPhysicist

x-sqrt(x-1/x)=sqrt(1-1/x)
x^2-2xsqrt(x-1/x)+x-1/x=1-1/x
x^2+x-1=2xsqrt(x-1/x)
x^4+x^2+1+2x^3-2x^2-2x=4x^3-4x
x^4-2x^3-x^2+2x+1=0
x^2(x^2-2x-1)+2x+1=0
The only way to prove this is by using a method of solving a 4th degree equation, check it in mathworld.

3. Feb 16, 2009

epenguin

Yes but you do not have to use the full machinery of the quartic, because your
x^4-2x^3-x^2+2x+1=0 (if that is right)

is a 'reciprocal equation' (if x is a root, 1/x is too) I think.
Divide by x^2, and express as a quadratic in z, z = (x + 1/x), you can solve that and thence solve in the original x.

4. Feb 16, 2009

dirk_mec1

I don't think its reciprocal because a4 = a1 but a3= -a2. I suspect that a smart substitution is the key.

5. Feb 16, 2009

epenguin

You're right. I think take out x^2 like I said but z = (x - 1/x) ?

6. Feb 16, 2009

dirk_mec1

No, also doesn't work. Note that this is an exercise that should be doable by high school students. So using the quartic formula is out of the question.

7. Feb 16, 2009

Tom Mattson

Staff Emeritus
I got the same polynomial equation that LQG got. Not sure of how to solve it without the quartic formula, but here are some thoughts. Note that the only possible rational roots are $x=\pm 1$. Since neither of those work there are no rational roots. Since this is a polynomial equation with rational coefficients, irrational roots must occur in conjuguate pairs.

8. Feb 16, 2009

Tom Mattson

Staff Emeritus
I got it. I considered the function $f(x)=x^4-2x^3-x^2+2x+1$ and proceeded to find it's relative extrema. Since *I* took calculus in high school I figured this was acceptable. :tongue: I did this in the hopes that the graph has no x-intercepts. What I found instead is that one of the critical numbers of $f(x)$ is actually a root of $f(x)$. Once you have one (irrational) root, you get its conjugate for free. And once you have 2 roots of a quartic you have defeated it, because you have reduced the problem to factoring a quadratic.

Try to follow my steps and see if you can't get it.

9. Feb 16, 2009

epenguin

I think you can do it my suggested way because I have now done it, and the result also agrees with what you say about there being two double roots.

10. Feb 16, 2009

Tom Mattson

Staff Emeritus
You suspected rightly: let $x=u+1/2$. I came across that by noticing that the function $f(x)=x^4-2x^3-x^2+2x+1$ has its relative minima at $x=(1\pm\sqrt{5}/2$ and a relative maximum right smack in the middle of them at $x=1/2$. That means the function is symmetric about the line $x=1/2$.

I'd be surprised if more than 1% of high school students could get this!

11. Feb 17, 2009

dirk_mec1

If I use $x = u +\frac{1}{2}$ I get something nasty:

$$(u-\frac{1}{2})^{3/2} - \sqrt{u - \frac{1}{2}} = \sqrt{u^2+u-\frac{3}{4}}$$

Are you sure that's the right substitution?

12. Feb 19, 2009

Tom Mattson

Staff Emeritus
You don't substitute $x=u+\frac{1}{2}$ into the original equation, you sub it into the polynomial equation that we derived from it:

$$x^4-2x^3-x^2+2x+1=0$$

It works out very nicely!

13. Feb 19, 2009

dirk_mec1

I cheated and used matlab :tongue:

and got: $$\frac{1}{16}(4u^2-5)^2 =0$$

Does this agrees with what you found?

14. Feb 20, 2009

Tom Mattson

Staff Emeritus
Yes, that agrees with what I got. If you can't see what the substitution should be then try to generate the graph of the function using Calculus I techniques (that's what I did). If $x=a$ is an axis of symmetry of the graph of $y=f(x)$ then $u=x-a$ is a good substitution. I saw that the graph of the function $f(x)=x^4-2x^3-x^2+2x+1$ has the line $x=\frac{1}{2}$ as its axis of symmetry, so presto. The graph is attached for your viewing pleasure. :tongue:

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15. Feb 21, 2009

dirk_mec1

16. Feb 23, 2009

MathematicalPhysicist

But Tom, the method of translating the equation is similar to the one of finding the roots of the general equation:
x^4+ax^3+bx^2+cx+d=0

is it not?

17. Feb 24, 2009

epenguin

Yes it is. (One of the ways of solving the quartic, to transform it into a reciprocal equation.) So you were sure you were going to get a solution that way, but on the way you get lucky in this case and it works out simpler than it usually would.

The way I proposed above was suggested by its resemblance to a reciprocal equation, and the substitution z = (x - 1/x) gives you (z - 1)2 = 0 , and finally the same solution as everybody else.

Last edited: Feb 24, 2009