# Algebra - Values of x

1. Oct 15, 2008

### tornzaer

The problem statement, all variables and given/known data
(x+6/x)^2-2(x-6/x)-35=0

Fine the values of x.

The attempt at a solution
Let y=x+6/x

y^2-2y-35=0
(y-7)(y+5)=0
y=7 or y=-5

Check:
x+6/x=7
x^2-7x+6=0
(x-6)(x-1)=0

and

x+6/x=7
x^2+5x=0
(x+3)(x+2)=0

Therefore, x=6 or x=1 or x=-3 or x=-2

Now, my teacher says there is another way to do it. He says by expanding it, the values of x can be found through algebra. I've the the way I know, but can someone tell me how to do this other way? Thank you.

2. Oct 15, 2008

### Staff: Mentor

Expanding it just means carrying out the multiplication in the factored forms.
(x+6/x)^2-2(x-6/x)-35=0
x^2 + 12 + 36/x^2 -2x + 12/x - 35 = 0

You'll need to multiply both sides by x^2 to get rid of the x terms in the denominators, which means you'll end up with a 4th degree equation, which could be hard to factor. In this case, however, you already know the roots of the equation, so the factors will be (x - 6) and so on.

3. Oct 15, 2008

I tried to delete my first post - it didn't work. Sorry.

If the original equation truly is

$$\left(x + \frac 6 x \right)^2 - 2 \left(x - \frac 6 x \right) - 35 = 0$$

then the given substitution will not solve it, as the two variable terms in are not identical.

Notice that none of $$6, 1, -3, -2$$ actually solve the equation in the form given above.

the rest of my original post was wrong and has been deleted. I just played around with this in Sage (a very neat program, by the way, that runs on Linux and Mac OS 10). If both the terms of the original equation are $$x - \frac 6 x$$ then the submitted method works.

Last edited: Oct 15, 2008
4. Oct 15, 2008

If the original equation truly is

$$\left(x + \frac 6 x \right)^2 - 2 \left(x - \frac 6 x \right) - 35 = 0$$

then the given substitution will not solve it, as the two variable terms in are not identical.

Notice that none of $$6, 1, -3, -2$$ actually solve the equation in the form given above. I don't believe they solve it in the form that seems to be intended, and that is why the teacher gave the hint about there being another method of solution.

5. Oct 15, 2008

### tornzaer

Guys, I figured it out. I just expanded it and then did long polynomial divisions to get the factors along the way. Thanks for the responses. :)