Proof of existence of y and z for x>1

[playboy]

Prove: For every real number x>1, there exists two distinct positive real numbers y and z such that

x = (y^2 +9)/(6y) = (z^2 +9)/(6z)

Okay.. this has real got me beat. Firstly (this sounds stupid and obvious), when they give us a proof, is it really true? Do we just naturally believe that its true and with this in mind, prove it? Or could it be wrong and we eventually find that out if we do the proof?

I first tried to isolate y and z in terms of x and set them equal to each other, but that led me to a dead end.

Then i tried the contrapositive, but that didn't make it any easier.

So.. I tried contradiction.

For every real number x>1, there DOES NOT exists two distinct positive real numbers y and z such that

x = (y^2 +9)/(6y) = (z^2 +9)/(6z)

and now I am out of ideas?

Can somebody please help me out?

Thanks

 

[benorin]

Consider solving the equation

$$x=\frac{w^2+9}{6w},$$ where x>1 is a constant.

If $w\neq 0$, then

$$w^2-6xw+9=0$$

so by the quadratic equation, we have

$$w=\frac{6x\pm\sqrt{36x^2-36}}{2}=3x\pm 3\sqrt{x^2-1}$$

which are both real (since x>1) and thus we note that:

$$y=3x+3\sqrt{x^2-1},z=3x-3\sqrt{x^2-1}$$

are certianly distinct real numbers satisfing said conditions.

 

[matt grime]

There is of course another way of showing existence without actually finding the answers y aand z in terms of x.

Consider f(w)=(w^2+9)/6w - x

when w tends to zero this is positive, when w tends to infinity this is positive. If there exists any point where it is negative then the solution follows. We can finid the minimum: f'(w)= 1/6 - 3/(4w^2), and the minimal value there will be negative (for x>1) and we are done.