# Solving the equation

1. Jul 4, 2013

### Saitama

1. The problem statement, all variables and given/known data
Find all real numbers $x$ such that,
$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}=x$$

2. Relevant equations

3. The attempt at a solution
Moving the second term on LHS to the right and squaring both the sides,
$$x-\frac{1}{x}=x^2+1-\frac{1}{x}-2x\sqrt{1-\frac{1}{x}}$$
Rearranging,
$$x^2-x+1=2x\sqrt{1-\frac{1}{x}}$$
Squaring again leads to terms containing 4th and 3rd power of x and I don't find a way to proceed further. Here's the expression I get after squaring and rearranging,
$$x^4+2x^3-x^2-2x+1=0$$

Any help is appreciated. Thanks!

2. Jul 4, 2013

### LCKurtz

$x^4+2x^3-x^2-2x+1=(x^2+x-1)^2$

3. Jul 4, 2013

### Staff: Mentor

$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}=x$$

Multiply both sides by $\left(x-\frac{1}{x}\right)^{1/2}-\left(1-\frac{1}{x}\right)^{1/2}$ to get

$$\left(x-\frac{1}{x}\right)^{1/2}-\left(1-\frac{1}{x}\right)^{1/2}=1-\frac{1}{x}$$
Now, add the two equations to get:

$$2\left(x-\frac{1}{x}\right)^{1/2}=\left(x-\frac{1}{x}\right)+1$$

Let $y=\sqrt{\left(x-\frac{1}{x}\right)}$

Then $$y^2-2y+1=0$$

4. Jul 5, 2013

### Saitama

That was too obvious to be missed. :tongue:

Thank you!
Nice one Chet, thanks a lot!

5. Jul 5, 2013

### Staff: Mentor

Your final equation is incorrect. You need to redo the squaring and rearranging. The correct relationship is:
$$x^4-2x^3-x^2+2x+1=0$$
This is the same as:
$$(x^2-x-1)^2=0$$

6. Jul 5, 2013

### Saitama

Sorry about the typo, I knew the correct expression. I redid the squaring and rearranging when I was following LCKurtz's hint. :)

I shouldn't be posting questions in the midnight. :tongue: