# Solving the equation

## Homework Statement

Find all real numbers ##x## such that,
$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}=x$$

## 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!

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LCKurtz
Homework Helper
Gold Member

## Homework Statement

Find all real numbers ##x## such that,
$$\left(x-\frac{1}{x}\right)^{1/2}+\left(1-\frac{1}{x}\right)^{1/2}=x$$

## 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!
##x^4+2x^3-x^2-2x+1=(x^2+x-1)^2##

• 1 person
Chestermiller
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 $$y^2-2y+1=0$$

• 1 person
##x^4+2x^3-x^2-2x+1=(x^2+x-1)^2##
That was too obvious to be missed. :tongue:

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

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 $$y^2-2y+1=0$$
Nice one Chet, thanks a lot! Chestermiller
Mentor
1.
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!

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:
$$(x^2-x-1)^2=0$$

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:
$$(x^2-x-1)^2=0$$
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: