Finding Real Solutions for a Complex Equation

[mathdad]

Find all the real solutions of the equation.

Let rt = root

x - rt{x^2 - x} = rt{x}, where x > 0

[x - rt{x^2 - x}]^2 = [rt{x}]^2

(x - rt{x^2 - x})(x - rt{x^2 - x}) = x

x^2 -2x•rt{x^2 - x} + x^2 - x = x

2x^2 - 2x•rt{x^2 - x} = 2x

Where do I go from here?

 

[MarkFL]

We are given:

\(\displaystyle x-\sqrt{x^2-x}=\sqrt{x}\) where $0<x$

Since, $0<x$, I would first divide through by $0<\sqrt{x}$ to get:

\(\displaystyle \sqrt{x}-\sqrt{x-1}=1\)

Next, arrange as:

\(\displaystyle \sqrt{x}-1=\sqrt{x-1}\)

Square:

\(\displaystyle x-2\sqrt{x}+1=x-1\)

Combine like terms and arrange as:

\(\displaystyle 2\sqrt{x}=2\)

Divide through by 2:

\(\displaystyle \sqrt{x}=1\)

Square:

\(\displaystyle x=1\)

Checking, we find this is a valid solution. :D

 

[Prove It]

Since, $0<x$, I would first divide through by $0<\sqrt{x}$ to get:

You're really going to divide by 0?

 

[MarkFL]

You're really going to divide by 0?

I am dividing through by a value great than zero, since if \(\displaystyle 0<x\) then \(\displaystyle 0<\sqrt{x}\).

 

[Prove It]

I am dividing through by a value great than zero, since if \(\displaystyle 0<x\) then \(\displaystyle 0<\sqrt{x}\).

I know lol, I was pointing out how odd your wording was hahaha.

 

[HallsofIvy]

I am dividing through by a value great than zero, since if \(\displaystyle 0<x\) then \(\displaystyle 0<\sqrt{x}\).

Yes that was what you meant but what you wrote was "Divide by 0...". You should have said "Divide by $$\sqrt{x}> 0$$".

 

[mathdad]

Check:

Let x = 1

x - rt{x^2 - x} = rt{x}

1 - rt{(1)^2 - (1)} = rt{1}

1 - rt{1 - 1} = 1

1 - rt{0} = 1

1 - 0 = 1

1 = 1

It checks to be true.

 

[MarkFL]

Sorry for the confusing wording...from now on I will write something like

Since \(\displaystyle 0<x\implies0<\sqrt{x}\), we may divide through by $\sqrt{x}$ to obtain...

 

[mathdad]

Sorry for the confusing wording...from now on I will write something like

Since \(\displaystyle 0<x\implies0<\sqrt{x}\), we may divide through by $\sqrt{x}$ to obtain...

I did not get confused.

 

[MarkFL]

I did not get confused.

Once I thought about how what I posted could legitimately be interpreted after considering the posts made, I had to admit my choice of wording was not ideal and should be amended in the future for the sake of clarity. :D

 

[mathdad]

Once I thought about how what I posted could legitimately be interpreted after considering the posts made, I had to admit my choice of wording was not ideal and should be amended in the future for the sake of clarity. :D

So far, I understand all your replies.