Finding Real Solutions for a Complex Equation

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SUMMARY

The discussion focuses on solving the equation \( x - \sqrt{x^2 - x} = \sqrt{x} \) for \( x > 0 \). The solution process involves manipulating the equation through squaring and isolating terms, ultimately leading to the conclusion that \( x = 1 \) is a valid solution. Participants clarify the importance of proper wording when discussing mathematical operations, particularly when dividing by terms that are guaranteed to be positive.

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mathdad
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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?
 
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We are given:

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

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

$$\sqrt{x}-\sqrt{x-1}=1$$

Next, arrange as:

$$\sqrt{x}-1=\sqrt{x-1}$$

Square:

$$x-2\sqrt{x}+1=x-1$$

Combine like terms and arrange as:

$$2\sqrt{x}=2$$

Divide through by 2:

$$\sqrt{x}=1$$

Square:

$$x=1$$

Checking, we find this is a valid solution. :D
 
MarkFL said:
Since, $0<x$, I would first divide through by $0<\sqrt{x}$ to get:

You're really going to divide by 0?
 
Prove It said:
You're really going to divide by 0?

I am dividing through by a value great than zero, since if $$0<x$$ then $$0<\sqrt{x}$$.
 
MarkFL said:
I am dividing through by a value great than zero, since if $$0<x$$ then $$0<\sqrt{x}$$.

I know lol, I was pointing out how odd your wording was hahaha.
 
MarkFL said:
I am dividing through by a value great than zero, since if $$0<x$$ then $$0<\sqrt{x}$$.
Yes that was what you meant but what you wrote was "Divide by 0...". You should have said "Divide by \sqrt{x}&gt; 0".
 
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.
 
Sorry for the confusing wording...from now on I will write something like

Since $$0<x\implies0<\sqrt{x}$$, we may divide through by $\sqrt{x}$ to obtain...
 
MarkFL said:
Sorry for the confusing wording...from now on I will write something like

Since $$0<x\implies0<\sqrt{x}$$, we may divide through by $\sqrt{x}$ to obtain...
I did not get confused.
 
  • #10
RTCNTC said:
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
 
  • #11
MarkFL said:
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.
 

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