Finding Real Solutions for a Complex Equation

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Discussion Overview

The discussion revolves around finding all real solutions for the equation \( x - \sqrt{x^2 - x} = \sqrt{x} \) under the condition that \( x > 0 \). Participants explore various algebraic manipulations and check the validity of proposed solutions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equation and seeks guidance on how to proceed after initial manipulations.
  • Another participant simplifies the equation to \( \sqrt{x} - \sqrt{x-1} = 1 \) and derives \( x = 1 \) as a solution, checking its validity.
  • There is a discussion about the appropriateness of dividing by \( \sqrt{x} \) and clarifying that it is valid since \( \sqrt{x} > 0 \) when \( x > 0 \).
  • Several participants engage in clarifying the wording used in the division step, with some humor about the phrasing "divide by 0".
  • One participant acknowledges the potential for misinterpretation of their wording and expresses a commitment to clearer communication in future posts.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the solution \( x = 1 \), but there is some debate regarding the clarity of mathematical expressions and the appropriateness of certain steps taken in the derivation process.

Contextual Notes

Some participants express concern over the wording used in mathematical steps, indicating a need for clarity in communication. The discussion does not resolve whether other solutions exist beyond \( x = 1 \).

Who May Find This Useful

Readers interested in algebraic manipulation, solution verification, and the importance of clear mathematical communication may find this discussion relevant.

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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