-12.1 find x w\ eq w\ 3 variables

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Discussion Overview

The discussion revolves around isolating the variable \(x\) in the equation \(x^2 + m^2 = 2mx + (nx)^2\), with the condition that \(n \neq 1\) and \(n \neq -1\). Participants explore different methods for solving this quadratic equation, including factoring and using the quadratic formula.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant rewrites the equation to \(m^2 - 2mx - n^2x^2 + x^2 = 0\) and expresses difficulty in isolating \(x\).
  • Another participant identifies the equation as a quadratic in \(x\) and provides a formula for the roots: \(x = \frac{2m \pm \sqrt{4m^2 - 4(1-n^2)m^2}}{2(1-n^2)}\).
  • A later reply simplifies the quadratic formula to show that it leads to the solutions \(\{\frac{m}{1-n}, \frac{m}{1+n}\}\).
  • One participant acknowledges the correctness of the derived solutions but suggests a potential typo in the expression for \(x\), proposing \(x = \frac{m(1 \pm n)}{1 - n^2}\) instead.
  • Another participant proposes an alternative approach by rewriting the original equation as \((x - m)^2 = (nx)^2\) and suggests proceeding from there.

Areas of Agreement / Disagreement

There is some agreement on the correctness of the derived solutions, but there are also differing views on the expressions used and potential typos. The discussion remains somewhat unresolved regarding the best approach to isolate \(x\).

Contextual Notes

Participants express uncertainty about the correctness of certain expressions and the potential for typos, indicating that the discussion is still open to interpretation and refinement.

karush
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$x^2+m^2=2mx + (nx)^2$, $n\neq1$, $n\neq-1$
isolate $x$

rewriting
$m^2-2 m x-n^2 x^2+x^2 = 0$
$m(m-2x)-x^2(n-1)(n+1)=0$

I set this to 0 thinking I could factor by grouping but not..
So seem to be stuck on how to to isolate x

the books answere to this is $\frac{m}{1-n},\frac{m}{1+n}$

thnx ahead, :cool:
 
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karush said:
$x^2+m^2=2mx + (nx)^2$, $n\neq1$, $n\neq-1$
isolate $x$

rewriting
$m^2-2 m x-n^2 x^2+x^2 = 0$
$m(m-2x)-x^2(n-1)(n+1)=0$

I set this to 0 thinking I could factor by grouping but not..
So seem to be stuck on how to to isolate x

the books answere to this is $\frac{m}{1-n},\frac{m}{1+n}$

thnx ahead, :cool:

Hi karush, :)

\[m^2-2 m x-n^2 x^2+x^2 = 0\]

\[\Rightarrow (1-n)(1+n)x^2-2mx+m^2=0\]

This is a quadratic equation of \(x\) and the roots of this equation is given by,

\[x=\frac{2m\pm\sqrt{4m^2-4(1-n^2)m^2}}{2(1-n^2)}\]

Hope you can simplify this and see if you get the given solutions. :)

Kind Regards,
Sudharaka.
 
Sudharaka said:
\[x=\frac{2m\pm\sqrt{4m^2-4(1-n^2)m^2}}{2(1-n^2)}\]

Hope you can simplify this and see if you get the given solutions. .

how this

$\frac{2m\pm \sqrt{4m^2n^2}}{2(1-n^2)} \Rightarrow \frac{m(1 \pm n^2)}{(1-n^2)} \Rightarrow \{\frac{m}{1-n},\frac{m}{1+n}\}$
 
karush said:
how this

$\frac{2m\pm \sqrt{4m^2n^2}}{2(1-n^2)} \Rightarrow \frac{m(1 \pm n^2)}{(1-n^2)} \Rightarrow \{\frac{m}{1-n},\frac{m}{1+n}\}$

Yes that is correct. (Yes)

Edit: Maybe this is just a typo since you have obtained the answer correctly. But the second expression you have written should be,

\[x=\frac{m(1\pm n)}{1-n^2}\]

Kind Regards,
Sudharaka.
 
Last edited:
Sudharaka said:
Yes that is correct. (Yes)

thot it might quadratic but couldn't compose it.

Much Mahalo...:cool:
 
karush said:
$x^2+m^2=2mx + (nx)^2$, $n\neq1$, $n\neq-1$
isolate $x$

rewriting
$m^2-2 m x-n^2 x^2+x^2 = 0$
$m(m-2x)-x^2(n-1)(n+1)=0$

I set this to 0 thinking I could factor by grouping but not..
So seem to be stuck on how to to isolate x

the books answere to this is $\frac{m}{1-n},\frac{m}{1+n}$

thnx ahead, :cool:

Why not this?

\[ \displaystyle \begin{align*} x^2 + m^2 &= 2mx + (nx)^2 \\ x^2 - 2mx + m^2 &= (nx)^2 \\ (x - m)^2 &= (nx)^2 \\ |x - m| &= |nx| \\ |x - m| &= |n||x| \end{align*} \]

Go from here...
 

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