MHB -12.1 find x w\ eq w\ 3 variables

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The discussion focuses on isolating the variable x from the equation x^2 + m^2 = 2mx + (nx)^2, with the conditions n ≠ 1 and n ≠ -1. Participants rewrite the equation into a quadratic form and explore methods to factor or simplify it. The correct solutions for x are identified as x = m/(1-n) and x = m/(1+n). An alternative approach using absolute values is also suggested, leading to the same conclusions. The thread concludes with participants confirming the correctness of the derived solutions.
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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...
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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