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[SOLVED] find x with equation with 3 variables

karush

Well-known member
Jan 31, 2012
2,778
$x^2+m^2=2mx + (nx)^2$, $n\neq1$, $n\neq-1$
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:
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$x^2+m^2=2mx + (nx)^2$, $n\neq1$, $n\neq-1$
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.
 

karush

Well-known member
Jan 31, 2012
2,778
\[x=\frac{2m\pm\sqrt{4m^2-4(1-n^2)m^2}}{2(1-n^2)}\]

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

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

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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:

karush

Well-known member
Jan 31, 2012
2,778

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
$x^2+m^2=2mx + (nx)^2$, $n\neq1$, $n\neq-1$
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?

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

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