An equation with moduli

Aug 2018
128
7
România
Hello all,

A problem at the high school level:

Solve the equation \(\displaystyle |x-2|+|x-1|+|x|=-m^2\) where \(\displaystyle m\) is a some parameter.

All the best,

Integrator
 

romsek

Math Team
Sep 2015
2,773
1,552
USA
The only non-negative value on the right hand side is when $m=0$

The left hand side is the sum of 3 non-negative values that are not all
simultaneously zero.

There is no solution for any parameter value $m$

Unless you mean for $m$ to be imaginary, in which case why not just get rid of the negative sign.
 
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Aug 2018
128
7
România
The only non-negative value on the right hand side is when $m=0$

The left hand side is the sum of 3 non-negative values that are not all
simultaneously zero.

There is no solution for any parameter value $m$

Unless you mean for $m$ to be imaginary, in which case why not just get rid of the negative sign.
Hello,

If \(\displaystyle m=\mp k\cdot i\) where \(\displaystyle k\in \mathbb R\)* and \(\displaystyle i^2=-1\) , then does the equation have solutions? If yes, then what are the solutions of the equation?
Thank you very much!

All the best,

Integrator
 
Last edited by a moderator:

romsek

Math Team
Sep 2015
2,773
1,552
USA
Hello,

If \(\displaystyle m=\mp k\cdot i\) where \(\displaystyle k\in \mathbb R\)* and \(\displaystyle i^2=-1\) , then does the equation have solutions? If yes, then what are the solutions of the equation?
Thank you very much!

All the best,

Integrator
Sure. Plot the thing and you'll see it's a pretty simple function symmetric about $x=1$

Other than $m=0$ there are going to be two solutions for any given $m$
 
Last edited by a moderator:

skipjack

Forum Staff
Dec 2006
21,321
2,386
The right-hand side must be real.
There are no solutions if the right-hand side is less than 2.
If the right-hand side equals 2, $x = 1$ is the unique real solution.
If the right-hand side exceeds 2, there are two real solutions (that are easily found, but you didn't find them).
 

romsek

Math Team
Sep 2015
2,773
1,552
USA
Sure. Plot the thing and you'll see it's a pretty simple function symmetric about $x=1$

Other than $m=0$ there are going to be two solutions for any given $m$
Oops... The minimum $m$ is 2, not zero.
 
Aug 2018
128
7
România
Hello all,

If \(\displaystyle m=\mp k\cdot i\) where \(\displaystyle k\in \mathbb R\)* and \(\displaystyle i^2=-1\) , then the equation becomes \(\displaystyle |x-2|+|x-1|+|x|=k^2\) and so what are the solutions of the initial equation?Thank you very much!

All the best,

Integrator
 

skipjack

Forum Staff
Dec 2006
21,321
2,386
For $k^2 = 2$, $x = 1$.

For $2 < k^2 \leqslant 3$, $x = 3 - k^2$ or $k^2 - 1$.

For $k^2 > 3$, $x = 1 \pm (k^2)/3$.
 
Aug 2018
128
7
România
For $k^2 = 2$, $x = 1$.

For $2 < k^2 \leqslant 3$, $x = 3 - k^2$ or $k^2 - 1$.

For $k^2 > 3$, $x = 1 \pm (k^2)/3$.
Hello,

What are the solutions of the initial equation \(\displaystyle |x-2|+|x-1|+|x|=-m^2\) where \(\displaystyle m\) is a some parameter?Thank you very much!

All the best,

Integrator