My Math Forum An equation with modulus

 Complex Analysis Complex Analysis Math Forum

 April 28th, 2019, 09:40 PM #1 Member   Joined: Aug 2018 From: România Posts: 84 Thanks: 6 An equation with modulus Hello all, Solve the equation $\displaystyle |(x^2-3x+2)'|+|x^2-3x+2|'=0$. All the best, Integrator
 April 29th, 2019, 01:29 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 Start with $x^2-3x+2=(x-1)(x-2)$ to get interval divisions for second term.
April 30th, 2019, 09:24 PM   #3
Member

Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

Quote:
 Originally Posted by mathman Start with $x^2-3x+2=(x-1)(x-2)$ to get interval divisions for second term.
Hello,

What can be the nature of the equation solutions?Thank you very much!

All the best,

Integrator

May 1st, 2019, 07:47 AM   #4
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,002
Thanks: 1587

Quote:
 Originally Posted by Integrator Solve the equation $\displaystyle |(x^2-3x+2)'|+|x^2-3x+2|'=0$.
$\displaystyle |2x-3| + \frac{x^2-3x+2}{|x^2-3x+2|} \cdot (2x-3) = 0$

$\displaystyle |2x-3| = (3-2x) \cdot \frac{(x-1)(x-2)}{|(x-1)(x-2)|}$

note:

$\displaystyle |2x-3| = 3-2x$ for $x \le \frac{3}{2}$

$\displaystyle \frac{(x-1)(x-2)}{|(x-1)(x-2)|} = 1 \, \text{ if } \, x \in (-\infty,1) \cup (2, \infty)$

and

$\displaystyle |2x-3| = -(3-2x)$ for $x \ge \frac{3}{2}$

$\displaystyle \frac{(x-1)(x-2)}{|(x-1)(x-2)|} = -1 \, \text{ if } \, x \in (1,2)$

... can you reason out the solution set from this point?

Last edited by skipjack; May 4th, 2019 at 02:29 AM.

May 3rd, 2019, 09:17 PM   #5
Member

Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

Quote:
 Originally Posted by skeeter $\displaystyle |2x-3| + \frac{x^2-3x+2}{|x^2-3x+2|} \cdot (2x-3) = 0$ $\displaystyle |2x-3| = (3-2x) \cdot \frac{(x-1)(x-2)}{|(x-1)(x-2)|}$ note: $\displaystyle |2x-3| = 3-2x$ for $x \le \frac{3}{2}$ $\displaystyle \frac{(x-1)(x-2)}{|(x-1)(x-2)|} = 1 \, \text{ if } \, x \in (-\infty,1) \cup (2, \infty)$ and $\displaystyle |2x-3| = -(3-2x)$ for $x \ge \frac{3}{2}$ $\displaystyle \frac{(x-1)(x-2)}{|(x-1)(x-2)|} = -1 \, \text{ if } \, x \in (1,2)$ ... can you reason out the solution set from this point?
Hello,

I did not understand your reasoning! Some say the equation has real but also complex solutions. How do we find all the solutions of the equation? Thank you very much!

All the best,

Integrator

Last edited by skipjack; May 4th, 2019 at 02:31 AM.

 May 4th, 2019, 02:36 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 If $x$ needn't be real, how is $|x^2 - 3x + 2|\,'$ defined?
May 4th, 2019, 06:17 AM   #7
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,002
Thanks: 1587

Quote:
 Originally Posted by skipjack If $x$ needn't be real, how is $|x^2 - 3x + 2|\,'$ defined?
... isn't $|f(x)|' = \dfrac{f(x)}{|f(x)|} \cdot f'(x)$, or am I mistaken?

Quote:
 Originally Posted by integrator Some say the equation has real but also complex solutions.
The solution set I determined ...

$x \in (-\infty,1) \cup \left[\frac{3}{2}, 2\right)$

I'd be interested in how "some" determined non-real solutions to this equation.

May 4th, 2019, 09:46 PM   #8
Member

Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

Quote:
 Originally Posted by skeeter ... isn't $|f(x)|' = \dfrac{f(x)}{|f(x)|} \cdot f'(x)$, or am I mistaken? The solution set I determined ... $x \in (-\infty,1) \cup \left[\frac{3}{2}, 2\right)$ I'd be interested in how "some" determined non-real solutions to this equation.
Hello,

Indeed $\displaystyle |f(x)|' = \dfrac{f(x)}{|f(x)|} \cdot f'(x)$.
Some say that one solution would be $\displaystyle x=a\mp \sqrt{\frac{6a^2-18a+13}{2}}\cdot i$ where $\displaystyle a\in [2 , +\infty)$ and $\displaystyle i^2=-1$.How can we find this solution?Thank you very much!

All the best,

Integrator

May 5th, 2019, 05:18 AM   #9
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,002
Thanks: 1587

Quote:
 Originally Posted by Integrator Hello, Indeed $\displaystyle |f(x)|' = \dfrac{f(x)}{|f(x)|} \cdot f'(x)$. Some say that one solution would be $\displaystyle x=a\mp \sqrt{\frac{6a^2-18a+13}{2}}\cdot i$ where $\displaystyle a\in [2 , +\infty)$ and $\displaystyle i^2=-1$.How can we find this solution?Thank you very much! All the best, Integrator
... where did the value, $a$, come from?

May 5th, 2019, 09:30 PM   #10
Member

Joined: Aug 2018
From: România

Posts: 84
Thanks: 6

Quote:
 Originally Posted by skeeter ... where did the value, $a$, come from?
Hello,

I suppose we have to consider the possibility that $\displaystyle x = a + k(a)i$ where $\displaystyle a \in \mathbb R$ and $\displaystyle i^2=-1$ , which means that the initial equation will result an equation in the form $\displaystyle u + vi=0$ where $\displaystyle u=f(a)$ and $\displaystyle v=g(a)$.How do we solve the equation $\displaystyle f(a)+g(a)i=0$ where $\displaystyle i^2=-1$?Thank you very much!

All the best,

Integrator

 Tags equation, modules, modulus

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post NAC54321 Physics 0 November 13th, 2018 05:03 AM bilano99 Calculus 2 March 15th, 2013 12:48 PM dpwheelwright Number Theory 5 August 5th, 2012 01:36 PM omega Complex Analysis 2 December 17th, 2011 02:51 PM Hemi08 Abstract Algebra 1 August 20th, 2009 07:54 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top