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 April 28th, 2019, 09:40 PM #1 Member   Joined: Aug 2018 From: România Posts: 45 Thanks: 2 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: 6,755 Thanks: 695 Start with $x^2-3x+2=(x-1)(x-2)$ to get interval divisions for second term.
April 30th, 2019, 09:24 PM   #3
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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
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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
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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: 20,623 Thanks: 2076 If $x$ needn't be real, how is $|x^2 - 3x + 2|\,'$ defined?
May 4th, 2019, 06:17 AM   #7
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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
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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
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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
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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

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