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April 28th, 2019, 09:40 PM   #1
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An equation with modulus

Hello all,

Solve the equation $\displaystyle |(x^2-3x+2)'|+|x^2-3x+2|'=0$.

All the best,

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April 29th, 2019, 01:29 PM   #2
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Start with $x^2-3x+2=(x-1)(x-2)$ to get interval divisions for second term.
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April 30th, 2019, 09:24 PM   #3
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Quote:
Originally Posted by mathman View Post
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,

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May 1st, 2019, 07:47 AM   #4
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Quote:
Originally Posted by Integrator View Post

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.
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May 3rd, 2019, 09:17 PM   #5
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Quote:
Originally Posted by skeeter View Post
$\displaystyle |2x-3| + \frac{x^2-3x+2}{|x^2-3x+2|} \cdot (2x-3) = 0$

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

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.
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May 4th, 2019, 02:36 AM   #6
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If $x$ needn't be real, how is $|x^2 - 3x + 2|\,'$ defined?
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May 4th, 2019, 06:17 AM   #7
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Quote:
Originally Posted by skipjack View Post
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.
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May 4th, 2019, 09:46 PM   #8
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Quote:
Originally Posted by skeeter View Post
... 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,

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May 5th, 2019, 05:18 AM   #9
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Quote:
Originally Posted by Integrator View Post
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?
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May 5th, 2019, 09:30 PM   #10
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Quote:
Originally Posted by skeeter View Post
... 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,

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