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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 $\displaystyle (x^23x+2)'+x^23x+2'=0$. All the best, Integrator 
April 29th, 2019, 01:29 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,755 Thanks: 695 
Start with $x^23x+2=(x1)(x2)$ to get interval divisions for second term.

April 30th, 2019, 09:24 PM  #3 
Member Joined: Aug 2018 From: România Posts: 45 Thanks: 2  
May 1st, 2019, 07:47 AM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 2,923 Thanks: 1518  Quote:
$\displaystyle 2x3 = (32x) \cdot \frac{(x1)(x2)}{(x1)(x2)}$ note: $\displaystyle 2x3 = 32x$ for $x \le \frac{3}{2}$ $\displaystyle \frac{(x1)(x2)}{(x1)(x2)} = 1 \, \text{ if } \, x \in (\infty,1) \cup (2, \infty)$ and $\displaystyle 2x3 = (32x)$ for $x \ge \frac{3}{2}$ $\displaystyle \frac{(x1)(x2)}{(x1)(x2)} = 1 \, \text{ if } \, x \in (1,2)$ ... 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: 45 Thanks: 2  Quote:
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  
Math Team Joined: Jul 2011 From: Texas Posts: 2,923 Thanks: 1518  ... isn't $f(x)' = \dfrac{f(x)}{f(x)} \cdot f'(x)$, or am I mistaken? Quote:
$x \in (\infty,1) \cup \left[\frac{3}{2}, 2\right)$ I'd be interested in how "some" determined nonreal solutions to this equation.  
May 4th, 2019, 09:46 PM  #8  
Member Joined: Aug 2018 From: România Posts: 45 Thanks: 2  Quote:
Indeed $\displaystyle f(x)' = \dfrac{f(x)}{f(x)} \cdot f'(x)$. Some say that one solution would be $\displaystyle x=a\mp \sqrt{\frac{6a^218a+13}{2}}\cdot i$ 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: 2,923 Thanks: 1518  Quote:
 
May 5th, 2019, 09:30 PM  #10 
Member Joined: Aug 2018 From: România Posts: 45 Thanks: 2  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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equation, modules, modulus 
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