Hey there. So I have ran into an equation I can't solve. It involves absolute values and powers of x.
I've tried going to online calculators but found no aid, except in wolfram alpha, but I don't have access to their explanations. If anybody knows how to solve this, I'd be all ears.
Thanks. OP.
>>7944182
have you tried 1
>>7944182
Let [math]x = i^0[/math]
>>7944185
Yeah, it works, but I'd like to know a system for resolution of the problem, instead of just trial and error.
>>7944187
same thing I said to
>>7944182
consider your four cases
|x^3 - 1| and |2 - x^3| > 0
|x^3 - 1| < 0, |2 - x^3| > 0
|x^3 - 1| > 0, |2 - x^3| < 0
|x^3 - 1| and |2 - x^3| < 0
If some of the answers (or all) match then thats nice, but nobody says that a single solution must hold over the entire domain
>>7944182
Just make some assumptions. Break it down into cases.
>>7944182
When ever there is a variable in an absolute value bar, you break the equation up into two new ones, one where the absolute value is removed, and the other where its removed but also multiplied by -1. Figure it put to get both solutions, then plug back in to make sure you remove any extraneous solutions.
>>7944194
>|x^3 - 1| < 0
>|2 - x^3| < 0
by definition that's not possible
>>7944205
Yeah, something like this. I've worked like this with inequalities. It's new for me to wrap my mind around this, considering it's an equation,
>>7944208
ok so that makes your life easier
i put not thought into listing those. the first step is to consider, the next is to think about them
Problem SOLVED. I worked as I had, previously, with inequalities. I broke the domain into 3 main parts, and calculated possible values for X in each one.
Thanks for your help, everyone. My answer seems to be correct, according to wolfram alfa.
I'm writing in spanish, so the "De" before the intervals means "From", for those who don't know.
Anyway, here's how I did it :D
