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| I was not expecting an easy answer and I'm really glad that someone takes the time to answer. This is the code that I ran in Mathematica: > q1 = Root[(48 a^2 + 16 b) #^4 - (40 a^3 + 168 a b)#^3 + (-45 a^4 + 225 a^2 b + 72 b^2) #^2 + (27a^3 b... |
| Hello! Consider the quartic polynomial $$(48a^2+16b)x^4-(40a^3+168 ab)x^3+(-45 a^4+225a^2 b+ 72b^2)x^2+(27a^3 b - 162 ab^2)x+27b^3, \qquad (*)$$ where $a |
| That was the first thing I tried, but when I just type (for instance) x = 4 without using the symbol '=' then I don't get the typical in[3]:= x = 4 out[3]: =4 I assume you mean this: ![enter image description here][1] [1]:... |
| Hello! I have solved this system of equations (see below) in *Mathematica* for $x$ where the coefficients of the equations/inequalities are functions of $a,b$ and $c$. *Mathematica* then displays real solutions $x$ with constraints on $a,\,b$ and... |
| I'm using the online version on wolframcloud.com. I noticed that I added x 0 obviously works to only extracts real solutions. So, since I do not have a legitimate version of Mathematica I guess I should hold on to that? Could you perhaps show me... |
| If you want I will explain a little bit more background. In fact, the original problem was to solve the system of equations $$\begin{cases} 16\mu^4 -40 a \mu^3+(15a^2+24b)\mu^2-18ab\mu + 3b^2 = 0 \\ 5a\mu-4\mu^2-b >0 \\ 15a \mu-20 \mu^2 - 3b 3,... |
| Thanks for the answer! That is indeed the system that I had to solve. The constraints on $a$ and $b$ are easy to handle. The constraint on $c$ is harder to handle. if Mathematica gives as output $\mbox{Root}[\ldots,1]$ and... |