Problem
Find such value for constant $a$ that equation: $$ x^3+ax^2+3ax-10x=0 $$ has two solutions. Show that this equation has three solutions with any other value of $a$.
Attempt to solve
If we assume that polynomial $ax^2+bx+cx=0$ has solution of: $$ x=\frac{- b \pm \sqrt{b^2-4ac}}{2a} $$ (Quadratic formula)
$$ x^3+ax^2+3ax-10x=0 $$ $$ x(x^2+ax+3a-10)=0 $$
It is visible that this equation has solutions: $x=0$ or $x^2+ax+3a-10=0$
$$ x^2+ax+3a-10=0 $$ with Quadratic formula we have:
$$ x=\frac{-a \pm \sqrt{a^2-4(3a-10)}}{2} $$
If discriminant $>0$ equation has two real solutions, if discriminant $=0$ equation has one real solution, if discriminant $<0$ equation has no real solutions.
Since we already had one solution of $x=0$ and if $a^2-4(3a-10)=0$ that would make total of two solutions.
$$a^2-4(3a-10)=0$$ $$a^2-12a+40=0$$ $$a=\frac{12\pm \sqrt{(-12)^2-4\cdot 40}}{2}$$ $$ a=\frac{12 \pm \sqrt{-16}}{2} $$ $$ a_1=6-2i $$ $$ a_2=6+2i $$
This would indicate the discriminant cannot be $0$ with real inputs. The polynomial $a^2-12a+40$ is parabola that is opening upwards meaning all of it's values are positive. The global minima for this polynomial is when $a=6$.
* NON relevant info * I don't know if this happened by accident that local minima happens to be in same as in real part of both complex solutions.
If I mark the polynomial as function of $f(a)$
$$ f(a)=a^2-12a+40=0 $$ $$ f'(a)=\frac{d f(a)}{da}= \lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h} $$ $$ f'(a)=2a-12 $$ $$ f'(a)=0 $$ when $$ 2a-12=0 $$ $$ a=6 $$ $$ f''(a)=2 $$ which indicates $f(6)$ is minima If there is a way to determine local minima using complex solution of this equation please let me know. Also if this was just coincidence please let me know.
If $a^2-12+40 > 0$ for all real $a$. This means equation $x^3+ax^2+3ax-10x=0$ has always three solutions independent from value of constant $a$.
Now if I try to solve it in different way:
I have $x(x^2+ax+3a-10)=0$ meaning $x=0$ or $x^2+ax+3a-10=0$. $$ x^2+ax+3a-10=0 $$ if I set $x=0$ I have: $$ (0)^2+a(0)+3a-10=0 $$ $$ 3a-10=0 $$ $$ 3a=10 $$ $$ a=\frac{10}{3} $$
Now if I insert this in original equation:
$$ x^3+\left(\left(\frac{10}{3}\right)x^2+3\left(\frac{10}{3}\right)\right)x-10x=0$$ $$ x^3+\frac{10x^2}{3}+10x-10x=0$$ $$ x^3+\frac{10x^2}{3}=0 $$ $$ x^2\left(x+\frac{10}{3}\right) $$ which makes $x^2=0$ or $x+\frac{10}{3}=0$ $$ x=-\frac{10}{3} $$
Now we have total of two solutions from this equation when $a=\frac{10}{3}$. These solutions are $x=0$ and $x=-\frac{10}{3}$
If I try to compute discriminant with this value:
$$ \left(\frac{10}{3}\right)^2-4\left(3\cdot \frac{10}{3}-10\right)= \frac{100}{9} $$ which is $>0$ which would make it $3$ solutions in total ? But as I computed we had total of two solutions.
Now this would indicate that there is flaw in my method for testing if this equation has desired amount of solutions. Now if someone can tell me the reason why this test with discriminant is not working as intended ?
The problem in the first method is that there really no solutions to $a^2-4(3a-10)=0$ in the real numbers! After noticing this you need to remember that you already have one solution, namely: $x=0$, so if $(-a\pm \sqrt{a^2-4(3a-10)})/2$ will give $2$ solutions such that one of them is $0$ you'll have a total of $3$ solutions where $2$ of them are the same, hence you'll be left with only $2$ solutions.
For example if I say I have $3$ variables: $x_1,x_2,x_3$ and I ask you to give me values to those $3$ such that I'll have only $2$ different values, setting $x_1=x_2\ne x_3$ will answer my question, in your case $x_1,x_2,x_3$ are the roots and the task is the find the value of $a$ such that the set of $x_1,x_2,x_3$ will have exactly $2$ different values, you found $x_1$ by factoring $x$ out so now you left with finding $a$ such that $x_2=x_1$ and $x_3\ne x_1$. This is why the second method worked, you searched for $a$ that will give you root for $x^2+ax+3a-10$ at $x=0$, i.e. $\overbrace{x_1}^0=\overbrace{x_2}^0$
If you put $10/3$ to the quadratic formula you get $0,-10/3$, combining with the original $0$ you get $0,0,-10/3$ so only 2 different solutions