It seems pretty simple, but it's keeping me busy for some time:
x^2-a-√(x-a)=0
Find roots of X in a dependency of A.
Thanks a lot.
Here's a kicker
$\displaystyle \displaystyle x^2-a-\sqrt{x-a}=0$
$\displaystyle \displaystyle x^2-a= \sqrt{x-a}$
Now square both sides
$\displaystyle \displaystyle (x^2-a)^2= x-a$
Your turn, expand the left hand side, what do you get?
True, I jumped pretty quick on that, if the RHS side was independant of x you could sub $\displaystyle u = x^2$ and make it a quadratic.
Other than that, have you seen the quartic formula?
Quartic function - Wikipedia, the free encyclopedia
x^2-a-√(x-a)=0 - Wolfram|Alpha. The solutions are ridiculos so make sure you have correct equation.
In cases like this is better to spend a minute in thinking before to connect to Wiki or Wolfram Alpha...
The relation...
$\displaystyle x^{2}-a - \sqrt{x-a}=0$ (1)
... define implicity an $\displaystyle x= f(a)$ or equivalently an $\displaystyle a=f^{-1}(x)$. Clearly finding $\displaystyle f(a)$ requires the [impratical...] solution of a quartic and finding $\displaystyle f^{-1}(x)$ requires the [easy enough...] solution of a quadratic, so that the second alternative is preferable. Ther result is...
$\displaystyle a= \frac{2\ x^{2} -1 \pm \sqrt{1+4\ x -4\ x^{2}}}{2}$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
In cases like this it is better a little of knowledge of elementary theory!...
The first branch of the function...
$\displaystyle a=f^{-1}(x)= \frac{2\ x^{2}-1 + \sqrt{1 + 4\ x -4\ x^{2}}}{2}$ (1)
... has the properies that $\displaystyle a(0)=0$ so that exists a neighborhood of $\displaystyle x=0$ where its inverse $\displaystyle x(a)$ exists and can also be effectively computed. The same is for the other branch...
$\displaystyle a=f^{-1}(x)=\frac{2\ x^{2}-1 - \sqrt{1 + 4\ x -4\ x^{2}}}{2}$ (2)
... for which is $\displaystyle a(0)=-1$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Wasn't satisfied from this "Quartic" solution myself, had no need to break the equation down as mentioned in Wiki (i've got the main idea). After couple of minutes was on my way to perform the same, didn't finished but the idea above was clear enough, shame i saw your post just now
Thanks.