For what a values does the equation has equal roots? Equation is:
$x^2+2a\sqrt{a^2-3x}+4=0$
Also, can someone please recommend me some good book or textbook about this subject? Something easy to read and understand.
Thanks!
 A: There are no clean solutions for generic $a$ if $x$ is indeed squared.  You can write the problem as
$$x^2+4 = -2a\sqrt{a^2-3x},$$
square both sides and use the solutions for $x$ noting that if $x$ is to be real (as opposed to complex) then there are no solutions for $a$ nonnegative or small negative.
A: This is not a full solution, but it might help get you closer.  At the very least it may help provide some visual intuition.
Consider the two functions $f(x)=x^2+4$ and $g(x)=-2a\sqrt{a^2-3x}$.  Your problem is equivalent to finding a value for $a$ such that $f(x)=g(x)$ has a unique solution.
Now the graph of $f(x)$ is, of course, a parabola (concave up) with vertex at $(0,4)$.
The graph of $g(x)$ is, for $a>0$, contained entirely in the 3rd and 4th quadrants (below the $x$-axis), so if $a>0$ the two graphs do not intersect at all.
So we consider only the case where $a<0$.  In this case the graph of $g(x)$ is defined on the domain $x \leq a^2/3$.  On that domain it is concave-down and strictly decreasing.  From these considerations we can see that $f(x)$ and $g(x)$ intersect in either two, one, or zero points.  Moreover if there is one solution, then at that point of intersection the two graphs are tangent.
So take the derivatives of both functions and set them equal to each other.  You get
$$2x = \frac{3a}{\sqrt{a^2-3x}}$$
or equivalently $\sqrt{a^2-3x}=\frac{3a}{2x}$.  Square this and cross-multiply; you will find that it has now been reduced to a 3rd degree polynomial equation.
Hope this helps!
