For which values of $a$, equation $\sqrt{x-3}+ax=2a+3$ has one solution for $x$? For which values of $a$, euation $\sqrt{x-3}+ax=2a+3$ has one solution for $x$?
According to my student solution book, the answer is $a\in \left \{ \frac{3-\sqrt{10}}{2} \right \}\cup \left [ 0;3 \right ]$ but I cannot figure out how they did it. And I'm more interested in how to arrive at the correct answer than I am in the answer itself.
 A: Hint:
take the square to get
$x-3=(2a+3)^2+a^2x^2-2ax(2a+3)$
which becomes
$a^2x^2-( 2a(2a+3)+1)x+(2a+3)^2+3=0$
it has one solution if the discriminant is zero.
A: $\sqrt{x-3}$ is defined in $\Bbb R$ only when $x\ge 3$ and the corresponding value of $a$ for $x=3$ is $3$. For other values of $x$, making the equality gives for $x=3+\delta\gt 3$ 
$$a=\frac{3}{x-2}=\frac{3}{1+\delta}\Rightarrow a\lt3$$
On the other hand $a$ can not be negative (in this context!) and tends to $0$ when $\delta \to \infty$ hence the values of $a$ form the closed interval $[0,3]$.
However there are negative values of $a$, other than $\frac{3-\sqrt{10}}{2}$, for which one has equality, for example $a\approx-0.05$ which with $x\approx 17.1$ gives the equality (obviously equivalent to the given one)
$$\sqrt{x-3}-3=a(2-x)\\\sqrt{17.1-3}-3\approx -0.05(2-17.1)\approx 0.755$$
Notice that $\frac{3-\sqrt{10}}{2}\approx-0.081\lt -0.05$ and we can ensure that for other negative values of $a$ one has equality also. In the figure below it is show the intersection point of curves $\sqrt{x-3}-3$ and $a(2-x)$ for $a=-0.07$. There is a little interval of negative numbers to add to the interval $[0,3]$ and not only the negative point $\frac{3-\sqrt{10}}{2}$

A: Set $\sqrt{x-3}=t$, so $x=t^2+3$ and the equation becomes
$$
t+at^2+3a=2a+3 \tag{*}
$$
or
$$
at^2+t+a-3=0
$$
This equation should have a single nonnegative solution. The case $a=0$ gives $t=3$, hence $x=12$; so we can assume $a\ne0$. First of all it must have one, so the discriminant should be nonnegative:
$$
1-4a^2+12a\ge0
$$
so $4a^2-12a-1\le0$, which holds for
$$
\frac{3-\sqrt{10}}{2}\le a\le\frac{3+\sqrt{10}}{2}
$$
Now we can distinguish a few cases based on Descartes' rule of signs
Case $a=(3+\sqrt{10})/2$
Equation (*) has a double negative solution.
Case $3<a<(3+\sqrt{10})/2$
Equation (*) has two distinct negative solutions.
Case $a=3$
Equation (*) has a negative solution and a null solution
Case $0<a<3$
Equation (*) has one negative and one positive solution
Case $a=0$
Equation (*) has a single positive solution, as already seen.
Case $(3-\sqrt{10})/2<a<0$
Equation (*) has two positive solutions
Case $a=(3-\sqrt{10})/2$
Equation (*) has a double positive solution
Summary
The cases where our equation has a single positive solution are when (*) has a unique positive solution (double or not). So we find
$$
a\in\left\{\frac{3-\sqrt{10}}{2}\right\}\cup[0,3]
$$
