11.44 Solve the equation $\sqrt{3 a-2 x}+x=a$ and investigate for which values of the parameter $a$ it has roots (and how many) and for which values it does not have roots.
11.44 Let $\sqrt{3 a-2 x}=t \geq 0$. Then $3 a-2 x=t^{2}$, from which $x=\frac{3 a}{2}-\frac{t^{2}}{2}$; the given equation will take the form $t^{2}-2 t=a$. Let's construct the parabola $y=t^{2}-2 t$ and the line $y=a$ (in Fig. 11.16, the part of the parabola corresponding to $t \geq 0$ is highlighted with a bold line). Using the graph, we establish that for $a < -1$ there are no roots, for $a = -1$ there is one root, and for $-1 < a < 0$ there are two roots. For $a = 0$ there is one root, and for $a > 0$ there is one root.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-362.jpg?height=477&width=664&top_left_y=1090&top_left_x=318)

Fig. 11.16

It remains to find these roots. For this, we solve the equation $t^{2}-2 t-a=0$, from which we get $t_{1,2}=1 \pm \sqrt{a+1}$, or, returning to the original variable, $x_{1,2}=a-1 \pm \sqrt{a+1}$. Thus, for $-1 < a < 0$ there are two roots $x_{1,2}=a-1 \pm \sqrt{a+1}$; for $a = -1$ there is one root $x = a - 1 = -2$; for $a = 0$ there is one root $x = a - 1 - \sqrt{a+1} = -1$; finally, for $a > 0$ there is one root $x = a - 1 - \sqrt{a+1}$; and for $a < -1$ the equation has no roots.