$2 \cdot 133$ If $a \geqslant 1$, then the sum of all real roots of the equation $\sqrt{a-\sqrt{a+x}}=x$ is
(A) $\sqrt{a}-1$.
(B) $\frac{\sqrt{a}-1}{2}$.
(C) $\sqrt{a-1}$.
(D) $\frac{\sqrt{a-1}}{2}$.
(E) $\frac{\sqrt{4 a-3}-1}{2}$.
(32nd American High School Mathematics Examination, 1981)
[Solution] Since $x$ is an arithmetic root, we have $x \geqslant 0$. Given $x \geqslant 0$, the given equation is equivalent to
$$
a-\sqrt{a+x}=x^{2} .
$$

The left side of the above equation is a decreasing function of $x$, while the right side is an increasing function of $x$. Therefore, it is easy to verify that this equation has only one solution. For this purpose, let $y=\sqrt{a+x}$. Then we have:
$a-y=x^{2}$, thus $a-y-y^{2}=x^{2}-y^{2}$.
Noting that $y^{2}=a+x$, we have $a-y-(a+x)=x^{2}-y^{2}$,

which gives
$$
-(x+y)=(x+y)(x-y), \\
0=(x+y)(x-y+1) .
$$

Since $a \geqslant 1$ and $x \geqslant 0$, we have $y>0$, so $x+y \neq 0$. Therefore,
$$
x-y+1=0 \text {, i.e., } x+1=y \text {. }
$$

Since $y=\sqrt{a+x}$, we have
$$
x+1=\sqrt{a+x} \text {, or }(x+1)^{2}=a+x \text {. }
$$

Solving this, we get
$$
x=\frac{-1 \pm \sqrt{4 a-3}}{2} .
$$

Given $x \geqslant 0$, we have $x=\frac{\sqrt{4 a-3}-1}{2}$. Therefore, the answer is $(E)$.