A XVI. secondary school student competition, Problem 1. (L. 1. o.)

The following equation can be solved with a suitable choice of signs for the square roots involved:

$$
\pm \sqrt{x+2} \pm \sqrt{x-1}= \pm \sqrt{4 x-7}
$$
Squaring both sides, we get:

$$
x+2+x-1 \pm 2 \sqrt{(x+2)(x-1)}=4 x-7
$$

After simplification and combining like terms:

$$
\pm \sqrt{(x+2)(x-1)}=x-4 \ldots
$$

At this point, $x \gtrless 4$ depending on whether we take the square roots on the left side with the same or opposite signs! Squaring both sides of (2):

$$
x^{2}+x-2=x^{2}-8 x+16, \quad \text { or } \quad 0 \cdot x^{2}+9 x-18=0 \ldots
$$

One root of (3) is $x_{1}=2$; the other root has become infinite. $x_{1}=24, x=+\infty$ satisfies the

$$
\sqrt{x+2}-\sqrt{x-1}=\sqrt{4 x-7} \quad \text { or } \quad-\sqrt{x+2}+\sqrt{x-1}=-\sqrt{4 x-7}
$$

equation.

Freud Géza (Berzsenyi Dániel g. VII. o. Bp. V.).

Note. The square roots in the equation are all real if the

$$
x \geqq-2, \quad x \geqq 1, \quad x \geqq \frac{7}{4}
$$

conditions are simultaneously satisfied. Therefore, $x \geqq \frac{7}{4}$ must hold, in other words, the root of the equation can only be in the domain of $x \geqq \frac{7}{4}$.