4.51 If $x$ is a real number, and $4 y^{2}+4 x y+x+6=0$, then the set of all $x$ values for which $y$ is real is
(A) $x \leqslant-2$ or $x \geqslant 3$.
(B) $x \leqslant 2$ or $x \geqslant 3$.
(C) $x \leqslant-3$ or $x \geqslant 2$.
(D) $-3 \leqslant x \leqslant 2$.
(E) $-2 \leqslant x \leqslant 3$.
(17th American High School Mathematics Examination, 1966)
［Solution 1］First, consider the given equation as an equation in $y$:
$$
4 y^{2}+(4 x) y+(x+6)=0,
$$

If $y$ is to be real, then $\Delta=(4 x)^{2}-4 \cdot 4 \cdot(x+6) \geqslant 0$.
Solving this, we get $x \geqslant 3$ or $x \leqslant-2$.
Therefore, the answer is $(A)$.
[Solution 2] From the given, we have $4 y^{2}+4 x y+x^{2}-\left(x^{2}-x-6\right)=0$.
This simplifies to
$$
(2 y+x)^{2}=(x+2)(x-3),
$$

The left side is always $\geqslant 0$ (since $x, y$ are real numbers), thus $(x+2)(x-3) \geqslant 0$,
which gives $\quad x \geqslant 3$ or $x \leqslant-2$.
Therefore, the answer is $(A)$.