, 2014

Let $a$, $b$, $c$, $d$ be the quadratic polynomials defined by
$$\begin{aligned} a(x) &= 2x^2 + 2x + 3 & b(x) &= 2x^2 + 2 \\ c(x) &= 3x^2 + 2x - 1 & d(x) &= x^2 + 6.\end{aligned}$$
The equation to be solved is writable in the form
$$\sqrt{a(x)} + \sqrt{b(x)} = \sqrt{c(x)} + \sqrt{d(x)}.$$

Observe that $a$, $b$, $d$ are strictly positive for all real numbers $x$, and that $c(x) = (3x - 1)(x + 1) \ge 0$ iff $x \in (-\infty, -1) \cup (1/3, \infty) \equiv S$. Thus the solution set of the equation is a subset of $S$.

Note too that $a + b = c + d$, i.e., $a - d = c - b \equiv p$, say. Thus the equation to be solved is
$$\sqrt{d(x) + p(x)} + \sqrt{b(x)} = \sqrt{b(x) + p(x)} + \sqrt{d(x)}.$$
If the real number $x$ satisfies this equation, then, by squaring and cancelling common terms, we see that
$$\sqrt{(d(x) + p(x))b(x)} = \sqrt{(b(x) + p(x))d(x)},$$
whence squaring once more and cancelling the common term $b(x)d(x)$ we arrive at the equation $p(x)(b(x) - d(x)) = 0$. Hence, either $p(x) = 0$ or $b(x) = d(x)$. Equivalently,
$$0 = p(x) = x^2 + 2x - 3 = (x + 3)(x - 1), \quad x \in \{-3, 1\}.$$
Or
$$0 = b(x) - d(x) = x^2 - 4 = (x + 2)(x - 2), \quad x \in \{-2, 2\}.$$
Thus the solution set is a subset of
$$T = \{-3, -2, 1, 2\} \cap S = \{-3, -2, 1, 2\}.$$
By direct substitution it can be verified that each member of $T$ satisfies the given equation. In other words, $T$ is the required solution set.

By squaring twice it can be seen that if $x$ satisfies the given equation then it must satisfy the equation
$$(2x^2 + 2x + 3)(2x^2 + 2) = (3x^2 + 2x - 1)(x^2 + 6).$$
Expanding both sides we see that this is equivalent to the equation
$$4x^4 + 4x^3 + 10x^2 + 4x + 6 = 3x^4 + 2x^3 + 17x^2 + 12x - 6.$$
In other words,
$$x^4 + 2x^3 - 7x^2 - 8x + 12 = 0,$$
or
$$(x^2 - 3)(x^2 - 4) + 2x(x^2 - 4) = 0.$$
Hence $x^2 = 4$ or $x^2 + 2x - 3 = 0$. By inspection of this and verification of the original equation it can be seen that $T = \{-3, -2, 1, 2\}$ is the solution set.