Solutions of $\sqrt{x^2+5x-14} + |x^2+4x-12|=0$

Question

My attempt:

Given,

$$\sqrt{x^2+5x-14} + |x^2+4x-12|=0 \tag{1}$$

Since $|a|=\sqrt{a^2}$, $$\sqrt{x^2+5x-14}=-\sqrt{(x^2+4x-12)^2}$$

Squaring both sides, $$x^2+5x-14=(x^2+4x-12)^2$$

When I simplify the above, I get two real solutions: $x=2$ and $x=2.138 \text{ (approximately)}$. However, there is only one solution to equation $(1)$ according to Wolfram Alpha, $x=2$.

Is my solution incorrect? If so, where did I go wrong?

Answer 1

When you square, you may get additional solutions. You need to test those solutions in the original equation.

Alternatively,

$\sqrt{x^2+5x-14} + |x^2+4x-12|=0$

$\implies x^2+5x-14 = 0 \ \text { and } x^2+4x-12 = 0$

$\implies (x-2) (x+7) = 0 \ \text { and } (x-2)(x+6) = 0$

So only solution that works is $x = 2$.

Answer 2

Why not factorise first, since both quadratics clearly very easily factorise, into $\sqrt{(x+7)(x-2)} + |(x+6)(x-2)| = 0 \implies$
$\sqrt{(x+7)(x-2)} = -|(x+6)(x-2)| \implies $
$(x+7)(x-2) = ((x+6)(x-2))^2$

This very clearly only has one solution, being $x=2$, since both terms must be zero

Answer 3

Both $\sqrt{x^2+5x-14}\geq 0$ and $|x^2+4x-12|\geq 0$

So, both of them must equal to zero, since equation is equal to zero.

$$\sqrt{x^2+5x-14}= 0$$

$$x^2+5x-14=0$$

$x=-7$, or $x=2$

$$|x^2+4x-12|= 0$$

$$x^2+4x-12=0$$

$x=-6$, or $x=2$

Finally $x=2$

Answer 4

When you squared both sides, you generated candidate solutions that each had to be checked against the original equation. As a simpler example:

$\sqrt{4} = x.$

Squaring both sides yields the equation $4 = x^2$, which has the two candidate solutions of $x = \pm 2.$ Each candidate solution must be checked against the original problem to see if it satisfies the problem.

More formally, you have that if a value satisfies the original equation, and you square both sides, then the value will satisfy the new equation. However, this is (in general) a one-way implication. This means that the new equation may have (candidate) solutions that don't satisfy the original equation.