Problems in $\sqrt{5x+4}=x-2$ So when I solve $\sqrt{5x+4}=x-2$, I end up with $x(x-9)=0$. Yet only when $x=9$ is the original inequality satisfied. Can somebody give me some details on what exactly goes wrong here? 
 A: I guess you started by squaring both sides:
$$\sqrt{5x+4} = x+2 \tag{A}$$
$$5x+4 = (x+2)^2 \tag{B}$$
While it is true that $(A)$ implies $(B)$, the other direction does not hold. Maybe a simple example is following: While $3^2 = 9 = (-3)^2$ is true, $3 = -3$ is clearly not true.
So if some $x = x_0 $ satisfis equation $(A)$ it must also satisfy the second equation $(B)$, but if some $x = x_1$ satisfies equation $(B)$ it does not necessarily satisfy $(A)$.
A: If you insert $x=0$, you can see what goes on:
$$
\sqrt{5\cdot0+4}=2\\
0-2=-2
$$
These are different. However, the moment you square both sides, you don't see this difference any more, as both sides become $4$.
In general, when you square an equation, in addition to the actual solutions where the left-hand side is equal to the right-hand side of the original equation, you suddenly include the cases where the left-hand side is the negative of the right-hand side.
A: It is just because$$\sqrt{5x+4}=x-2\implies5x+4=(x-2)^2,\tag1$$but this is not an equivalence. However, it is clear that, if $x<2$, then the equality $\sqrt{5x+4}=x-2$ cannot hold (since $\sqrt{5x+4}\geqslant0$ and $x-2<0$). And if you restrict $x$ to $[2,\infty)$ then, yes, $(1)$ is an equivalence.
A: At $x=0$, the LHS is $2$, while the RHS is $-2$. So the new solution come from squaring both sides, because $(2)^2=(-2)^2$.
A: When you square both sides to solve, you create an extra incorrect answer. See, at x=0, the squares of both sides of the equation are equal, but the sides of the original equation are not. The constraint that x-2 cannot be negative is lost in the process of squaring, and thus an extra solution is created. 
A: As a general rule, when you are trying to solve an equation with radicals in the form $\sqrt{P(x)}>Q(x)$, you have to solve the following system:
$$\left\{\begin{matrix}
P(x)\geq0
\\Q(x)\geq0
\\P(x)=Q(x)^2
\end{matrix}\right.$$
So, I have:
$$\left\{\begin{matrix}
5x+4\geq0
\\x-2\geq0
\\5x+4=x^2-4x+4
\end{matrix}\right.$$
The solutions of the system are: $x(x-9)=0$ but $x=0$ isn't acceptable, so the only correct is $x=9$.
