<s>[INST] Why is the solution to $\sqrt{6-5x}=x$ only $x=1$ and not $x=-6$? I solved the equation $\sqrt{6-5x}=x$ as follows:
$$(\sqrt{6-5x})^2=x^2$$
$$6-5x=x^2$$
$$0=x^2+5x-6=(x+6)(x-1)$$
$$x=-6 \quad \text{or} \quad x=1$$
If I plug in $x=-6$ into the original equation, I get $\sqrt{6+30}=\sqrt{36}=\pm 6$ and if I plug in $x=1$, I get $\sqrt{6-5}=\sqrt{1}=\pm 1$.
To me it seems that both values satisfy the original equation. I am using an online education system for my class called MyMathLab and the solution is only $x=1$. Why is that?
Thank you in advance.
 [/INST] You may not select all solutions from an expression $B$ derived (calculated) from expression $A$ when $B$ is not equivalent to $A$, and have to refer to the initial expression to consider the valid values.
Here (notice the / within the $\Leftrightarrow$)
$$\sqrt{6-5x}=x\quad\not\Leftrightarrow\quad6-5x=x^2$$
because removing the $\sqrt{ }$ removes a condition on the expression as well, as $\sqrt{x}$ is valid for $x \geqslant 0$, thus $\sqrt{\text{expression}} = x$ implies $x \geqslant 0$. And you may solve
$$6-5x \geqslant 0$$
that gives $x \leqslant \frac{6}{5}$.
From now on, you know the requirements: $x \geqslant 0$ and $x \leqslant \frac{6}{5}$.
$$\sqrt{6-5x}=x\quad\implies\quad6-5x=x^2,\quad 0\leqslant x \leqslant \frac{6}{5}$$
Then after solving $6-5x=x^2$ that gives 
$$x=-6 \quad \text{or} \quad x=1$$
you refer to the requirements to remove the invalid $x = -6$, and may safely give the only solution
$$x = 1$$
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