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.
 A: The confusion you are having comes from the concept of finding the solutions to $z^2 = 36$. Definitely $z = \pm 6$ are solutions because $(-6)^2 = 36 = 6^2$, but this is not the same thing as $y = \sqrt{36}$. Otherwise, we get nonsense like $-6 = 6$ which isn't true.
If we plug the answer $x = -6$ back into the original equation, we have $$\sqrt{6 - 5(-6)} = \sqrt{36} = 6 = -6.$$ Again, we get nonsense of $-6 = 6$. Even it were positive and negative $6$, you'd have $-6 = \pm 6$. This is true for only one of the values, which shows it's not valid to assume that $\sqrt{36} = \pm 6$.
A: Because the equation $\sqrt{6-5x}=x$ is not equivalent to $6-5x = x^2$ but to
$6-5x = x^2$ and $x \geq \frac{6}{5}$.
Squaring can change the set of solutions. For example, consider $x = 4$ and $x^2 = 16$. We also have $\sqrt 1$, which have the solutions $x_1 = 1, x_2 = -1$, but $1 \neq -1.$ If you square a radical equation you'll always get a real and a false root.
For this very same reason, dividing by $x$ also means you can "lose" solutions.
A: You are confused a little bit. 
Let me clear you. 
Consider these two equations:
$y^2=36$ and $z=\sqrt{36}$
See the degree of $y$ which is 2. that's why you get two solutions here in this case, that is
$$y=\pm 6$$
Se the degree of $z$ which is 1 . that's why one must get a single root. Moreover, sqrt of a number can't be negative.
Therefore ,
$$z=6$$
Now again satisfy your results and now you'll know the answer.
A: 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
$$x = 1$$
A: The square root, denoted by $$y(x)=\sqrt x$$ is a function of the variable $x$. As such it can take only one value.
Naturally, one always chooses the positive solution of
$$y^2=x.$$
This is called the main branch.

Remember:
$$y^2=x$$ has two solutions in $y$, but
$$y=\sqrt x$$ only one.
