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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.

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    $\begingroup$ No, $\sqrt{36}$ can only be 6, not -6. The square root of a non-negative number is uniquely defined: it is a non-negative number. (Put differently, $\sqrt{} $ is a function: each number in its domain has a unique image.) $\endgroup$ – Clement C. Feb 2 '16 at 3:17
  • $\begingroup$ @ClementC. By definition, $\sqrt{b}=a$ if $b=a^2$, so how can you say that $\sqrt{36}$ only equals $6$ if $36=(-6)^2$? $\endgroup$ – Allison Feb 2 '16 at 3:19
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    $\begingroup$ No, by definition $\sqrt{b}$ is the unique non-negative number $a$ such that $b=a^2$. $\endgroup$ – Clement C. Feb 2 '16 at 3:20
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    $\begingroup$ Because then you are not looking for the square root: you are looking for any number satisfying the equation, which are the square root and its negative counterpart. $\endgroup$ – Clement C. Feb 2 '16 at 3:24
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    $\begingroup$ It's confusing. And those of us who like to treat this as a "gotcha!" don't help. But basically if we write an expression such as $\sqrt{foo}$, an expression must have a single value for a statement to be consistent or meaningful. By convention we define $\sqrt$ to be the non-negative value. $x^2 = c$ has two solutions (if c > 0) but $\sqrt c$ is the positive one. And $-\sqrt c$ is the negative one. The tricky think is $\sqrt{ 6 -5x} = x$ does not actually mean $6 - 5x = x^2$. But it means $6 - 5x = x^2$ AND $x \ge 0$. $\endgroup$ – fleablood Feb 2 '16 at 3:41
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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 $\sqrt{36} = \pm 6$.

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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.

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  • $\begingroup$ $x\geq \frac{6}{5}$ is incorrect. If you take an $x$ that is larger than $\frac{6}{5}$, you will have a negative number underneath of the square root, which has no real output. As mentioned to the other person who made the same mistake, $\sqrt{6-5x}=x$ implies $6-5x=x^2$, $x\geq 0$, and $x\color{red}{\leq}\frac{6}{5}$. What is the problem with the OP's solution is the fact that $x\geq 0$ for the reason that the output of the (principle) square root must be non-negative (as already mentioned several times above). $\endgroup$ – JMoravitz Feb 2 '16 at 3:39
  • $\begingroup$ Was $x \ge 6/5$ a typo for $x \ge 0$. 1 < 6/5. $\endgroup$ – fleablood Feb 2 '16 at 3:46
  • $\begingroup$ I'd say $\sqrt {6 -5x}$ being expressed implies $6-5x \ge 0$ but it isn't equivalent to it. After all, we can write $\sqrt{-5}$ but that doesn't meant it's a valid real number. $\sqrt{6 -5x} = x$ is equivalent to $6-5x = x^2;x\ge0$. It also implies $6-5x \ge 0$ but that isn't a defining aspect. Also $x = -6$ satisfies $6 -5x \ge 0$ ($6 - 5(-6) = 36 \ge 0$.) $\endgroup$ – fleablood Feb 2 '16 at 3:51
  • $\begingroup$ @JMoravitz But now we are debating semantics. Yes, the square root strictly speaking only implies that the root is positive. However, in our case, it doesn't matter because no roots in that interval exists anyway. $\endgroup$ – Markus Klyver Feb 12 '16 at 3:29
  • $\begingroup$ @fleablood See my comment above to JMoravitz. $\endgroup$ – Markus Klyver Feb 12 '16 at 3:29
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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.

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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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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.

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