Why is the solution to $\sqrt{6-5x}=x$ only $x=1$ and not $x=-6$?

Question

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.

Answer 1

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 $\sqrt{36} = \pm 6$.

Answer 2

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.

Answer 3

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.

Answer 4

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 $$x = 1$$

Answer 5

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.

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Remember:

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