Square root confusion: Why am I getting an answer if it doesn't work?

Alright, so I have $\sqrt{x-15} = 3-\sqrt{x}$. I first square both sides to get $x-15 = (3-\sqrt{x})(3-\sqrt{x})$ which simplifies to $x-15 = 9 -6\sqrt{x} + x$.

I solved for $x$ and got $x = 16$, however, when I plug it in, the equation doesn't work. Why does this happen?

Solution 1:

You did good work in doing what many students forget to do - checking that your proposed solution actually is a solution.

Whenever you square both sides of an equation, you potentially introduce new solutions that may or may not be solutions of the original equation. Consider the simple example $x=1$. Obviously $x=1$ is the solution, but squaring both sides gives $x^2 = 1$, which has solutions $x=\pm 1$. Over $\mathbb{R}$ it is most definitely not true that $-1=1$, so $x=-1$ cannot be a solution here.

You should always be careful and check your proposed solutions by plugging them back into the original equation like you have done here.

Indeed, your equation actually has no solutions, which is what you should conclude when asked to solve it.

Solution 2:

Before doing some computation, you should check the condition of variable. Here, you have the conditions $\begin{cases} x - 15 &\geq& 0, \\ 3 - \sqrt{x} &\geq& 0, \\ x \geq 0. \end{cases}$

You will get that $x \geq 15$ and $0 \leq x \leq 9$. So, you can conclude that the equation has no solution.

Solution 3:

What you have done with your work is prove:

(1) If $x$ satisfies that equation, then $x=16$.

This is not the same thing as

(2) $x$ satisfies that equation if and only if $x = 16$.

Typically, "solving" an equation means to produce a statement like (2).

However, generally, the method of starting with the thing you're trying to solve and making algebraic simplifications only winds up proving something like (1), and additional work is needed to determine the correct statement like (2).

A typical form of that additional work is to check each possibility of the conclusion (here, $x=16$) to see which are actually solutions. Here, we eliminate the only possibility and get

$x$ satisfies that equation if and only if (contradiction).

or, more naturally phrased,

That equation has no solutions for $x$.

One case where additional work is not needed is if every step is "reversible"; i.e. your work is a chain of if and only if steps: e.g. subtraction is reversible, and

$3x + 1 = 5x - 2$ if and only if $1 = 2x - 2$

A lot of basic algebra exercises happen to be solved through reversible steps, which is why you can get the right answer without actually learning how to reason correctly.

Squaring is (generally) not reversible.