Why does $\sqrt{2x+15}-6=x$ have an "imposter" solution?

Question

If you solve $\sqrt{2x+15}-6=x$ for $x$, you get: $$ \sqrt{2x+15} = x+6 $$ $$ 2x+15=(x+6)^2 $$ $$ 2x+15=x^2+12x+36 $$ $$ x^2 + 10x + 21 = 0 $$ $$ (x+7)(x+3) = 0 $$ $$ x=-3,-7 $$ But, while $x=-3$ works, $x=-7$ does not. I don't think there are any flaws in the arithmetic, or at least I can't find any (dividing by zero accidentally, etc.).

Why would this method of solution yield one answer that works and one that is completely false? What's happening here? Is there a better way I could have solved this to get only answers that are true?

Answer 1

The "imposter" solution cropped up when you squared. Squaring both sides of an equation is not an equivalence operation; while it is true that from $x=y$ it follows that $x^2=y^2$, the reverse is not true, as $(-x)^2 = x^2$ but (except for $x=0$) we have $-x\ne x$.

In your calculation, the step generating the extra solution was when you moved from $\sqrt{2x+15} = x+6$ to $2x+15=(x+6)^2$. You can easily verify that the second equation is solved both by $-3$ and $-7$.

Answer 2

The very first change to the equation you make does not give you an equivalent equation. Squaring both sides of the equation does not preserve the solution set. All solutions of the original solution still solve the new one but some solutions of the new one don't solve the old.

Consider $$x-2=4$$ and $$(x-2)^2=16$$ The first only has the solution $x=6$ but the second one also has the solution $x=-2$.

Answer 3

The issue is with squaring. As Ritz points out in the comments, $x=1$ has one solutions where $x^2=1$ has two. The number of solutions increases when you square.

This is because squaring is not invertible over $\Bbb{R}$. If $a=b$, THEN $a^2=b^2$ but the opposite doesn't hold, so when you plug the solution and go back, you get some extra solutions.

Answer 4

If you rewrite your equation after setting $t=x+6$, you get $$ \sqrt{2t+3}=t $$ Now you should clearly see where the problem lies: the left-hand side is a non-negative number (for $t\ge-3/2$, otherwise it is undefined). But, for $-3/2\le t<0$, the right-hand side is negative. Thus solutions of the equation must be non-negative.

When you square you “accept the risk” of introducing spurious results: indeed, from the false relation “$1=-1$” you get, after squaring, the true relation “$1=1$” which, however, doesn't make the original one true.

In this case, squaring gives $t^2-2t-3=0$, which has $t=3$ and $t=-1$ as roots; since $-3/2\le -1<0$, the root $t=-1$ must be discarded.

You can remain on the safe side if you always accompany the equation with the conditions under which squaring does not introduce rogue solutions: $$ \begin{cases} 2x+15=(x+6)^2\\[3px] x+6\ge0\\[3px] \color{gray}{2x+15\ge0} \end{cases} $$ The last one is the condition for the existence of the radical and I have greyed it out because it's implicit in the first equation.

After solving the equation, finding the roots $x=-7$ and $x=-3$, you discard the first one because it doesn't satisfy $x+6\ge0$.