For $x \neq 0$, solve $\sqrt{x+4}-2 =x$
Attempt: $x+4=x^2+4x+4 \implies x^2+3x=0$, $x=0$ or $x=-3$, but doing $x+2=-1$, then we just find $x=0$. So for nonzero $x$ is this an absurd fact?
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$\begingroup$ It might be worth asking if you’re also looking at complex solutions? I’m not sure what topics are being covered but I have seen complex numbers in parts of algebra like this. $\endgroup$Clayton– Clayton2021-10-29 16:23:01 +00:00Commented Oct 29, 2021 at 16:23
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$\begingroup$ $x=a$ implies($\implies$) $f(x) = f(a) $ however the reverse implication $f(x) = f(a) $ implies($\implies$) $x=a$ is generally not true; functions for which the reverse implication always works are called injective (1 to 1) functions. Here $f(x) = x^2$ is not injective and so there can be "extra" solutions which you found there is. $\endgroup$kingW3– kingW32021-10-29 17:32:06 +00:00Commented Oct 29, 2021 at 17:32
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5 Answers
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Squaring both sides of an equation (as you validly did in your first step) potentially creates extraneous solutions. So, you did the right thing to plug in the candidate solutions into the original equation.
And yes, the candidate solution $x=-3$ is extraneous (violating the given equation's implicit condition that $x+2$ is nonnegative) and ought to be discarded, leaving $x=0$ as the only actual solution.
Extraneous solutions aren't “absurd facts” though, and are in fact commonplace. In general, when solving equations—unless you have ensured that every step is “reversible”— you ought to filter away any extraneous solution.
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Instead of "absurd fact" we simply say that the equation has no solution (for $x\ne 0$).
One implicit condition from the equation is that $x+2\geq 0$.
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2$\begingroup$ Also be aware that we take $\sqrt{x + 4} \geq 0$ if $x$ is a real number. Therefore if $x = -3$, the LHS becomes $\sqrt{1} - 2 = - 1$ while the RHS becomes $-3$ which are not equal. $\endgroup$Yuki.F– Yuki.F2021-10-29 16:05:46 +00:00Commented Oct 29, 2021 at 16:05
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The conventional meaning of the square root symbol is the non-negative square root.
The problem comes about when you square (or raise to even powers) both sides of an equation when trying to solve it. If you start with $x = a$, but then square it, you end up with $x^2 = a^2$. That second equation - solve it like so: $x^2 - a^2 = 0 \implies (x+a)(x-a) = 0$ - has roots of $x = \pm a$, and clearly one of them is not acceptable since you started with $x = a$ (single root). It's just a simple artefact of the process.
These "extra" roots are called redundant or extraneous roots and they should be rejected. You should test any roots you get after squaring with the original equation and discard any that don't satisfy the original equation.
Now you should be able to see what's going on here. $x =-3$ "satisfies" the original equation only if you take the non-standard approach of allowing the square root to return a negative root (which is not the typical meaning).
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There is no solution (for $x\neq 0$), and your reasoning is correct.
The reason this arises is because squaring both sides of the equation is not a reversible step: we can't square root both sides of an equation, because we would need both the positive and negative square root.
You can make a little bit more sense out of $x=-3$ as an answer by noticing that $(x+4)^2=(-1)^2$. If $\sqrt{}$ meant "negative square root" then $x=-3$ would be a solution. However, that's not what the symbol indicates.
A more trivial example of doing something irreversal to an equation which will allow invalid solutions that need to be checked against the original equation is multiplying everything by $0$. For instance, if we tried to solve $2x=4$ by multiplying by $0$, we get $0=0$, which is true for any value of $x$, but we have just allowed in infinitely many solutions (every value of $x$ other than $2$) that do not solve the original equation.
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We can boil this down to a very simple example. $$ \sqrt{x} = -1 $$
Now, we know for sure that this isn't possible, since that symbol $\sqrt{x}$ is defined to be the positive number that squares to $x$. Since this is an "absurd fact", we can say that this equation has no solutions.
However, if we naively attempt to isolate $x$ by squaring both sides (because we learned the square and square root are opposites*), something happens. The equation becomes $$x = 1$$
which is no longer absurd, and no longer has no solutions. In fact, $x=1$ is (obviously) a solution to this equation! What's the deal?
When we squared both sides of the equation, we actually changed the equation in some fundamental way; when we do something like add or subtract a number, or multiply or divide by a non-zero number, we don't change the equation all. So, it seems pretty valuable to know which operations will change an equation (and how) so that we can adjust for it.
If an operation can turn two different values into the same value, then it may change the equation. The squaring function has this property, because $$(-x)^2 = (-x)\cdot (-x) = x\cdot x = x^2$$
Using our "solution" above in the absurd equation, we get a pretty concrete absurdity $$ 1 = -1 $$
but squaring that equation gives the true statement
$$ 1 = 1 $$
and this is where our false solution came from. The two sides of the equation were different, but squaring can turn different numbers into the same number, making an equality where there was none.
Luckily, it's easy to adjust for this: just check all of our solutions in the original equation. If we get something absurd, just toss that solution, and the ones that work are the ones we keep.
* There are a lot of functions which are "inverses" of one another in some limited sense which is not always explained well. Even powers and even roots, exponential and logarithm, trig and trig inverse, etc. are all examples of this, and you should be careful with how you use any of these in the process of solving an equation, lest you create or destroy solutions without realizing it!
