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I was solving a relatively difficult algebraic equation and I had to square both sides of the equation multiple times. I got a few solutions. However, when I inputed them back into the original equation some of them didn't work. How did I get them as a solution then?

James Chadwick
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    That's the thing about this stuff. You unknowingly divide by $0$ sometimes. You cancel a variable and lose solutions. You square something and end up generating false solutions. There are ways of getting around this problem. It's just practicing and familiarising yourself. – Bumblebee Apr 11 '23 at 07:03
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    If say $x = 5$ and $y = -5$, then $x^2 = y^2$ but it would be wrong to conclude that therefore $x = y$. So maybe you made a mistake along those lines, but it’s hard to tell without more detail about what you did. – NikS Apr 11 '23 at 07:13
  • See Why can't you square both sides of an equation?. – dxiv Apr 11 '23 at 07:16

3 Answers3

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$A=B$ implies $A^2=B^2$ (because $A=B$ means the expressions $A$ and $B$ denote the same number!), so squaring both sides of a true equation gives you another true equation.

However, $A^2=B^2$ doesn't imply $A=B$, because squaring is not injective. For example, $(-1)^2=1^2$ but $-1\ne1$. Therefore, a solution to $A^2=B^2$ is not necessarily a solution to $A=B$.

Karl
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You can't simply take the square and expect to get all correct result, as in the following obvious example:

$x=-x$ // Take the square:
$x^2=(-x)^2$ // Simplify:
$x^2=x^2$

=> all numbers are a solution, which is clearly wrong!

Why doesn't this work? Well, simply because when taking the square, you might get a negative number "being equal to" a positive number.

However, in the following case you might actually do that:

$\sqrt{x+5}=\sqrt{1-x}$ // Take the square:
$x+5=1-x$ // Simplify:
$x=-2$

... but even then, it's not that simple:
$\sqrt{x-3}=\sqrt{1-x}$ // Take the square:
$x-3=1-x$ // Simplify:
$x=2$

Nope, because $\sqrt{x-3}=\sqrt{-1}$ which is not defined (for real numbers).

Dominique
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Let x = 2.

Square both sides to get $x^2 = 4$

Now take the square root: $x = \pm2$.

But we said at the start that $x = 2$, so $x = -2$ is a false solution.

The trouble is that there is ambiguity in the square root function because for every value there are two legitimate square roots, all else equal. However, one of those solutions maybe false in the larger context of the problem you are trying to solve.

Ryan Pierce Williams
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    The trouble is not with the square root function (which by definition assigns one and only one value to each possible input) but rather with $x\mapsto x^2$ not being injective on $\mathbb R$. – Hirshy Apr 11 '23 at 07:22
  • @Hirshy taking the square root gives two possible solutions: one positive and one negative. – Ryan Pierce Williams Apr 11 '23 at 07:25
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    „Taking the square root“ is different from the „square root function“ you mentioned. The square root function gives you one and only one value for each input, simply because it is a function. – Hirshy Apr 11 '23 at 07:27
  • If you want to nit pick then there is no square root “function” because it is inherently ambiguous. Nevertheless, it is referred to as a function often enough that it’s a moot point - ppl know what you are talking about – Ryan Pierce Williams Apr 11 '23 at 07:30
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    @RyanPierceWilliams Keep in mind the difference between the square root function $,y=\sqrt{x},$ defined on $,\mathbb R_{\ge 0},$, and the solution set of the equation $,y^2 = x,$. They are not one and the same, see for example Why is $\sqrt{x}$ a function? with answers and linked duplicates. – dxiv Apr 11 '23 at 07:37
  • Sure there is a square root function, as soon as you define what the square root of a (positive real) number $x$ is - usually one decides on the uniquely determined positive value $a>0$ with $a^2=x$. The ambiguity that arises with $-a$ also being a solution to the equation $a^2=x$ is a different and closely related question, hence I would argue that we need to really distinguish between functions and equations when talking about this problem. The problem does not lie within the ambiguity of taking the square root and is more concerned what the usual transformations we do when solving equations – Hirshy Apr 11 '23 at 07:38
  • [...] actually are, what allows us to conclude that two equations share the exact same set of solutions, what goes wrong when we can't make this conclusion and what that means for the underlying transformation. So I would not call it nit pick when the question itself is based on "why can't we do that". – Hirshy Apr 11 '23 at 07:40
  • If you contextually limit the result set to either the positive or negatives, then yes you can call it a true function. However, it is referred to as a function- albeit in a looser sense- even if not technically one in the mathematical sense. But I will agree that you are technically correct - which according to Hermes is the best kind of correct ;) – Ryan Pierce Williams Apr 11 '23 at 07:59