Starting with $(x+1)^2 = x^2 + 2x+1$ we can find new identities by substituting. However if I sub $x=2x^2 - 5$ into the equation, giving $(x+1)^2 = (2x^2 -5)^2 +2x +1%$ I get a true statement which holds for all $x$ such that $x + 5 - 2x^2 = 0$ just as my substitution said. However as you can see I only substituted for one of the $x$ in the equation (in particular the one being squared), am I allowed to choose which parts of the identity I want to substitute for, or do I have to substitute for all the $x$ in the identity? (Making what I did forbidden). Thanks in advance.
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To not get confused, keep all your equations together. You had $(x+1)^2=x^2+2x+1$ or the system $\begin{cases}(x+1)^2=x^2+2x+1\end{cases}$. You can think of a system as representing all equations that you can get out of its elements/generators by multiplying by arbitrary polynomials on both sides and adding. If you introduce $x=2x^2-5$, now you have the system $\begin{cases}(x+1)^2&=x^2+2x+1\x&=x^2-5\end{cases}$. Since now you have more generators to multiply and add, the system potentially changed, and in fact it did. – NDB Jul 13 '23 at 14:15
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Now, when you obtained $(x+1)^2=(2x^2-5)^2+2x+1$, what really happened is that you took $x-(x^2-5)=0$ (the second equation of the second system), multiplied it by $x+(x^2-5)$ on both sides, and subtracted it from $(x+1)^2=x^2+2x+1$. To get the new equivalent system $\begin{cases}(x+1)^2&=(x^2-5)^2+2x+1\x&=x^2-5\end{cases}$. – NDB Jul 13 '23 at 14:17
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So, introducing the equation $x=x^2-5$ did change the system, but producing $(x+1)^2=(x^2-5)^2+2x+1$ did not change the second system. It is still different from the first equation alone. – NDB Jul 13 '23 at 14:19
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Yeah so the final equation you wrote is just a "special case" of the $(x+1)^2$ identity (special as it holds for only some $x$). – Nav Bhatthal Jul 13 '23 at 14:22
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The first equation is satisfied by all real $x$. The second equation is satisfied by only two real $x$. When you combine them, the equation is satisfies by the intersection of the two solution sets, which is exactly the same as the solution set of the second equation. – Matthew Leingang Jul 13 '23 at 14:23
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And this idea of "partial substitution" where we only put the condition on $x$, ($2x^2 -5)$ into some instances of $x$ is allowed? (As we keep substituting ,the $x$ for which the equation is valid does not change, as we have already modified our domain) – Nav Bhatthal Jul 13 '23 at 14:24
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Allowed vs not allowed depends on what you wanted to do with those equations.. What is important is to understand what changed and what stayed the same. When the equation $x=x^2-5$ was introduced, the system got "richer" (has more equation in it and therefore, the set of solutions got smaller. When the last equation was introduced, the system stayed the same. Purpose matters. If you intended to find particular solutions, it was ok making the system richer, if you intended to find all solutions, then it was not ok. – NDB Jul 13 '23 at 14:29
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As I showed in my second comment. If "partially substituting" makes you unsure, note that you don't need the notion of "substitution" to get that last equation. What you did was multiply one equation by a polynomial on both sides and subtract it from another equation. Both operations allowed by the $=$. – NDB Jul 13 '23 at 14:31
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Take $x-(x^2-5)=0$, multiply both sides by $x\mathbf{+}(x^2-5)$, to get $x^2-(x^2-5)^2=0$, and add this equation to $(x+1)^2=x^2+2x+1$. You get your equation. No substitution needed. – NDB Jul 13 '23 at 14:34
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Alright, thanks for the detail. I guess all the work you did validates the substitution and shows it is consistent with another approach. – Nav Bhatthal Jul 13 '23 at 14:46
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There are two main substitution rules in algebra:
-If an equation holds for all values of $x$, you may substitute all $x$s in the equation with a valid expression. As you point out, that all here is essential.
Be careful with your notation here! When you describe "substituting $x$ with $2x^2-5$ you usually should write it as "substituting $x \to 2x^2-5$" instead of "substituting $x = 2x^2-5$" to avoid mistakenly assuming $x$ is actually equal to $2x^2-5$, since there is another rule that applies when things are actually equal:
-If two expressions are equal, you may substitute one for the other. In this case, you do not need to replace all instances.
--This applies if two expressions are equal always:
$x^2+2 = x\cdot x+2$ since $x^2 = x\cdot x$ always.
--Or if two expressions are equal just in a particular context:
Assuming $x = 2$ what is $x^2+5$? In this case we're within a context where $x=2$ is a true equation, so we can replace the $x$s in $x^2+5$ with $2$.
It is this latter situation you find yourself in with your question. You've created a context where $x$ is genuinely equal to $2x^2-5$ and your results will only hold for values of $x$ where $x = 2x^2-5$ (i.e. $-1.3508\ldots$ and $1.8508\ldots$).
What you did is not forbidden at all: you have correctly concluded that $(x+1)^2 = (2x^2-5)^2 + 2x + 1$ for all $x$ such that $x = 2x^2-5$. Just be careful what you do with this equation next, since you only know that it works for very few values of $x$!
TomKern
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Sorry! I misread your question. Your new equation does indeed hold for all $x$ such that $x = 2x^2-5$ (i.e. $-1.3508\ldots$ and $1.8508\ldots$). – TomKern Jul 13 '23 at 13:44
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Yes I know. What I am asking is that is it allowed to "chose" which instances of $x$ to swap for $2x^2 -5$? For example I did not swap all instances of $x$ in my example, just one of them. – Nav Bhatthal Jul 13 '23 at 13:45
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Starting from the identity
$$(x+1)^2 = x^2 + 2x+1$$
with the substitution $x=2x^2 - 5$ in this way
$$(x+1)^2 = (2x^2 - 5)^2 + 2x+1$$
yes we obtain a new identity for the $x$ values (real or complex) such that $2x^2-x - 5=0$ indeed for these particular values $x_0$ the quantities $x_0$ and $2x_0^2 - 5$ are the same (real or complex number) and therefore all the following are identities
$$(x_0+1)^2 = x_0^2 + 2x_0+1$$
$$(2x_0^2 - 5+1)^2 = x_0^2 + 2x_0+1$$
$$(x_0+1)^2 = (2x_0^2 - 5)^2 + 2x_0+1$$
$$(x_0+1)^2 = x_0^2 + 2(2x_0^2 - 5)+1$$
$$(2x_0^2 - 5+1)^2 = (2x_0^2 - 5)^2 + 2x_0+1$$
$$(x_0+1)^2 = (2x_0^2 - 5)^2 + 2(2x_0^2 - 5)+1$$
$$(2x_0^2 - 5+1)^2 = x_0^2 + 2(2x_0^2 - 5)+1$$
$$(2x_0^2 - 5+1)^2 = (2x_0^2 - 5)^2 + 2(2x_0^2 - 5)+1$$
Note that if we consider these as equations we can obtain exactly the $x_0$ values as solutions as for example for
$$(2x^2 - 5+1)^2 = (2x^2 - 5)^2 + 2x+1 \iff 2x^2-x-5=0$$
or possibly we can obtain other extraneous solutions as for example for
$$(2x^2 - 5+1)^2 = x^2 + 2(2x^2 - 5)+1 \iff 4x^4-21x^2+25=0$$
which leads to four real solutions.
As an extreme case consider the last one, that is
$$(2x^2 - 5+1)^2 = (2x^2 - 5)^2 + 2(2x^2 - 5)+1 \iff 0=0$$
which is true for any $x$ and therefore, in some sense, leads to infinitely many extraneous solutions.
user
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All those identities are valid, and only true for some $x$. I see. So partial substitution is fair game and allowed? – Nav Bhatthal Jul 13 '23 at 14:23
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@NavBhatthal Yes of course they works as identities for the $x_0$ values. If we consider them as equations we can obtain more (extraneous) solutions. – user Jul 13 '23 at 14:27
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Can you make an edit on where the extraneous solutions come from? Thank you. – Nav Bhatthal Jul 15 '23 at 13:57
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@NavBhatthal Yes of course! I just realize now that I've given the bad example! I edit to fix it and explain better – user Jul 15 '23 at 16:09
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Thanks for the further help, what I really meant was explain how the solutions arise, not which ones give the extra solutions, if that makes sense. Everyone keeps telling me "they just are" but I know there must be a reason (Perhaps to do with injectivity?) – Nav Bhatthal Jul 15 '23 at 16:23
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In the first case we obtain an equivalent quadratic equation whereas in the second case we obtain a fourth order equation which leads to two extraneous solutions. – user Jul 15 '23 at 16:36