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I seem to be stuck on some seemingly simple algebra; The aim is to rearrange $$y = 3 + \sqrt{x + 2}$$ for $x$.

My working is as follows, I just square everything: $$y^2 = 3^2 + x + 2$$ It is clear to me that this is incorrect, however I don't understand why one has to take the 3 over to the other side before squaring everything. Correct answer: $$(y-3)^2 = x+2$$

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In squaring the RHS, you have done $(a+b)^2 = a^2+b^2$, which does not hold in general. In fact $(a+b)^2 = a^2+2ab+b^2$. – Macavity May 17 '14 at 17:58
up vote 2 down vote accepted


in general $$\ne3^2+x+2$$ unless $$\sqrt{x+2}=0\iff x+2=0$$

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Ah yes of course. I should get more sleep. Really stupid mistake. Thanks! – Cobbles May 17 '14 at 17:59

You have squared it wrong: $$ (3+\sqrt{x+2})\cdot(3+\sqrt{x+2})=3^{2}+3\sqrt{x+2}+\sqrt{x+2}\cdot3+(\sqrt{x+2})^{^{2}} $$

To solve the question I suggest first moving $3$ to the RHS to get $$ y-3=\sqrt{x+2} $$

this way the squaring won't add give you an expression with a root

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I believe you misunderstood the operational properties of the square root, that's why you make a wrong manipulation. When we talk about the square root of some $x$, we are just talking about the set of solutions to the equation $y^2=x$, but that involves set theory, let's not talk about that now. Consider, instead, that when your teacher or every one writes down something like this $x=y$ or $\phi(x)=\phi(y)$ we are talking about an atomic formula, which, by it's identity, preserves all the operations on the both sides. For example, WHY exactly do need to "switch sides" to solve an equation like $x-2=4$? It's just that we have $4$ exactly equals to $x-2$, e.g, whenever $4$ occurs, we can replace it by $x-2$. But since you want to know the value of the variable $x$, you can't work with the term $x-2$, so, you can't work with the atomic formula $x-2=4$, you need to find some other. That's where the reasoning process begins: we want $x$, what we have to do it's just work with some atomic formula that has some "derivative" term from the original one, so, $x-2+2$ which is $4+2$. Another way of seing is that, since, for example $x=6$, if I put $x+2$, that's automatically $6+2$, since every occurrence of $x$ can be replaced by $6$, because of $x=6$.

When we talk about square roots, we assume that it has the general manipulational form: $\sqrt[2]{x} = x^{1/2}$, because $(x^{1/2})^2=x$. But there's a problem when you work with addition and exponentials at the same time: you can't solve, for example, $(x+3)^7$ in a direct way, you need a formula for that (exposed in the binomial theorem). So, what you did wrong was to assume that, whenever $a$ and $b$, $(a+b)^n=a^n + b^n $. So, when you start working with $y^2$ insted of $y$, you could replace the occurrence of $3+ \sqrt[2]{x+2}$ by $(3+ \sqrt[2]{x+2})^2$ which can be just writen in the form $(a+b)^2 =(a+b)(a+b)$ and not $a^2 + b^2$. It's a lot easier if you just start working with $y-3$ and then go to $(y-3)^2$, which gives $(\sqrt[2]{x+2})^2=((x+2)^{1/2})^2=(x+2)$. Have a nice day.

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Thank you but I think just $(a+b)^n=a^n+b^n$ would have sufficed! – Cobbles May 17 '14 at 19:47
You're welcome. It's a normal mistake for beginning students, I made a lot of it in the old days :}. Anyway, if there's any other question that's beeing puzzling you, feel free to ask me directly, if I know I would be glad to help you. – Ricardo May 17 '14 at 21:44

When rearranging an equation, you should pay attention to the order of operations. When solving for a particular variable, one should also pay attention to the order in which those operations are applied to that variable. To help you see what this means, consider the expression $$y = 3 + \sqrt{x+2}.$$ If we look at the right-hand side, we see that the following operations are done on $x$, in this exact order:

  1. $2$ is added to $x$.
  2. The square root is taken of the result from step (1).
  3. $3$ is added to the result of step (2).
  4. $y$ is equated to the result of step (3).

So, in order to solve for $x$, we must do the inverse operation in the reverse order that they were applied to $x$:

  1. (reverse of step 3): Subtract $3$ from both sides.
  2. (reverse of step 2): Square both sides.
  3. (reverse of step 1): Subtract $2$ from both sides.

Of course, when the given equation is more complicated (especially if the variable to be solved for appears more than once), then this simple method doesn't really apply. You'll have to do other kinds of simplifications, depending on the situation. For example, if we have $$\frac{x+1}{x-1} = y^2,$$ and we wish to solve for $x$, then one approach is to multiply both sides by $x-1$ to get $$x+1 = y^2(x-1),$$ expand the RHS to get $$x+1 = y^2 x - y^2,$$ then collect like terms in $x$: $$x - y^2 x = -1 - y^2.$$ Now we factor the LHS and get $$(1-y^2)x = -1-y^2,$$ and dividing both sides by $1-y^2$ gives the desired solution $$x = \frac{-1-y^2}{1-y^2} = \frac{y^2+1}{y^2-1}.$$ As you continue to practice algebraic manipulations like these, you will develop more techniques and an intuition for what types of rearrangements and operations will work.

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