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Given this equation (containing one variable $x$, and one parameter $a$):

$$\frac{(1 - a)^2}{x - a} = \frac{( 1 + a )^2}{x + a}$$

I need to solve this equation and "discuss its solutions" depending on the parameter $a$. The proposed correct answer is:

  • If $a = 0, a = 1, a = -1$, then the equation doesn't have a solution
  • Otherwise, the solution is $x = \dfrac{a^2 + 1}2$

I don't know how to get to this answer. Could you enlighten me, please?

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Have you tried seeing what happens if you make the suggested replacements of $a$? – J. M. Dec 7 '11 at 2:55
@J.M. For a = 1 and a = -1 the equation is indeed unsolvable. For a = 0 the solution is undefined ( x = x ). The problem is that I don't know how to get to this answer. – Šime Vidas Dec 7 '11 at 2:59
The first bulleted statement isn’t quite correct: if $a=0$, every real number except $0$ is a solution. – Brian M. Scott Dec 7 '11 at 3:02
@Brian Yes, I agree. I got the proposed answer from a friend over the telephone, so it could be incomplete / partially wrong. If $a = 0$, then $x = x$ (the solution is "not defined", at least that's how we refer to this case here in Croatia). – Šime Vidas Dec 7 '11 at 3:06

The solution process breaks down into two steps:

  • "Solve" the equation (push symbols until you isolate $x$ on one side);
  • Figure out what your solution means and check its validity.

The reason I put "solve" in scare quotes will become clear shortly.

The equation itself involves only ratios of linear functions of $x$, so I'll let you solve it yourself. Your strategy will be to clear the fractions, solve the resulting quadratic, then simplify.

The interesting part is what happens next, when you have a "solution." You have indeed pushed symbols around and gotten $x$ on one side, but what you've got now are candidate solutions, written in terms of the parameter $a$. You need to go back and check them. Why? Because in the process of solving the equation, you may have accidentally broken it by dividing by zero or taking the square root of a negative number. (This is why "solve" was in scare quotes, by the way: even though you've got $x= $stuff, you're not done yet.

So you put your solution back into the original equation to check that it actually makes sense and does solve the equation. (This is usually a good idea anyway, even if you're solving a straightforward problem, just to make sure you didn't mess up somewhere in the symbol-pushing.) Now you have an equation in $a$. Does this equation make sense for all values of $a$? Some? None? Which values 'break' the equation?

Here's a toy example. Solve $$ax = 5$$ in terms of $a$ and discuss the solutions.

First step: "solve" for $x$. Easy: divide both sides by $a$, to get $x=\frac{5}{a}$.

Now we have "solved" for $x$, we need to put it back into the original equation to double-check. This gives $$a\frac{5}{a} = 5,$$ which makes perfect sense ... unless $a=0$, in which case we're dividng by zero.

So the equation $ax=5$ has no solution when $a=0$, and when $a\neq 0$, it has the solution $x=\frac{5}{a}$.

I hope this clears things up! I know that this is a long wall of text, but don't let it psyche you out. Remember, all you're doing is solving the equation, then double-checking that you didn't accidentally break the rules and pick up spurious solutions while you were solving.

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