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© Copyright 2000, Jim Loy
You probably know that you cannot divide by zero. It's a trap. You often get the wrong answer. Well, there is another similar trap. You can square both sides of an equation (after all they are the same number). But sometimes (rarely) you get more answers than you need. And some of them may not be the right answers. Here are some examples.
First example:
sqr(x) + x = 2 (where sqr(x) means the positive square root of x)
sqr(x) = 2-x
x = 4-4x+x^2 (squaring both sides)
x^2-5x+4 = 0
x = 4 and 1 (using the quadratic formula)
See Quadratic Formula. If you plug these two solutions back into the original equation, you will see that x=4 doesn't work. It is a solution of the quadratic equation in the next to the last step. But it is not a solution to the original equation. Why is that? Every step looked legal. Well, every step was legal. But squaring both sides can give us extra solutions.
Second example: This is simpler.
-1 = 1 (not true)
1 = 1 (squaring both sides)
The last step is true, 1=1. Did we just prove that the first step is also true? No, proofs don't work like that. We may try to do proofs like that: Assume something and see if we get true consequences out of it. But that is a backward proof, which is no proof at all. In the first example, we could have started sqr(4)+4=2, which is false. But the third step (squaring both sides) produces something that is true (4=4-16+16), just as in our -1=1 example.
Third example: Here is a more complicated example: 3sqr(x)+x+2=0. Solving for x in the same way as our first example, we again get x=4 and 1. This time neither "solution" works. The original equation has no solution. There are plenty of equations with no solutions (like x=x+1). In essence, we find that the statement 3sqr(x)+x+2=0 is false, just as -1=1 is false.
Of course most numbers have two square roots, one positive and one negative. Above, we were dealing with the positive square root. In the third example, the negative square root would work. But then the original equation would be -3sqr(x)+x+2=0, which is not the same equation. As written, the third example has no solutions.