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What makes some equations so much easier to solve than others?

Quick description

Solving an equation can frequently be thought of as determining x given that f(x)=y, for some function f and some number y. For example, we might want to find x when (3x+5)^2=12, or when x^2+2x=8. If you think about these two examples, you will see that the first is much easier than the second. One sign that it is easier is that you can work out (3x+5)^2 on a calculator using no memory and inputting x only once: after putting x in, all you have to do is multiply it by 3, add 5, and square it. (This assumes your calculator has an x^2 button.)

Example 1

Suppose you are asked to solve the equation (3x+5)^2=12. It is rather easy to do: taking square roots tells you that 3x+5=\pm\sqrt{12}, subtracting 5 gives 3x=-5\pm\sqrt{12}, and dividing by 3 gives x=\frac{-5\pm\sqrt{12}}3.

Now suppose you are asked to solve the equation 9x^2+30x+15=2. It isn't nearly so easy, and you have to use the full theory of quadratic equations to solve it.

General discussion

What is it that makes the first equation easier? It's that we can see quickly how to undo the operation of turning x into (3x+5)^2. That operation naturally splits into three stages: multiply by 3, add 5, and take the square. So if that gives us the number 12, then we can recover x by reversing each of the three stages of the calculation, starting with the last and ending with the first. So we take the square root, subtract 5, and divide by 3.

If we try to do the same thing with 9x^2+30x+15, we find we cannot. That is because replacing x by 9x^2+30x+15 doesn't naturally split up into simple stages in the same way, so we can't just do the reverse process to 2.

Perhaps you would like to dispute this. For example, here is a way that one might calculate 9x^2+30x+15. We could take x, multiply it by 3, add 10, multiply by 3 again, multiply by x, and add 15. Why isn't this a simple process?

It is quite simple (and is in fact quite a useful way of evaluating polynomials) but there is a big difference: to do this process we had to input x twice. So if you had a complicated number x and wanted to work it out on a calculator with no memory, then you would have to write x down and key it in at two points in the calculation.

Thus, the answer to the question in the title of this article is as follows. Some equations are particularly easy because they are of the form f(x)=y, where to work out f(x) you start with x and do a succession of simple operations on it without inputting x again. So if you are told that f(x)=y then you can work out x by applying the inverse operations to y in the reverse order.

Example 2: two more examples

Here is another example of an easy equation to solve. If e^{-x^2/2}=w, then -x^2/2=\log w, so x^2=-2\log w, so x=\pm\sqrt{-2\log w}. (This uses the fact that \log(e^t)=t for any t.)

The same basic principle applies to other sorts of equations too. Here, for example, is a differential equation where the unknown is a function g. We are told that \frac d{dx}(\log g(x))^2=4x+5. From this it follows that (\log g(x))^2=2x^2+5x+C, so \log g(x)=\pm\sqrt{2x^2+5x+C}, so g(x)=\exp(\pm\sqrt{2x^2+5x+C}). The reason this was easy is that to obtain the left-hand side we started with g, took its logarithm, squared the result, and differentiated. So to solve the equation we just applied the inverse operations in reverse order to the function 4x+5.

By contrast, if we want to solve the equation \frac d{dx}(\log g(x))^2-3g(x)=4x+4 then we will have a much harder problem, because this time we have had to "input g twice".