Dear Uncle Colin,
My teacher suggested that if you factorise a quadratic and add the brackets, you get the derivative. I am now too frightened to sleep.
-- Quadratic Understanding Is Not Easy
Hi, QUINE, and thanks for your message!
To take a simple example, $x^2 + 5x + 6 = (x+2)(x+3)$ and, indeed, $2x+5$ is the sum of the brackets and the derivative of the quadratic. However, it doesn’t always work. For example, consider $2x^2 + 5x + 2 = (2x+1)(x+2)$. The derivative of that is $4x + 5$, rather than $3x+3$ as the rule suggests. (Had the second bracket been $2x+4$, it would have worked.)
For quadratics of the form $(x-p)(x-q) = x^2 - (p+q)x + pq$, the derivative is clearly $2x - (p+q)$, which is the sum of the brackets. The reason it works is that the derivative has to be zero at the vertex, which is midway between the roots. Adding the brackets effectively gives you the mean of the roots (as long as the x-coefficient is 1.) You’ve also got the nice coincidence that adding the $x$s together gives you $2x$, which is the derivative; I think this is simply a coincidence.
So you can sleep a bit more soundly, QUINE: it does work, but only when the $x^2$ coefficient is 1, and even then only because of a coincidence. Sweet dreams!
-- Uncle Colin
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