2
$\begingroup$

Say we have an equation $ax^2 + bx - c = 0$ and want to find $x$. Obviously the way to solve would be to use the quadratic equation or factorize. I understand that saying $ax^2 + bx = c => x(ax + b) = c$ and then solving is wrong (the values of $x$ when subbed back in do not satisfy the eqn), but why is it wrong? Each step seems logical?

Many thanks.

  • 0
    All too often in your past experience, you have been given quadratic equations $ax^2+bx-c=0$ which have **integer** solutions. Then rewriting as $x(ax+b)=c$ and testing for $x$ various divisors of $c$ is a strategy that works. Of course, *some* divisors may not work, but that's no problem, you have found a solution. However, if the quadratic has no integer solutions (and most of them don't) then the procedure breaks down.2012-12-23

2 Answers 2

3

The equation $x(ax+b) = c$ is valid but does not help. There is a general fact that $AB = 0$ implies $A = 0$ or $B=0$ and this allows us to solve a product expression by reducing it to easier equations. So we need 0 on the right hand side of the product to be useful.

e.g. we want to rewrite your equation $ax^2 + bx - c = 0$ as $a(x+\alpha)(x+\beta) = 0$.

In order to find $\alpha, \beta$ to do this, note that we ensured the quadritic term is already OK: $ax^2$ in both. The linear term is $a(\alpha+\beta) = b$ and the constant term is $a\alpha\beta = -c$. So you need to find $\alpha$ and $\beta$ with known sum $\frac{b}{a}$ and known product $\frac{-c}{a}$, and this can sometimes be seen by inspection for concrete $a,b$ and $c$.

1

The factorization is wrong. We need to reach at $(ax-\alpha)(x-\beta)=0$ so that we can say either $ax-\alpha=0$ or $x-\beta=0$

  • 0
    Yes, I believe (if I recall correctly) I have used it when solving diophantine equations.2012-12-23