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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.

CAF
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  • 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. – André Nicolas Dec 23 '12 at 21:34

2 Answers2

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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 $c$.

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

lab bhattacharjee
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  • That is clear, but is there a reason why the method I showed does not work? – CAF Dec 23 '12 at 15:24
  • If the product of two terms involving $x$ is some given constant, how can we get the value of $x?$ – lab bhattacharjee Dec 23 '12 at 15:25
  • Multiplying out those terms and rearranging such that we have $0$ on the RHS. I realise this is the way to solve, but why does the method I give not work? Consider the simple case $x^2 + 4x + 3 = 0$. By factoring as a product of two binomials gives $x = -3, -1$. Now say $x^2 + 4x = -3 $ and write $x(x+4) = -3$ and then $x = -3, -7$ Where is the error in this method? – CAF Dec 23 '12 at 15:38
  • @CAF, we need to replace $x$ in both places in the left hand side with the value like $-3$ etc. $-3,-1$ would make $-3=-3$ which is true, but $-7$ would make $(-7)(-1)=-3$ which is false.Hence $-3,-1$ are solutions unlike any other value. – lab bhattacharjee Dec 23 '12 at 15:42
  • I understand this is wrong, but is there an underlying reason? As I said, each step seems logical and there seems to be no a priori assumptions about $x$ at each stage. – CAF Dec 23 '12 at 15:45
  • In general there is no easy way to solve $x(x+4) = -3$. What would the method be: there are infinitely many ways to write -3 as a product, and how would you find one with the two factors 4 apart as we have here. While writing $(x+3)(x+1) = 0$ allows for an easy way, as we know that $x+3$ must be $0$ or $x+1$ must be $0$. – Henno Brandsma Dec 23 '12 at 15:45
  • Only if you know beforehand that the solutions are whole numbers it can be useful to write $x(x+4) = 3$, because then $x$ must be a divisor of $3$, so equal to one of $-1,-3,1,3$ and then we can check by hand. But if the solutions are general reals then this will not help you at all! – Henno Brandsma Dec 23 '12 at 15:48
  • This makes sense, thanks! – CAF Dec 23 '12 at 15:52
  • @CAF, the method you are talking about, is general of trial & error unless any other constraint(s) specified. So it can be time consuming and not elegant either. – lab bhattacharjee Dec 23 '12 at 15:52
  • Yes, I believe (if I recall correctly) I have used it when solving diophantine equations. – CAF Dec 23 '12 at 15:54