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We can solve the equation $12x^2-5x-3>0$ with the pq-formula, but I think this is very error-prone without a calculator. Now I find a simpler way to solve it. I rewrite the equation as $(3x+1)(4x-3)>0$. Then I solve both multiplicands for $x$:

$(3x+1)>0$

$x>-1/3$

and

$(4x-3)>0$

$x>3/4$

It seems to work. But the right result is

$x<-1/3$ : Here the inequality is reversed. Why? Where is my mistake?

$x>3/4$ : That's right.

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2 Answers

up vote 4 down vote accepted

$(3x+1)(4x-3)>0$

($3x+1>0$ and $4x-3>0$) or ($3x+1<0$ and $4x-3<0$)

($x>-\frac{1}{3}$ and $x>\frac{3}{4}$) or ($x<-\frac{1}{3}$ and $x<\frac{3}{4}$)

$x>\frac{3}{4}$ or $x<-\frac{1}{3}$

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If you remember that in the real field we have

$$(x-a)(x-b)>0\Longleftrightarrow x<\min\{a,b\}\,\,\vee\,\,x>\max\{a,b\}$$

Then you can directly do as follows:

$$(3x+1)(4x-3)>0\Longleftrightarrow 12\left(x+\frac{1}{3}\right)\left(x-\frac{3}{4}\right)>0\Longleftrightarrow$$

$$\Longleftrightarrow \left(x+\frac{1}{3}\right)\left(x-\frac{3}{4}\right)>0\Longleftrightarrow x<-\frac{1}{3}\,\,\vee\,\,x>\frac{3}{4}$$

After understanding and remembering the above one understands that the really hardest part is to reach the decomposition $\,12x^3-5x-3=(3x+1)(4x-3)\,$ , and this is one of the reasons why in Junior High School they're (usually) so insistent in this kind of stuff.

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Exists an simple algorithm to reform a quadratic equation in (x+a)(x-b) ? –  bob morane Oct 27 '12 at 11:50
    
Sorry, I don't understand what can "reform a quadratic equation" possibly mean. –  DonAntonio Oct 27 '12 at 11:51
    
an algorithm to do kind of this: 12x2−5x−3>0 = (3x+1)∗(4x−3) Currently I try it with mental arithmetic. –  bob morane Oct 27 '12 at 11:54
    
There are several, the best well known ones being the famous roots equation $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$for the roots of a quadratic equation, but you can also try *sometimes* (this works fine when $\,a=1\,$) the formula that stems from Viete formulae: find two numbers $\,\alpha\,,\,\beta\,$ s.t. $\,\alpha\beta=\frac{c}{a}\,\,,\,\,\alpha+\beta=-\frac{b}{a}\,$ , when all the time we talk about the quadratic equation $\,ax^2+bx+c=0\,\,,\,a\neq 0\,$ –  DonAntonio Oct 27 '12 at 12:03
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