Tips and Tricks: Inequalities

Hi guys,
I created this topic to share with you a quick and easy way to solve inequalities(second-degree).

Straightaway here's my tip:

This works with \(\geq{}\) and \(\leq{}\) as well, you just add the = sign, that's it.



If the sign of \(a\) and the operator are "the same" (>,+) or (<,-) we take the ESTERNAL values. Otherwise the INTERNAL values.

I find this method very easy to remember and to use: I'm sure it will save you time.
This is it, but for those of you interested in what goes on behind the scene I have a mathematical explanation.

\(ax^2+bx+c\) is a parabola.

I took two parabolas one with a positive \(a\), one with a negative \(a\).
\(a\) in the formulas defines 2 characteristics of the parabola:
1) its "slope" ( greater \(a\) greater the slope)
2) where the parabola looks at ( \(a\) +ve the parabola looks up, \(a\) -ve the parabola looks down).



With this little theory, and looking to the graphs we can conclude that:
if \(a\) is +ve and we want the values >0 we have to take the esternal values, if we want the values <0 the internal values.
This same principle can be applyied to a negative \(a\)

Some examples:
\(f(x)=x^2+2x-3>0\)
Step 1) replace > with = : \(x^2+2x-3 = 0\)
Step 2) find x1 and x2: \(x1=1, x2=-3\)
Step 3)use the "tip": \(a\) is +ve and the operator is > solution \(x>1\) and \(x<-3\)
\(f(x)=x^2+2x-3<0\) same function diff operator
all the steps are the same
Step 3)use the "tip": \(a\) is +ve and the operator is < solution \(-3<x<1\)

More examples: tips-and-tricks-inequalities-150873.html#p1225182

If you like the tip and you're gonna use it, give it a KUDOS!

Hope you guys find it useful, regards