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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
I created this topic to share with you a quick and easy way to solve inequalities(second-degree).
Straightaway here's my tip:
Attachment: eq.jpg [ 27.79 KiB | Viewed 60477 times ] |
The METHOD I) replace the\(>,<\) sign with a more friendly \(=\) II) find \(x1\) , \(x2\) as you would do normally III) use my trick to define the interval. |
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).
Attachment: a.jpg [ 51.24 KiB | Viewed 60561 times ] |
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
15 May 2013, 03:31
Kudos
Bookmarks
connexion
Hi Zarrolou ,
Thanks for your tip. Can you please elaborate it with more examples? I have difficulty in understanding it.
Thanks.
Thanks for your tip. Can you please elaborate it with more examples? I have difficulty in understanding it.
Thanks.
Sure!
Here are all possible cases:
1)\(x^2-3x+2>0\)
2)\(x^2-3x+2<0\)
The roots of \(x^2-3x+2=0\) are x=1 and x=2
In 1 the sign of x^2 is + and the operator is > => they are "the same" so we take the external values
\(x<1\) and \(x>2\)
In 2 the signs are not the same => we take the internal values
\(1<x<2\)
in these the sign of x^2 is -
3)\(-2x^2+3x+2>0\)
4)\(-2x^2+3x+2<0\)
The roots of \(-2x^2+3x+2 =0\) are \(x=2\) and \(x=-\frac{1}{2}\).
In 3 the sign of x^2 is - and the operator is > => the are not the same so we take the internal values
\(-\frac{1}{2}<x<2\)
In 4 they are the same (-,<) external values
\(x<-\frac{1}{2}\) and \(x>2\)
Hope it's clear now. Let me know
19 Sep 2013, 04:58
Kudos
Bookmarks
violetsplash
This and many more tricks which I discovered here on GMAT Club off late are turning out to be a life saver. GMAT in 10 days and am wondering why wasn't I here earlier 
Theory on Inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189
700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189
700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html
The following posts might also be helpful:
new-to-the-math-forum-please-read-this-first-140445.html
new-project-review-discuss-and-get-kudos-points-153555.html
