Hello
I get very confused in solving inequlities questions and finding ranges, so i went online to understand the concept..This question is taken from purple math. I would like someone to help me understand if i am doing it right and can proceed this way for all the question like this.
-2x^2 + 5x +12 <= 0 solve for the range.
step 1: factor it
(-2x - 3) (x-4)
step 2: take them both as positive
-2x-3 > 0
therefore x< - 3/2
x-4 > 0
therefore x> 4
Now plot them all in the ranges of
1. (-infinity to -3/2)
2. (-3/2 to 4)
3. (4 to + infinity)
not test (-2x-3) (x-4) in these plots and see where the answer is <= 0. That would give the range.
Here range is (-infinity to -3/2 inclusive) (4 inclusive to positve infinity)
Am i right in doing it this way. can i follow this method for all the equations where i need to find ranges.
also, i havent seen much of this tested on GMAT in OG problems, please tell me if this is relevant or not to spend lot of time on ....as i sure find it very confusing.
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- KK
Ron/Stacey help me with inequality
In that case Stacey or Ron, please help me how to solve this question
- KK
inequality
I am sorry but i get really confused in Inequality. Please someone explain.
Stacey/Ron. do you think such questions are expected in GMAT as it takes so much time.
Stacey/Ron. do you think such questions are expected in GMAT as it takes so much time.
- RonPurewal
- Students
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- Joined: Tue Aug 14, 2007 8:23 am
you can use factoring to help you find the range, but only indirectly. here's how it works:
if you take the SOLUTIONS of the quadratic (i.e., the numbers that "solve" the two factors when they're set to zero), then the MAX/MIN of the quadratic will occur at the point HALFWAY BETWEEN those two solutions.
example:
let's say you want the range of the quadratic y = (x - 2)(x - 6). we know this quadratic is going to have a minimum value, not a maximum, because the square term is positive (which means that "the parabola opens upward" or "it can get infinitely large if x is big").
the solutions would be 2 and 6.
the number halfway between those two is 4, so the minimum will occur at 4.
plug in to find it: (4 - 2)(4 - 6) = -4
therefore, the range is y > -4.
--
normally, the gmat won't make you do something like this - too much computation, and too much like a "homework problem". what they're much more likely to do is to give you the quadratic in VERTEX FORM, which means something containing a PERFECT SQUARE (such as y = (x - 5)^2 + 3).
finding the range of a quadratic written in that way is pretty easy: the max/min will occur when the perfect square is equal to 0. (this happens because 0 is the smallest possible value of a perfect square; perfect squares can't be negative.)
it will be a min if the perfect square term is positive; it will be a max if the perfect square term is negative.
for the above function, y = (x - 5)^2 + 3, the minimum will occur when the perfect square term is 0, which happens when x = 5. plugging in 5 yields y = 3, so the range is y > 3.
if you take the SOLUTIONS of the quadratic (i.e., the numbers that "solve" the two factors when they're set to zero), then the MAX/MIN of the quadratic will occur at the point HALFWAY BETWEEN those two solutions.
example:
let's say you want the range of the quadratic y = (x - 2)(x - 6). we know this quadratic is going to have a minimum value, not a maximum, because the square term is positive (which means that "the parabola opens upward" or "it can get infinitely large if x is big").
the solutions would be 2 and 6.
the number halfway between those two is 4, so the minimum will occur at 4.
plug in to find it: (4 - 2)(4 - 6) = -4
therefore, the range is y > -4.
--
normally, the gmat won't make you do something like this - too much computation, and too much like a "homework problem". what they're much more likely to do is to give you the quadratic in VERTEX FORM, which means something containing a PERFECT SQUARE (such as y = (x - 5)^2 + 3).
finding the range of a quadratic written in that way is pretty easy: the max/min will occur when the perfect square is equal to 0. (this happens because 0 is the smallest possible value of a perfect square; perfect squares can't be negative.)
it will be a min if the perfect square term is positive; it will be a max if the perfect square term is negative.
for the above function, y = (x - 5)^2 + 3, the minimum will occur when the perfect square term is 0, which happens when x = 5. plugging in 5 yields y = 3, so the range is y > 3.
6 posts Page 1 of 1