GMAT Math

One often encounters expressions such as:
(x - a)(x - b)(x - c)...(x -k) > 0 or < 0
in which we have to solve for x.

Critical Points Method is the easiest way to solve such questions once you learn it.
Let me teach you the method here.

for the function f(x) = (x - a)(x - b)(x - c)...(x - k), Critical Points are those points at which f(x) = 0. So a, b, c.. k are the critical points of f(x).

Example 1:
Solve
(x + 1).(x - 2) < 0
Step 1. Find out the critical points
The critical points are -1 and 2 in this case.
Step 2. Plot them on a number line in a proper order.



Step 3: Consider the region above the number line to be positive and below it to be negative. Start drawing a curve starting from the right most - top side as shown below



And then complete the curve in an alternate up-down fashion as shown below




Step 4: Mark '+' in the region above the number line and '-' in the region below the number line as shown below.



In the '+' region f(x) > 0 and in the '-' region f(x) < 0.

Since the question wants us to solve for f(x) < 0, the values of x for which the curve goes below the number line will be chosen.

The shaded region shown below is our answer.



Answer: -1 < x < 2

Example 2: Lets take another question.
(x - 3)(x - 2)(x - 1) > 0

Step 1: The critical points are 1, 2 and 3.
Step 2: Mark them on a Number Line.
Step 3: Plot the curve starting from top-right side in an alternate fashion.
Step 4: Mark alternate '+' and '-' and choose the appropriate region
In this case the region in which f(x) > 0 is favourable to us is shown below



Answer: 1 < x < 2 , x > 3

Note:
(i) If the expression is in the for (a - x)(x - b)(x - c) form, first convert it into (x - a)(x - b)(x - c) form by multiplying both sides of the inequality by -1 (and flipping the inequality sign). Apply the Critical Points method only after that.

For example:
(1 - 2x)(1 + x) < 0
has to be first converted into
(2x - 1)(x + 1) > 0 form.

(ii) If the inequality contains an 'equal to' sign as well such as (x - a)(x - b)(x - c) >=0 or =<0,
we have to include the critical points in our region as a part of our solution.

aneesh.kg wrote:One often encounters expressions such as:
(x - a)(x - b)(x - c)...(x -k) > 0 or < 0
in which we have to solve for x.

Critical Points Method is the easiest way to solve such questions once you learn it.
Let me teach you the method here.

for the function f(x) = (x - a)(x - b)(x - c)...(x - k), Critical Points are those points at which f(x) = 0. So a, b, c.. k are the critical points of f(x).

Example 1:
Solve
(x + 1).(x - 2) < 0
Step 1. Find out the critical points
The critical points are -1 and 2 in this case.
Step 2. Plot them on a number line in a proper order.



Step 3: Consider the region above the number line to be positive and below it to be negative. Start drawing a curve starting from the right most - top side as shown below



And then complete the curve in an alternate up-down fashion as shown below




Step 4: Mark '+' in the region above the number line and '-' in the region below the number line as shown below.



In the '+' region f(x) > 0 and in the '-' region f(x) < 0.

Since the question wants us to solve for f(x) < 0, the values of x for which the curve goes below the number line will be chosen.

The shaded region shown below is our answer.



Answer: -1 < x < 2

Example 2: Lets take another question.
(x - 3)(x - 2)(x - 1) > 0

Step 1: The critical points are 1, 2 and 3.
Step 2: Mark them on a Number Line.
Step 3: Plot the curve starting from top-right side in an alternate fashion.
Step 4: Mark alternate '+' and '-' and choose the appropriate region
In this case the region in which f(x) > 0 is favourable to us is shown below



Answer: 1 < x < 2 , x > 3

Note:
(i) If the expression is in the for (a - x)(x - b)(x - c) form, first convert it into (x - a)(x - b)(x - c) form by multiplying both sides of the inequality by -1 (and flipping the inequality sign). Apply the Critical Points method only after that.

For example:
(1 - 2x)(1 + x) < 0
has to be first converted into
(2x - 1)(x + 1) > 0 form.

(ii) If the inequality contains an 'equal to' sign as well such as (x - a)(x - b)(x - c) >=0 or =<0,
we have to include the critical points in our region as a part of our solution.

Thanks for sharing Aneesh! Useful1

Pl. have a look at my posts for 'Cheeky approach to few tricky problems' series 1-4 & 'Must see for Maximum/Minimum value problems'. Hope you like it.

Please mention the scenarios when

1
--------------- > 0
(x-a)(x-b)(x-c)

If you notice, (x - 1)(x - 2)(x - 3) > 0 has been solved for x in Example 2 above.

Since this post is generating some interest now, let me give a few practice problems to solve by this method:

Solve for x in the following:
i) x^2 > x
ii) x^2 + 5x + 6 > 0
iii) (2 - x)(5x + 1)(x +2) < 0

I would like to see some attempts on these problems before giving away the solution. You can also validate your own answer by plugging-in numbers.

i) x^2 > x
X>1
X<0

ii) x^2 + 5x + 6 > 0
= (x+3)(x+2) > 0

x>-2
x<-3

iii) (2 - x)(5x + 1)(x +2) < 0

=(x-2)(5x+1)(x+2) > 0

X > 2

-2 < X< -1/5


Your critical point technique is very useful, hopefully i got the answers correct, and my exam is coming up soon, and it seems this technique could come very handy!

Quick question, do you always start drawing the line coming from the top right, could something like this ever exist?:

Dear Aneesh
Thanks very much. This is a very good trick.

Can we extend this trick for two variables i.e, (x+1)(y+2)>0
with your trick the soln will be x > -1 and y < -2

Hi zueswoods,

First of all, All the best for your GMAT!

Good question, and I'm glad you asked.
I asked you to convert everything into a
(x - a)(x - b)(x - c)... form
if it is in
(a - x)(x - b)(x - c).. form
so that you don't have to worry about the possibility of the curve starting from top bottom.

Once you've done the conversion (if required, by multiplying by a -1) then the curve will always start with a positive on the right side.

If the inequality is
(a - x)(b - x) < 0
then you've to multiply it twice by -1 to reduce it to:
(x - a)(b - x)< 0.
Yeah, you get the same thing. But, do it anyway so that you stay in the habit of bringing it to the required form.

Cheers!

Hi Krishna,

If you apply it to two variables, you will not get a complete solution so Avoid it!

Let's take the example presented by you:
(x + 1)(y + 2) > 0

This inequality holds true in these cases:
Case1: x + 1 > 0, or x > -1 and y + 2 > 0, or y > -2
Case2: x + 1 < 0, or x < -1 and y + 2 < 0, or y < -2

So, if you apply Critical Points to this inequality, you will miss out Case2.

Moral of the Story: Apply this method for one variable only.