Inequalities trick

In addition to my previous post, I'd like to understand:

If the inequality is:
f(x) = (x+a)(x-b)^2(x+c)(x-d)^3 < 0
We are still going to work with the same number line from left to right starting with "+" and factors a,b,c,d placed on it in ascending order?
Since the number of factors remain an even number (4 total). Am I on the right track?

Hi Guys,
I'm in the process of absorbing the fundamentals and neat tricks provided by the experts here. This inequality trick is phenomenal. Would it be safe to conclude that - for odd number of factors or roots, we start with "-" (cosine waveform) and for even number of roots we start with "+" (sine wave)?

As explained by Gurpreet & Karishma:
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

Similarly:
If f(x) has two factors then the graph will have + - +
If f(x) has seven factors then the graph will have - + - + - + - + ?

Thanks!

Can somebody please explain for me why this one doesnt work that way?

Which of the following is a value of x for which x^11 - x^9 > 0 ?

A. -2
B. -1
C. -1/2
D. 1/2
E. 1


x11−x9>0 --> x9(x2−1)>0 --> (x+1)x9(x−1)>0 --> roots are -1, 0 and 1 --> −1<x<0 or x>1. Only C fits.

Answer: C.

But it does. Everything you have done is correct (ignoring the formatting issues).

I knew the answer makes sense, but if I apply the inequalities trick into this, with 3 roots -1, 0, and 1 and for the f(x) >0, shouldn't it be x <-1 and 0 <x<1?
Thanks

I knew the answer makes sense, but if I apply the inequalities trick into this, with 3 roots -1, 0, and 1 and for the f(x) >0, shouldn't it be x <-1 and 0 <x<1?
Thanks

Dear All,

Is x a negative number?
(1) 9x > 10x
(2) is positive.

GMAT OG claims A is sufficient. But, what if x is a fraction number? Such as 2/90. In this case 2/90 also satisfies ST 1 ..
What is the exact solution for this?

Dear All,

Is x a negative number?
(1) 9x > 10x
(2) is positive.

GMAT OG claims A is sufficient. But, what if x is a fraction number? Such as 2/90. In this case 2/90 also satisfies ST 1 ..
What is the exact solution for this?

if (k – 5)(k – 1)(k – 6) < 0
or
(k – 5)(k – 1)(k – 6) > 0
what will be range for both? I have tried method suggested by gurpreetsingh, but got confused when to take ">" and "<".
can anyone explain? It will be more helpful if anyone can upload an image of the solution.

Thanks.

Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.

If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.



This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.

When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.

When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.

Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7

It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.

Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.

Please explain how will it be different when you have <= or >=

The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side.

e.g.
(x + 2)(x + 3) < 2
x^2 + 5x + 6 - 2 < 0
x^2 + 5x + 4 < 0
(x+4)(x+1) < 0

Now use the concept.

Dear Ma'am!
Please try to respond to my pm...
really needed that help, if possible then revert on forum so that others can benefit for it too.
thanks