Inequalities trick

Yes, absolutely but after you bring the factors in the form (ax +/- b)

Responding to a pm:



We are given that \((x-4)^2(x - 9)(x+11)\) is less than 0.
We need to find the range of values that x can take in that case.
Note that the expression IS LESS THAN 0. This means that it cannot be 0. So x cannot be 4 because that will make the expression 0. So we don't plot 4 on the number line. (x-4)^2 is positive only and hence doesn't affect our signs.

Answer = 11

Check solution as attached

Thanks a lot for this post , it's very helpful. Would be great if you clarify the below query:

Given:
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

Question: What if we have 2 factors? do we need to start out interpretation as + - + and for 5 factors, do we need to have it in this way? - + - + - +

Thanks in advance,
Uma

Yes, that is correct. Usually, n factors will divide the number line into (n+1) regions. You start out by giving + to the rightmost region (provided all your factors are of the form (ax + b) or (ax - b)) and then alternating the signs on the left.

It will be good if you take a look at this post which gives links to explanations on why this happens:
https://gmatclub.com/forum/inequalities ... l#p1753431

Hi karishma, abhimahna, mikemcgarry, Bunuel, Skywalker18, and other experts

I am not able to understand the graph. How do we know which one is negative/positive? How does this graph help us in solving inequalities?

Thanks

Hi Shiv2016 ,

The concept is very easy.

1. Find out the zero points.
2. Arrange them in ascending order.
3. Draw them on a number line.
4. Take the right most as +ve and proceed towards left taking alternate signs.
5. If the inequality is of > form, then take all +ve ranges.
6. if the inequality is of lesser form, then take all -ve ranges.

E.g.

(x-2) (x-3 ) > 0

Equality form is greater than(>).

Zero points = 2 and 3

Draw them on number line. You will get 3 ranges. x<2; 2<x<3; and x >2.

Here, right most will be +ve, or x>2 will be +ve.
then 2<x<3 will be -ve
then x<2 will be +ve.

Since inequality is of (>) form, we will take all the ranges which have +ve sign.

Hence, the answer will be x>2 and x<2.

I hope it makes sense.

The links of the three posts given here should clear all your doubts: https://gmatclub.com/forum/inequalities ... l#p1753431

Dear Shiv2016,

I'm happy to respond.

I see that abhimahna already gave I wonderful explanation. I will just add a few points.

The post at the beginning of the thread is not 100% accurate and it is way to general for the GMAT. Inequalities with powers of x is a exceeding rare topic on the GMAT, and when there is a power in an inequality, it almost always is just x. I think you could take 50 GMATs in a row and not see a power of x higher than 2 in an inequality.

You may find this blog helpful:
GMAT Quadratic Inequalities

I definitely agree that finding the roots, the x-values that make the function zero, is the first step. Find these, and put these on a number line.

If the highest power is simply 2, then you should know that graph is a parabola. If the coefficient of \(x^2\) is positive, then the parabola opens upward, like the letter U. If the coefficient of \(x^2\) is negative, then the parabola opens downward, like an upside-down U. Right there, that should tell you where the function is positive or negative.

Higher powers of x (\(x^3\), \(x^4\), etc.) appear infrequently. The alternating pattern doesn't always work: in particular, the signs don't alternate if there's a repeated root, e.g. \(x^3 - 4x^2 + 4x > 0\)). That's a caveat for higher mathematics, but I don't think anyone would ever need to know this for the GMAT.

My friend, part of the reason you were confused is because the person who opened the thread was talking about more advanced math than you need for the GMAT. People who study engineering learn way more math than is needed for the GMAT, and some of them like to show off, but this doesn't really help anyone.

Does all this make sense?
Mike

Responding to a pm:


Note that in case of division, the denominator cannot be 0.

So given \(\frac{(x+2)(x-1)}{(x-7)(x-4)} <= 0\)

We will use the same method to find that the solution will be -2 < x < 1 or 4< x < 7 but here, we have an equal to sign too. So the expression can be equal to 0 too. So x can be equal to -2 and it can be equal to 1 too. But x cannot be 4 and neither can it be 7 since that will make the denominator 0.

So the actual solution here will be -2 <= x <= 1 or 4< x < 7

[quote="gurpreetsingh"]I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.

Suppose you have the inequality

\(f(x) = (x-a)(x-b)(x-c)(x-d) < 0\)

Just arrange them in order as shown in the picture and draw curve starting from + from right.

now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.

Don't forget to arrange then in ascending order from left to right. \(a<b<c<d\)

So for f(x) < 0 consider "-" curves and the ans is: \((a < x < b)\), \((c < x < d)\)
and for f(x) > 0 consider "+" curves and the ans is: \((x < a)\), \((b < x < c)\), \((d < x)\)

If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +

Hi, if the function is as such (-x-a)(-x-b), would we just flip the direction of the curve, and hence the order of signs? Thanks!

Some further clarification on the role of Odd (i.e. x) vs Even (i.e. x^2) powers on the Wavy Curve Method

Thought I'll add some more explanation below for better understanding on this method.

Explanation:

Suppose, if we have "x^3(x-1)<0", then that can be transformed to "x.x^2.(x-1)<0". Notice that "x" and "x^2" work differently here. At "x^2", the line bounces back so the graph doesn't actually change the sign, and hence we can always omit the even powers such as x^2 or (x-1)^2 from the inequality, leading to this: "x(x-1)<0". Using the wavy curve plot, we get 0<x<1.

More examples for transformation needed for Wavy Curve Method:

Example 1:
(x-1)^3 <0 => (x-1).(x-1)^2 => (x-1) < 0 [We ignore even power (x-1)^2 for the reason explained above]
Example 2:
x^2-x^3>0 => x^2(1-x)>0 => (1-x)>0 => x<1 [We ignore even power x^2 for the reason explained above]

Hi,

I have a question.
(1-2x)(1+x)<0 solving by curve method gives x in the range -1 and 1/2 for <0

but (2x-1)(x-1)>0 gives x<-1 and x>1/2.

I am not sure what I am doing wrong. This changes the answer completely.

hello VeritasKarishma

with reference to your post here https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... ns-part-i/


there is an inequality with your solution ( see below)

\(-2x^3 + 17x^2 – 30x > 0\)

\(x(-2x^2 + 17x – 30) > 0 \) (taking x common)

\(x(2x – 5)(6 – x) > 0\) (factoring the quadratic)

\(2x(x – 5/2)(-1)(x – 6) > 0\) (take 2 common)

\(2(x – 0)(x – 5/2)(x – 6) < 0\) (multiply both sides by -1)


i dont get how you got \(x(2x – 5)(6 – x) > 0\)

here is what i did

given : \(-2x^3 + 17x^2 – 30x > 0\)

\(x(-2x^2 + 17x – 30) > 0 \)

\(x(-2x^2 -12x - 5x – 30) > 0 \)

\(x(-2x^2 -12x) - (5x – 30) > 0 \)

\((x) (-2x) (x+6) - 5(x-6)\)

now what do do ? kinda confused

and how you got " - 1 " here ---> \(2x(x – 5/2)(-1)(x – 6) > 0\)

and how you got "x-0 " here --- > \(2(x – 0)(x – 5/2)(x – 6) < 0\)


then i had watched this https://www.khanacademy.org/math/algebr ... grouping-6


so i thought i would follow similar pattern


so again given \(-2x^3 + 17x^2 – 30x > 0\)

\(-x(2x^2 - 17x + 30x > 0)\)

\(-x(2x^2-12x-5x+30)>0\)

\(-x((2x^2-12x)-(5x+30))>0\)

\(-x((2x)(x-6)-5(x-6))>0\)

\((-x)(2x-5)(x-6)>0\)

okay now what to do ? again stuck

thanks

The step in red is incorrect. It should be

\(-x((2x^2-12x)-(5x-30))>0\)

If you are taking -5 common, the sign inside the bracket becomes - too so that when you open the bracket, you get a + back.

\(-x((2x)(x-6)-5(x-6))>0\)

\((-x)(2x-5)(x-6)>0\)

Now multiply both sides by -1 to get

\(x(2x-5)(x-6) < 0\)

Now simply plot 0, 5/2 and 6 on the number line and the negative sections are x < 0 and 5/2 < x< 6

VeritasKarishma many thanks for explanation

one question, if i multiply whole inequity \((-x)(2x-5)(x-6)>0\) by -1

then i get \((x) (-2x+5) (-x+6)<0\)

how is it possible that signs inside of brackets didn`t change in your case after multiplying by -1

Think about this:


3 * 5 > 0

Multiply both sides by 2 to get:

2 * 3 * 5 > 0

Does it become 2*3 * 2*5 > 0? No.
They are all factors. When you multiply by 2, one more factor comes in.

You can write it as 2*3 * 5 > 0 or as 3 * 2*5 > 0 but not as 2*3 * 2*5 > 0

Similarly,
(-x) (2x-5)(x-6)>0 has three factors. When you multiply by -1, only one term needs to get multiplied by -1, not all.

You are confusing it with

(3 + 5) > 0
Now if you multiply by 2, you will multiply both terms with 2. This happens only when you have a + or a - sign.