2 sums with Modulus and Inequality both

@ Krushna

Yes, you have got the first one right
The answer is : 1<=x<=4

If u can solve the 2nd one too, that will be a great help. I've been pulling my hair because of the 2nd one. I am not in agreement with its answer.

Cheers,
R J

P.S: One more thing though, you gave me the answer as 1,2,3,4, though you musn't forget that the numbers have not been stated as integers.
Kudos+1 for the approach that is different than the one I have.

if |x^2 + 3x| + x^2 - 2 >= 0.

|x(x + 3)| + x^2 - 2 >= 0.

boundary points are 0,3

the ranges are:

1) When x >= 0, |x^2 + 3x| = x^2 + 3x

hence the equation becomes x^2 + 3x + x^2-2 >=0

2x^2 + 3x - 2 >=0

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

as x >= 0, x+1 > 0

So we need 2x-1>=0 means x >=1/2

So combining: x >=1/2 --- A


2) When -3 < x < 0, |x^2 + 3x| = -(x^2 + 3x)

hence the equation becomes -x^2 - 3x + x^2-2 >=0

- 3x - 2 >=0

3x+2 <=0, x <=-2/3

as -3 < x < 0, combining: -3 < x <=-2/3


3) When x <= -3, |x^2 + 3x| = x^2 + 3x ----- B

hence the equation becomes x^2 + 3x + x^2-2 >=0

2x^2 + 3x - 2 >=0

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

as x <= -3, x+1 <0

So we need 2x-1 <=0 means x <=1/2

So as x <= -3, combining: x <= -3 --- C

from A,B and C, either x >=1/2 or x <= -2/3

(gIVE ME A KUDO IF YOU LIKE IT)

Solving such equations is very convenient and quick using graphs. I have attached a pdf to show how to solve the first one using graphs. If you understand how it is solved, let me know and I will send the solution of the second one using graphs too. If it is not clear, I will give a quick recap of graph theory for mods.

I got it. Very nice approach.
Please provide for the second one

Given \(|x^2 + 3x| + x^2 - 2 >= 0\)

This implies \(|x^2 + 3x| >= 2 - x^2\)
We need to find values of x for which this relation holds. We will draw the graph of both the left side and the right side and find the answer by checking the values of x for which the graph of left side has higher values than graph of right side. Check the attachment for solution.

For another interesting use of graphs to solve a tough question, check out this link
https://gmatclub.com/forum/ps-triple-modul-1185.html#p807380

Aaahhh...and the truth shall set u free!!!

The answer given was indeed wrong... the answer given to the 2nd question is x>=1/2 and x<= -3....

Cheers
R J

All right!!.. i understand the explanation, though what I am confused about is that in the 1st question you basically take an INTERSECTION of the two ranges of the values of x and get to the answer; however, in the 2nd question you provide the answers as x>= 1/2 OR x <= -2/3....
Basically i had solved it in this manner -
Since it is |x^2 +3x| in the given inequality, alternately assume it to be positive or negative

Case - 1: Considering it +ve
Therefore,
x^2+3x + x^2 - 2 >=0
Solving this we get: x>=1/2 OR x<=-2 -------> A

Case - 2: Considering it -ve
Therefore,
-x^2 - 3x + x^2 - 2 >=0
Hence, x<=-2/3 -----------> B

Now here the problem I am facing is whether I find the intersection of the ranges or Union of the ranges.
Can you please explain this..

Cheers
R J

Also can you provide for a sum where the slope is not equal to 1, just to make it clear.

Thanks!!!! :D

Cheers,
R J

Try mod(2x -8) > 10

I got the answer to the above question as x>9 OR x<-1.

And Kudos+1 for the explanation earlier. :D :D

Cheers,
R J

Yes, your answer is correct. Well done!

Note that in this graph, the point of the graph that lies on the x axis will be at x = 4 not x = 8 because mod(2x - 8) = mod(2(x - 4)).
Typically, linear mod inequalities can be easily and quickly solved using just the number line but let us keep that for another day! Get comfortable using graphs and later perhaps we can shorten the time taken even further.

Also, these questions are 700 level and you will see them on GMAT only if you are close to 50/51. Even then, you may not see the second question.

Yup...that is where I intend to go, the 700s!!!

Many many thanks for your help.. :D

Cheers,
R J

thanks for the clarification.

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