Doubt in an inequality problem

AI Thread Summary
The discussion centers on solving the inequality (y+2)(y-3) <= 0. The correct interpretation of the inequality leads to the conclusion that -2 <= y <= 3, as both factors must be considered for their signs. The confusion arises from misapplying the conditions for the product to be non-positive, leading to incorrect assumptions about the values of y. Clarification emphasizes that one factor must be negative while the other is positive, necessitating a careful analysis of the intervals. Understanding these intervals resolves the doubts about the solution.
Thiru07
Messages
41
Reaction score
0

Homework Statement


Given : (y+2)(y-3) <= 0

Homework Equations

The Attempt at a Solution


Now, I have y-3 <= 0 or y+2 <= 0
Hence, y <= 3 or y <= -2
But how
-2 <= y <= 3
is correct?
I think
-2 <= y <= 3
is wrong because y <= -2.
Can someone please clarify?
 
Physics news on Phys.org
Thiru07 said:

Homework Statement


Given : (y+2)(y-3) <= 0

Homework Equations

The Attempt at a Solution


Now, I have y-3 <= 0 or y+2 <= 0
Hence, y <= 3 or y <= -2
But how is correct?
I think is wrong because y <= -2.
Can someone please clarify?

You could draw a graph of ##(y+2)(y-3)## to see what's happening.

What happens if both ##y+2 < 0## and ##y-3 < 0##?
 
PeroK said:
You could draw a graph of ##(y+2)(y-3)## to see what's happening.

What happens if both ##y+2 < 0## and ##y-3 < 0##?
In that case , we will keep y < -2 and ignore y < 3.
I think I got it. We have to ignore y+2 <= 0 and y-3 >=0 as
y <= -2 and y >= 3
is not possible and keep
y >= -2 and y <= 3

Thanks PeroK :)
 

Attachments

  • Inequality.JPG
    Inequality.JPG
    6.5 KB · Views: 385
Thiru07 said:
Given : (y+2)(y-3) <= 0

Now, I have y-3 <= 0 or y+2 <= 0
No, the above doesn't follow from the inequality. I think you might already have solved this inequality, but showing where you went wrong is worthwhile.
For the product of two expressions to be negative, one of them has to be negative, and the other has to be positive. Since this can happen in either of two ways, you need to examine two different cases.
Thiru07 said:
Hence, y <= 3 or y <= -2
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top