# Inequalities written separately or together?

WatchPage 1 of 1

Go to first unread

Skip to page:

I have some questions with inequalities but don’t know if I’m supposed to write the solution together or not.

I’ll give an example :

find the set of values of x for which

2x^2-5x-3<0

I’ll give an example :

find the set of values of x for which

2x^2-5x-3<0

0

reply

Report

#2

(Original post by

I have some questions with inequalities but don’t know if I’m supposed to write the solution together or not.

I’ll give an example :

find the set of values of x for which

2x^2-5x-3<0

**gerib17**)I have some questions with inequalities but don’t know if I’m supposed to write the solution together or not.

I’ll give an example :

find the set of values of x for which

2x^2-5x-3<0

IF on the other hand you had 2x^2-5x-3 > 0, then either "x < -0.5 or x > 3". Clearly it can't satisfy both inequalities at the same time, and you'd write it as I put in quotes.

0

reply

Essentially if it’s less than then we write it together and if it’s bigger than we split it up

0

reply

Report

#4

(Original post by

Essentially if it’s less than then we write it together and if it’s bigger than we split it up

**gerib17**)Essentially if it’s less than then we write it together and if it’s bigger than we split it up

Whether you can put the results together or not, is going to depend on what the solution is, rather than on whether its < or > in the question.

Having solved the inequality, if you end up with x having to satisfy two inequalites. If there are a < x AND x < b, then you can combine them into a < x < b (assuming a < b)

0

reply

(Original post by

I assume you're refering to the original question there with the "< 0".

Whether you can put the results together or not, is going to depend on what the solution is, rather than on whether its < or > in the question.

Having solved the inequality, if you end up with x having to satisfy two inequalites. If there are a < x AND x < b, then you can combine them into a < x < b (assuming a < b)

**ghostwalker**)I assume you're refering to the original question there with the "< 0".

Whether you can put the results together or not, is going to depend on what the solution is, rather than on whether its < or > in the question.

Having solved the inequality, if you end up with x having to satisfy two inequalites. If there are a < x AND x < b, then you can combine them into a < x < b (assuming a < b)

Attachment 770390

Attachment 770392

Question 9 please!

0

reply

Report

#6

(Original post by

Question 9 please!

**gerib17**)Question 9 please!

This does NOT imply 4k<0

__and__k-3< 0.

You have the product of two numbers ( 4k and k-3 ) and that product is negative. For the product of two numbers to be negative, one must be positive and the other must be negative.

So, either 4k is positive and k-3 is negative or vsv.

The general methodology,

**of which there should be a worked example in your textbook**, is to look at the "critical values", the values where the inequalities are equalities, i.e. when they equal 0 in this case.

If 4k=0, then k=0.

If k-3 = 0, then k=3

This gives you the two critical values 0 and 3.

We now consider the number line split up at those critical points, 0 and 3.

So we have three parts: k < 0, 0 < k < 3, and 3 < k

We now check each in term.

IF k< 0, then:

4k<0, and k-3 < 0.

So, their product is > 0, and this part won't form part of the solution.

if k > 3, then:

4k >0, and k-3 > 0.

So, their product is > 0, and this part won't form part of the solution.

If 0 < k < 3, then:

4k > 0, and k-3 < 0.

Their product is < 0, and so this will form part of the solution.

In this case the solution has just one part 0 < k < 3.

Finally you need to check the end points to this interval.

If k is 0 or 3, then the product of 4k,k-3 is zero, which is not less than zero, so they do not form part of the solution.

Hence the result 0 < k < 3.................(1)

This differs slightly from the given solution 0 <= k < 3, but if k satisfies (1), then it must satisfy the given solution too.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top