Inequalities written separately or together?

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gerib17
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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
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ghostwalker
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(Original post by 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
The solution is -0.5 < x < 3, and you would write it like that. "x" satisfies both inequalites at the same time, "-0.5 < x" AND "x < 3". It lies in the interval -0.5 to 3.

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.
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gerib17
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Essentially if it’s less than then we write it together and if it’s bigger than we split it up
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ghostwalker
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(Original post by gerib17)
Essentially if it’s less than then we write it together and if it’s bigger than we split it up
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)
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gerib17
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(Original post by 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!
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ghostwalker
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(Original post by gerib17)
Question 9 please!
So, you've got to 4k(k-3)&lt; 0

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.
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