Determine intervals for which (i) x(x+2)> 0 and (ii) x(x+4)≤ 5

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richiesmasher

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Hello again, I have a question here that I have no previous knowledge of, I turn to the internet for help because I am currently studying on my own for a test.

I have a question here that to be honest I have no idea about.

It states: Determine the interval of x for which (i) x(x+2)> 0
(ii) x(x+4)≤ 5

Now all I know is the basic outline of what an interval is, on a graph the regions where say a number x, is either increasing or decreasing, hence positive or negatives intervals of the number x.

But that's about it, I really do not know where to go from this point. Any help would be appreciated.
 
richiesmasher said:
Hello again, I have a question here that I have no previous knowledge of, I turn to the internet for help because I am currently studying on my own for a test.

I have a question here that to be honest I have no idea about.

It states: Determine the interval of x for which (i) x(x+2)> 0
(ii) x(x+4)≤ 5

Now all I know is the basic outline of what an interval is, on a graph the regions where say a number x, is either increasing or decreasing, hence positive or negatives intervals of the number x.

But that's about it, I really do not know where to go from this point. Any help would be appreciated.
Click to expand...

I tend to recommend you first solve x(x+2) = 0 and see if you can gain any insight.

Then, I might solve x(x+4) = 5 and see if any insight can be gained.
 
tkhunny said:
I tend to recommend you first solve x(x+2) = 0 and see if you can gain any insight.

Then, I might solve x(x+4) = 5 and see if any insight can be gained.
Click to expand...

Hello, thanks for the response, well I'm not sure how to go about this, when I try to expand it just gets more confusing, and my friend was telling me about splitting up each equation and solving each component, but I'm still unsure.
 
tkhunny said:
I tend to recommend you first solve x(x+2) = 0 and see if you can gain any insight.

Then, I might solve x(x+4) = 5 and see if any insight can be gained.
Click to expand...

Ok, I gained some insight, i used the quadratic formula for both of those, and the answers I got is the limits for the ranges of x, but my only problem now is how do I put that into proper notation?

The answer for (i) using the quadratic formula, I got -2, and 0, and for (ii) I got -5 and 1, so I'm assuming these are the ranges.

The answer my book has is this, : (i) {x: x < -2} ∪ {x: x > 0} = {x:-2 ≤ x ≤ 0}^1
(ii) {x: -5 ≤ x ≤ 1}

Can anyone tell me how to get the answers I got using the quadratic formula into that notation up there? I'm honestly perplexed.
 
Last edited:
richiesmasher said:
Hello, thanks for the response, well I'm not sure how to go about this, when I try to expand it just gets more confusing, and my friend was telling me about splitting up each equation and solving each component, but I'm still unsure.
Click to expand...

You don't need the quadratic formula for the first. You can use it if you want, I suppose.

x(x+2) =0 ==> x = 0 or x = -2

Plot this on a Number Line

<---------(-2)---------(0)-------->

See how the number line is now divided into three distinct sections?

1) x < -2
2) -2 < x < 0
3) 0 < x

x = 0 and x = -2 are where the value of the expression is zero. The value must be something other than zero everywhere else. Check it out.

Find (-3)*(-3 + 2) = ??
Find (-1)*(-1 + 2) = ??
Find (2)*(2 + 2) = ??

I just picked a value to represent EACH interval. You should now be able to answer the question if you reason it out.

Show us what you do and what you get on both. :)
 
richiesmasher said:
Ok, I gained some insight, i used the quadratic formula for both of those, and the answers I got is the limits for the ranges of x, but my only problem now is how do I put that into proper notation?

The answer for (i) using the quadratic formula, I got -2, and 0, and for (ii) I got -5 and 1, so I'm assuming these are the ranges.

The answer my book has is this, : (i) {x: x < -2} ∪ {x: x > 0} = {x:-2 ≤ x ≤ 0}^1
(ii) {x: -5 ≤ x ≤ 1}

Can anyone tell me how to get the answers I got using the quadratic formula into that notation up there? I'm honestly perplexed.
Click to expand...

It isn't necessary to use the quadratic formula; both can be solved by factoring. Ultimately, it doesn't matter much how you solve them, but factoring lends itself to several ways to understand the inequalities.

But my first question for you would be, doesn't your book teach how to solve them, in addition to just giving you answers? If you don't have a textbook that explains these things, you can't really say that you are "studying" the subject -- and it can be terribly inefficient. This site isn't really the place to learn something from scratch. Have you at least searched for lessons on the topic? Try this one. And this. Or search on your own for "quadratic inequalities".

As for the answers from your book, there is some odd notation there. Does it literally say "{x:-2 ≤ x ≤ 0}^1"? I can only guess that it is supposed to be a symbol for the complement (that is, all real numbers except this set). That is a valid way to express the solution, though not usual. It is very common to express the solutions using "interval notation", rather than the "set-builder notation" used in those answers. Is part of your difficulty not being familiar with either notation? If so, then search for information on the two names I used for them.

Very briefly (you'll see fuller explanations in the links, and there's probably a lot more that could be said than those introductions say), the solution to an inequality like x(x+2) > 0 will always be one or more intervals whose boundaries are the solutions to the corresponding equation x(x+2) = 0; you can determine which of the three intervals to include by several means, including just testing a value in each interval. More sophisticated methods include graphing y = x(x+2) and finding where it lies above the x-axis, using the multiplicities of the zeros to determine whether the sign changes at each x-intercept, and observing the sign of each factor in each interval.

Read the pages I linked to (and more like them, if you wish), and then you can write back with specific questions about what you've learned as applied to your problems.
 
Dr.Peterson said:
It isn't necessary to use the quadratic formula; both can be solved by factoring. Ultimately, it doesn't matter much how you solve them, but factoring lends itself to several ways to understand the inequalities.

But my first question for you would be, doesn't your book teach how to solve them, in addition to just giving you answers? If you don't have a textbook that explains these things, you can't really say that you are "studying" the subject -- and it can be terribly inefficient. This site isn't really the place to learn something from scratch. Have you at least searched for lessons on the topic? Try this one. And this. Or search on your own for "quadratic inequalities".

As for the answers from your book, there is some odd notation there. Does it literally say "{x:-2 ≤ x ≤ 0}^1"? I can only guess that it is supposed to be a symbol for the complement (that is, all real numbers except this set). That is a valid way to express the solution, though not usual. It is very common to express the solutions using "interval notation", rather than the "set-builder notation" used in those answers. Is part of your difficulty not being familiar with either notation? If so, then search for information on the two names I used for them.

Very briefly (you'll see fuller explanations in the links, and there's probably a lot more that could be said than those introductions say), the solution to an inequality like x(x+2) > 0 will always be one or more intervals whose boundaries are the solutions to the corresponding equation x(x+2) = 0; you can determine which of the three intervals to include by several means, including just testing a value in each interval. More sophisticated methods include graphing y = x(x+2) and finding where it lies above the x-axis, using the multiplicities of the zeros to determine whether the sign changes at each x-intercept, and observing the sign of each factor in each interval.

Read the pages I linked to (and more like them, if you wish), and then you can write back with specific questions about what you've learned as applied to your problems.
Click to expand...

tkhunny said:
You don't need the quadratic formula for the first. You can use it if you want, I suppose.

x(x+2) =0 ==> x = 0 or x = -2

Plot this on a Number Line

<---------(-2)---------(0)-------->

See how the number line is now divided into three distinct sections?

1) x < -2
2) -2 < x < 0
3) 0 < x

x = 0 and x = -2 are where the value of the expression is zero. The value must be something other than zero everywhere else. Check it out.

Find (-3)*(-3 + 2) = ??
Find (-1)*(-1 + 2) = ??
Find (2)*(2 + 2) = ??

I just picked a value to represent EACH interval. You should now be able to answer the question if you reason it out.

Show us what you do and what you get on both. :)
Click to expand...

OK, I think I'm finally here, so I watched Khan academy's video, and read all the things Dr Peterson posted, my final conclusion is this
(i) x(x+2)>0

x>0 and x+2<0 OR x<0 and x+2>0
x>0 and x<-2 x<0 and x>-2

As I watched the video, he said to do the signs like that, and see which range there would satisfy the equation, so I substituted values from those ranges into the equation and the 2 values on the left are clearly correct.

SO: I am using set notation as well, (I looked it up) the answer for (i){x: x >0}∪{x: x <-2}

Now for part (ii) x(x+4)≤5
x^2 + 4x ≤ 5
x^2-x+5x-5≤0
(x+5) (x-1) ≤ 0
SO:
x≤-5 and x≥1 OR x≥-5 and x≤1
From the two equations on the right {x: x ≥-5} ∪ {x : x ≤1}

Clearly the two equations on the right satisfy the inequality, as I substituted the values and tested them out, this method is kind of long but it does make sense now to me.

And to answer your question Dr. Peterson, yes my book continuously gives the complement solution for these types of questions. They give the normal solution followed by the complement, I don't know why but that's how it is.
 
richiesmasher said:
OK, I think I'm finally here, so I watched Khan academy's video, and read all the things Dr Peterson posted, my final conclusion is this
(i) x(x+2)>0

x>0 and x+2<0 OR x<0 and x+2>0
x>0 and x<-2 x<0 and x>-2

As I watched the video, he said to do the signs like that, and see which range there would satisfy the equation, so I substituted values from those ranges into the equation and the 2 values on the left are clearly correct.

SO: I am using set notation as well, (I looked it up) the answer for (i){x: x >0}∪{x: x <-2}
Click to expand...

Good work. There are quicker ways, as I mentioned, but this is a good, and very general, way.

richiesmasher said:
Now for part (ii) x(x+4)≤5
x^2 + 4x ≤ 5
x^2-x+5x-5≤0
(x+5) (x-1) ≤ 0
SO:
x≤-5 and x≥1 OR x≥-5 and x≤1
From the two equations on the right {x: x ≥-5} ∪ {x : x ≤1}

Clearly the two equations on the right satisfy the inequality, as I substituted the values and tested them out, this method is kind of long but it does make sense now to me.
Click to expand...

One problem here: you had two possibilities separated by "or", the first of which is empty (contradictory), so you chose only the second. But within that you have "and", not "or", so you should be using an intersection rather than a union. And such an intersection can be written instead as {x : -5 ≤ x ≤ 1}, which means x≥-5 and x≤1.

richiesmasher said:
And to answer your question Dr. Peterson, yes my book continuously gives the complement solution for these types of questions. They give the normal solution followed by the complement, I don't know why but that's how it is.
Click to expand...

My question was mainly about your notation, which I assume meant a superscript 1; is it actually a "prime", like {x : -2 ≤ x ≤ 0}' ? That is a common notation for complements, while what you wrote is not.
 
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