Harry_the_cat
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So, can you solve the equation to find the x-intercepts?
Stuck on chapter polynomial and rational inequalities
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Harry_the_cat
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OK. Earlier you said you could graph parabolas, but maybe you can't. I'm just trying to establish what you do know.
I follow the algebra 1 workbook for dummies. Maybe there exist books that are better? How did you learn algebra?
Through classroom instruction from "vary caring" instructor and practice and practice and practice.......
Dr.Peterson
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First, you clearly didn't mean this,
since above it you put a negative sign for this region.
Now, the basic idea of this work is that the sign of the expression [imath]x^2-x-20[/imath] can change only at places where its value is zero; this is because it is continuous (that is, it can't leap over zero from a positive to a negative value without being zero in between).
So you can just check a point in each of the three regions separated by zeros of the expression, knowing it will be negative throughout the region if it is negative at any one point within it.
So, in which interval(s) will it be negative, making the original inequality true?
And do you include or exclude the endpoints of that interval?
I really dont know man..
Dr.Peterson
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Please make an attempt, rather than depend on others to do it for you. We're not here to do the gymnastics routine for you, but to be spotters and coaches, keeping you from hurting yourself and guiding you as you learn the routines. (Yes, I've been watching the Olympics. And this is not the finals, but the beginners' gym.)
Dr.Peterson
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What three regions did you divide the number line into?
Which of them have a negative value?
That's all I asked you to say.
Dr.Peterson
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You haven't identified the regions, but you have at least seen the idea and made a statement about them.
The regions are the intervals [imath](-\infty,-4)[/imath], [imath](-4,5)[/imath], and [imath](5,\infty)[/imath]. You have correctly seen that the expression is negative in [imath](-4,5)[/imath].
Good.
Now that tells us that [imath]x^2-x-20<0[/imath] in [imath](-4,5)[/imath]; and we know that [imath]x^2-x-20=0[/imath] at -4 and 5.
Therefore, the solution to [imath]x^2-x-20\le0[/imath] includes those endpoints; the solution is [imath][-4,5][/imath]; that is, [imath]-4\le x\le5[/imath].
Now, most of what I said was meant to answer your question, "What makes that you know that this is the solution?" Did you follow that?
Dr.Peterson
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Can you say anything more specific about why that is not clear? Which symbol, or concept, are you unsure of? Is there a reason you made part of that bold?
One way to make things clearer, both to us and to you, might be to try stating for yourself (a) what that answer means, and (b) why each part of it is true (leaving blanks, or asking questions, where you have nothing to say). You may discover that when you try writing it, you have more to say than you thought you did.
No, there is no reason why i make it bold. I using a tablet and did t know how to make in undone.
In the algebra workbook they didnt say something about intervals. I think that i using a book that is lack of explanation. Its the algebra 1 workbook for dummies. More people experience troubles with that book?
In the algebra workbook they didnt say something about intervals. I think that i using a book that is lack of explanation. Its the algebra 1 workbook for dummies. More people experience troubles with that book?
Dr.Peterson
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I guesses that. My first thought had been Swedish, but I see "bladeren" is Dutch, too.
It may be that language issues are contributing in both directions.
The images you provide show that they do use interval notation. That's good.
I see that they use an alternative method that I personally prefer over "plugging in a test value in each interval", namely using the factored form to see that everywhere in, say, the middle interval here, the first factor will be negative and the second factor will be positive, so the product, the expression you are interested in, is negative.
Here is how I like to mark up the number line, showing the sign of each factor, then of the entire product:
Can you see what I'm doing there? It matches what they say in the paragraph under the picture.
Harry_the_cat
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I thought I'd share how I think about these quadratic inequalities.
To solve \(\displaystyle x^2-x\leq20\) first rewrite as \(\displaystyle x^2-x-20\leq0\).
Now I think of the graph of \(\displaystyle y=x^2-2x-20\) which we know is a parabola.
Consider:
1. the shape and
2. the x-intercepts.
1. The shape ie \(\displaystyle \cup\) or \(\displaystyle \cap\) is determined by the sign of the coefficient (number in front of) \(\displaystyle x^2\).
If it is positive the shape is \(\displaystyle \cup\). If it is negative, the shape is \(\displaystyle \cap\).
In this case, the coefficient is positive, so the parabola is \(\displaystyle \cup\) shaped.
2. The x-intercepts are found when \(\displaystyle y=0\) ie \(\displaystyle x^2-x-20=0\).
(In an earlier post, I asked you if you could find the x-intercepts. You said no, but you actually can as you have shown in Post#46.)
So, in this case, the x-intercepts are \(\displaystyle -4\) and \(\displaystyle 5\).
Now, putting those two things together the graph of \(\displaystyle y=x^2-x-20\) looks something like:
Now, solving \(\displaystyle x^2-x-20\leq0\) requires you to find the values of \(\displaystyle x\) where the parabola lies below (or on) the x-axis.
(When the parabola is below the x-axis \(\displaystyle x^2-x-20 <0\) and when the parabola is on the x-axis \(\displaystyle x^2-x-20=0\)
You can see at a glance that the parabola lies below or on the x-axis when \(\displaystyle -4\leq x \leq 5\). Then, if required, you can indicate that on a number line.
I find it can be helpful if you can visualise what is going on. Hope that helps.
To solve \(\displaystyle x^2-x\leq20\) first rewrite as \(\displaystyle x^2-x-20\leq0\).
Now I think of the graph of \(\displaystyle y=x^2-2x-20\) which we know is a parabola.
Consider:
1. the shape and
2. the x-intercepts.
1. The shape ie \(\displaystyle \cup\) or \(\displaystyle \cap\) is determined by the sign of the coefficient (number in front of) \(\displaystyle x^2\).
If it is positive the shape is \(\displaystyle \cup\). If it is negative, the shape is \(\displaystyle \cap\).
In this case, the coefficient is positive, so the parabola is \(\displaystyle \cup\) shaped.
2. The x-intercepts are found when \(\displaystyle y=0\) ie \(\displaystyle x^2-x-20=0\).
(In an earlier post, I asked you if you could find the x-intercepts. You said no, but you actually can as you have shown in Post#46.)
So, in this case, the x-intercepts are \(\displaystyle -4\) and \(\displaystyle 5\).
Now, putting those two things together the graph of \(\displaystyle y=x^2-x-20\) looks something like:
Now, solving \(\displaystyle x^2-x-20\leq0\) requires you to find the values of \(\displaystyle x\) where the parabola lies below (or on) the x-axis.
(When the parabola is below the x-axis \(\displaystyle x^2-x-20 <0\) and when the parabola is on the x-axis \(\displaystyle x^2-x-20=0\)
You can see at a glance that the parabola lies below or on the x-axis when \(\displaystyle -4\leq x \leq 5\). Then, if required, you can indicate that on a number line.
I find it can be helpful if you can visualise what is going on. Hope that helps.
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