If p(c) is negative, therefore less than 0, why is it neccesary to ascertain the number of times that p(t)=0 when there can't be negative times p(t)=0 and therefore even if p(t) never equals zero, it will still be more than a negative number?

Sherlock, you are right: there is no need for us to determine the exact number of times when p(t)=0 to answer the question. I believe Sal worked through the second part of the problem just to show us the intuition and to have some fun:)

Isn't it that the square root of (a^2+b^2) is a positive and a negative number?

That's the principal square root you saw there in the denominator. So it is always positive. You would be right if you saw the plus/minus sign before the square root. A single minus sign would also make it negative.

I'm not seeing the big picture. It doesn't seem like the solution here tells us anything important about the problem itself.

The point here is that by looking at the problem in a way where we are concerned with only certain characteristics of the variables (in this case, which of the terms in the numerator end up positive, and which end up negative), we can get some information about how the function behaves over certain parts of the domain of the function. Knowing that when using an input value that is larger than the variable a and the variable b in the function, the result is always going to be negative, might be a powerful bit of information, depending on what the problem is. In this example, where the function is a position function, I think it is telling us that when we take the position at a time that is bigger than a or b, the position has moved backwards, in the negative direction. We can also get information about which direction the position goes when using a time value smaller than the variable a, or between the variable a and b. I can't see the original problem as I'm writing this, but I think the numerator was (t-a)(a-t)(t-a)(t - b) with the constraint that b>a>0. So if we pick a c that is smaller than a, the 1st term is negative, the 2nd term is positive, the third term is negative, and the fourth term is negative, so overall the numerator is negative. But if we pick a value between a and b. The first term is positive, the second term is negative, the third term is positive, and the fourth term is negative, so the numerator is positive.

Why does this seem difficult and unsolvable when Sal asks to pause the video and tackle it ourselves, but then it's easily understood when he starts explaining it?! What is it of the mindset that I may be missing?

Practice, practice, practice - for a lot of people, the more you work on math, the more you learn number sense, inductive and deductive reasoning, and more intuitive approaches, so you will be able to do more and more as you go. When you first are learning a new concept, unless it is closely related to something you know, then it will be hard to solve, but I could probably give you a lot of 4th and 5th grade problems that you could work quickly without much deep thinking. Enjoy Math for all of its patterns and logic, and you will be able to start tackling it yourself.

You need to explain how c can be negative. It clearly states c > b > a > 0 This is just fanciful maths there is no logic to it! A number less than zero is negative not a number greater than 0.

c *is not* negative, but the function p, evaluated at c, that is, p(c), gives a negative number. This is fine. I suggest you try the problem with some numbers, eg, let a=1, b=2 and c=3, for which c > b > a > 0, that is, 3 > 2 > 1 > 0, correct? Now evaluate p(3) and see what happens.

At 4:30 isnt the value for number of P(t)= 0 4 as their are 4 brackets so they could be 0?

While there are 4 brackets, there are only 2 distinct possibilities because (c - a)( a - c) and (c-a) are all the same c = a.

Wait, what if c is negative? Then would the first expression be correct or closer to correct?

C cannot be negative because in the second statement it says that it is greater than 0.

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Algebra (all content)

Course: Algebra (all content) > Unit 13

Lesson 11: Modeling with rational functions

Structure in rational expression

Name: Structure in rational expression
Uploaded: 2016-03-03T01:11:26Z
Description: Analyzing an elaborate rational expression to determine what's bigger: the value of that expression at some unknown c, or the number of times the expression is equal to 0? This is algebraic reasoning at its best!

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Analyzing an elaborate rational expression to determine what's bigger: the value of that expression at some unknown c, or the number of times the expression is equal to 0? This is algebraic reasoning at its best! Created by Sal Khan.

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Sherlock
Posted 9 years ago. Direct link to Sherlock's post “If p(c) is negative, ther...”
If p(c) is negative, therefore less than 0, why is it neccesary to ascertain the number of times that p(t)=0 when there can't be negative times p(t)=0 and therefore even if p(t) never equals zero, it will still be more than a negative number?
Button navigates to signup pageComment on Sherlock's post “If p(c) is negative, ther...”
(55 votes)
Answer
- Alexander Kokin
  Posted 8 years ago. Direct link to Alexander Kokin's post “Sherlock, you are right: ...”
  Sherlock, you are right: there is no need for us to determine the exact number of times when p(t)=0 to answer the question. I believe Sal worked through the second part of the problem just to show us the intuition and to have some fun:)
  Button navigates to signup page
  (35 votes)
box 0f rox
Posted 10 years ago. Direct link to box 0f rox's post “I'm not seeing the big pi...”
I'm not seeing the big picture. It doesn't seem like the solution here tells us anything important about the problem itself.
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- KrisSKing
  Posted 9 years ago. Direct link to KrisSKing's post “The point here is that by...”
  The point here is that by looking at the problem in a way where we are concerned with only certain characteristics of the variables (in this case, which of the terms in the numerator end up positive, and which end up negative), we can get some information about how the function behaves over certain parts of the domain of the function. Knowing that when using an input value that is larger than the variable a and the variable b in the function, the result is always going to be negative, might be a powerful bit of information, depending on what the problem is. In this example, where the function is a position function, I think it is telling us that when we take the position at a time that is bigger than a or b, the position has moved backwards, in the negative direction. We can also get information about which direction the position goes when using a time value smaller than the variable a, or between the variable a and b. I can't see the original problem as I'm writing this, but I think the numerator was (t-a)(a-t)(t-a)(t - b) with the constraint that b>a>0. So if we pick a c that is smaller than a, the 1st term is negative, the 2nd term is positive, the third term is negative, and the fourth term is negative, so overall the numerator is negative. But if we pick a value between a and b. The first term is positive, the second term is negative, the third term is positive, and the fourth term is negative, so the numerator is positive.
  Comment on KrisSKing's post “The point here is that by...”
  (26 votes)
R2D2
Posted 8 years ago. Direct link to R2D2's post “Isn't it that the square ...”
Isn't it that the square root of (a^2+b^2) is a positive and a negative number?
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- Aperneto
  Posted 8 years ago. Direct link to Aperneto's post “That's the principal squa...”
  That's the principal square root you saw there in the denominator. So it is always positive. You would be right if you saw the plus/minus sign before the square root. A single minus sign would also make it negative.
  Comment on Aperneto's post “That's the principal squa...”
  (6 votes)
Emad Omar
Posted 6 years ago. Direct link to Emad Omar's post “Why does this seem diffic...”
Why does this seem difficult and unsolvable when Sal asks to pause the video and tackle it ourselves, but then it's easily understood when he starts explaining it?! What is it of the mindset that I may be missing?
Button navigates to signup pageComment on Emad Omar's post “Why does this seem diffic...”
(4 votes)
Answer
- David Severin
  Posted 6 years ago. Direct link to David Severin's post “Practice, practice, pract...”
  Practice, practice, practice - for a lot of people, the more you work on math, the more you learn number sense, inductive and deductive reasoning, and more intuitive approaches, so you will be able to do more and more as you go. When you first are learning a new concept, unless it is closely related to something you know, then it will be hard to solve, but I could probably give you a lot of 4th and 5th grade problems that you could work quickly without much deep thinking. Enjoy Math for all of its patterns and logic, and you will be able to start tackling it yourself.
  Button navigates to signup page
  (8 votes)
Rebecca Xi
Posted 8 years ago. Direct link to Rebecca Xi's post “I understand the concept ...”
I understand the concept and how to solve the problem, but how can this aid me in solving actual problems with rational expressions?
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(3 votes)
Answer
John Robinson
Posted 7 years ago. Direct link to John Robinson's post “You need to explain how c...”
You need to explain how c can be negative. It clearly states c > b > a > 0 This is just fanciful maths there is no logic to it! A number less than zero is negative not a number greater than 0.
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Stefen
  Posted 7 years ago. Direct link to Stefen's post “c *is not* negative, but ...”
  c is not negative, but the function p, evaluated at c, that is, p(c), gives a negative number. This is fine.
  I suggest you try the problem with some numbers, eg,
  let a=1, b=2 and c=3, for which c > b > a > 0, that is, 3 > 2 > 1 > 0, correct?
  Now evaluate p(3) and see what happens.
  Button navigates to signup page
  (2 votes)
Sami Ching
Posted 6 years ago. Direct link to Sami Ching's post “At 4:30 isnt the value fo...”
At
4:30
isnt the value for number of P(t)= 0 4 as their are 4 brackets so they could be 0?
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(1 vote)
Answer
- David Severin
  Posted 6 years ago. Direct link to David Severin's post “While there are 4 bracket...”
  While there are 4 brackets, there are only 2 distinct possibilities because (c - a)( a - c) and (c-a) are all the same c = a.
  Button navigates to signup page
  (2 votes)
John Spallin
Posted 6 years ago. Direct link to John Spallin's post “Wait, what if c is negati...”
Wait, what if c is negative? Then would the first expression be correct or closer to correct?
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(1 vote)
Answer
- mvo
  Posted 6 years ago. Direct link to mvo's post “C cannot be negative beca...”
  C cannot be negative because in the second statement it says that it is greater than 0.
  Button navigates to signup page
  (2 votes)
eli
Posted 6 years ago. Direct link to eli's post “how o we know the sqrt of...”
how o we know the sqrt of a^2+b^2 is positive? Two the sqrt of a positive can be either positive or negative
Button navigates to signup pageComment on eli's post “how o we know the sqrt of...”
(1 vote)
Answer
- Mr.Unicorn
  Posted 6 years ago. Direct link to Mr.Unicorn's post “no, anything squared will...”
  no, anything squared will be positive. Say for ex.-2^2 which equals 4. Because if a negative multiplies a negative it will be positive, same for positives. So if a positive adds to a positive it will be positive. Hope it helps.
  Button navigates to signup page
  (1 vote)
Sobhan.Bihan
Posted 8 years ago. Direct link to Sobhan.Bihan's post “Couldn't we have used ded...”
Couldn't we have used deductive reasoning to state that as the number of values that would make p(t) equal to 0 can't be negative (this value has a minimum of 0) this value is greater than c without even finding the number of values that would make p(t) equal to 0?
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(1 vote)
Answer

Video transcript

Let's say we we're told that the position of a particle can be figured out from, or the position of a particle as a function of time is given by this crazy expression. The position could be positive or negative. And this expression is t minus a, times a minus t, times t minus a, times t minus b, all of that over the square root of a squared plus b squared. And they also tell us that c is greater than b, is greater than a, is greater than 0. So given that information, we have two statements right over here. We have the position at times c on the left hand side. And over here, we have the number of times that our function is equal to 0, the number of times that p of t is equal to 0. What I want you to do is pause the video right now and think about which of these statements provides a larger value. Is p of c greater than or less than the number of times p of t equals 0? And I guess a third and fourth option would be that you maybe don't have enough information to figure this out or maybe that these are equal. So I encourage you to figure that out right now. Which of these are larger? Or do you not have enough information? Or are they equal? So I'm assuming you've given a go at it. So let's think about each of them. So let's think about what p of c is. So p of c is going to be equal to, right here p-- I don't want to arbitrarily switch colors, which I sometimes do. p of c is going to be equal to, let's see. It's going to be c minus a. I'll do this all in this one color. It's going to be c minus a, times a minus c, times c minus a, times c minus b, all of that over the square root of a squared plus b squared. So what do we know about this quantity? What do we know about this quantity right over here? Let me highlight all the Cs here. So c minus a, a minus c, c minus a, c minus b. Well, they tell us that c is larger than a and b and that they're all positive. So maybe we can come up with some statement about whether this thing is positive or negative, whether this expression is. So what's c minus a going to be? Well, c is greater than a. So this is going to be positive. What about a minus c? Well, a is less than c. So this is going to result in a negative number. c minus a, well, this is going to be a positive again. And then c minus b is also going to be a positive. c is greater than both b and a. And what do we have here in the denominator? Well, the square root of a squared plus b squared, well, this is just going to be a positive value. So what do we have going on here? Here in the numerator, I have a positive times a negative, times a positive, times a positive. So what's that going to be? Well, that's going to be a negative. A positive, times a positive, times a positive is positive. And then you throw that negative in there. So you're going to get a negative over a positive. And what's a negative divided by a positive? Well, that's going to be a negative. So we don't know what the actual value is. But all we do know is that this provides us with a negative value. So this ends up being a positive value. Then we could make a statement that this, we just say, hey, this is a negative value as well. But we might not have enough information. So let's think about the number of times p of t is equal to 0. Well, p of t is equal to 0 whenever the numerator right over here is equal to 0. And when would the numerator equal 0? Well, I have the product of this 1, 2, 3, 4 expression. So if any one of these expressions is 0, then the entire numerator is going to be 0. So let's think about how you can make these expressions 0. So the expression could be 0. This will be 0 if t is equal to a. This would also be 0 if t is equal to a. And this would also be 0 if t is equal to a. And this would be 0 if t is equal to b. So there's two values for t that will make this numerator equal to 0, t equals a or t equals b. So there's two times. So the number of times that p of t is equal to 0 is two. So now let's answer our question. What is larger, the number 2, is the number positive 2, versus some negative number? Well, 2 is larger than any negative number. So this is the larger quantity.