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## Algebra (all content)

### Course: Algebra (all content) > Unit 9

Lesson 12: Quadratic inequalities# Quadratic inequalities (example 2)

Sal solves -x(2x-14)≥24. Created by Sal Khan.

## Want to join the conversation?

- What would the "one positive, one negative" rule from2:30be like with an equation that goes up to x^3? Or x^4 or above?(17 votes)
- For x^3, it would be "two positives, one negative" or "three negatives" to keep the whole expression negative. For x^4, it's "three positives, one negative" or "three negatives, one positive". Hope I'm not missing anything else here.(13 votes)

- How is it that two expressions of the same inequality produce two different parabolas? I understand the mechanics of graphing the inequalities, but I don't have any good intuition about this. Two different expressions of a linear equation, for example, map to a single line. (Don't they?)

The discussion of this starts at6:20or so. One expression of the inequality is an upward opening parabola, but the other opens downward.(12 votes)- Because there's more than one way of getting the same answer.

You're not saying the 2 parabolas will have to look the same/ be identical, you're just saying that the graphs of both will be greater than or equal to 0 in the same range of x values (and it's these x values that we wanted to find out in our question).

If you graph them, you'll see that the answers for the inequality on the two parabolas ARE the same thing.

Meaning, the downward opening parabola (=2x^2+14x=24) is greater than or equal to 0 (rises above y-0) between x=3 and x=4 on the graph, and the upward opening parabola (x^2-7x+12) is ALSO greater than/ equal to 0 between x=3 and x=4.

So the 2 graphs are really just showing us the same answer, which is also the answer we got using just algebra.(11 votes)

- Sometimes when Sal factors out the coefficient "a" and other times he divides everything else by it to get rid of it completely, how do you know which one you have to do?(4 votes)
- I have a question since I tend to factor and drag it around instead of dividing. I got -2{(x-3)(x-4)} is greater or equal to 0. Because the answer is different than sal, does that mean I did it wrong? Do you have to divid?(2 votes)

- In the upcoming practice section, Modeling with one-variable equations and inequalities, I stumbled across a problem which, given the following equation, required us to solve for t:

1024*(1/2)^(t/29)=32

solve for t.

The solution given in the hints went:

(1/2)^(t/29)=32/1024

(1/2)^(t/29)=1/32

(1/2)^(t/29)=(1/2)^5

therefore

t/29=5

t=29*5=145

Now, although I fully understand the logic when I see it, the passage that went from (1/2)^(t/29)=1/32 to (1/2)^(t/29)=(1/2)^5 would never have occurred to me on my own, not in a million years. Is there any previous section I should study harder, any method of handling bases and exponents I should assimilate perhaps, or am I just not smart enough to see what should be obvious at this stage? Thanks in advance for any answer.(3 votes)- I had the same thoughts about finding that special pair of numbers when factoring a quadratic. "How on earth will I go through all the pairs, it will never occur to me!" The truth is, with exponent problems a common approach is to simplify the other side to have the same base. Of course everything comes with practice. Keep in mind though that you always want to try to have the same base - especially if you see simple fractions such as 1/2 1/4 or 1/8 etc. as the powers of these are very common and easy to remember.(5 votes)

- 3:38- why does he say: if x-3<=0 and x-4>=0 ?

the expression (x-3)(x-4)<=0 is valid if

1) (x-3)(x-4)<=0

or

2) -(x-3)-(x-4)<=0 wich is equivalent to: (x-3)(x-4)>=0

so answer 1) x<=3 and x<=4 >>> x<=3

or 2) x>03 and x>=4 >>>>x>=4(3 votes)- Essentially, you want a result that's smaller or equal to zero. The way to get that when you're multiplying 2 terms (here the terms are "x-3" and "x-4") is to have one of those terms be a negative number. So either x-3<=0
**or**x-4<=0. That's why you end up with these "one of these terms will be greater than zero/positive and one of these terms will be smaller than zero/negative" inequalities at the end.

The next video in the sequence, Quadratic inequalities (visual explanation), makes it more clear as to why you get this type of result. Hope this helps!(4 votes)

- I didn´t understand why Sal changes the simbol of the inequality. Would be wrong if we don´t change the symbol?(3 votes)
- Sal explains it very clearly leading up to1:30. Since he divides both sides by -2, he MUST change the direction of the symbol.(4 votes)

- We don't have discussion forums on the exercise below, but I have a question?

(1/2)^t/14 = (1/2)^4

What steps do we take in order to come to the following:

t/14 = 4

I have a gap here and I can't quite figure it out? Help anyone?(2 votes)- hey this is a good question, and I will try and explain as best I can.

lets start with some real numbers 2^2 = 2^2 ok so this is same as saying 4 = 4, think about

2^2 = 2^x so I read it like this, 2 to the power of 2 is = 2 to the power of some number, the only correct value for x is 2 so we can say that x=2, if we had 2^2 = 2^(3x) it takes one more step since we have a common base of 2, we can just say 2 = 3x so x = 2/3, this is same as saying 2^2 = 2^((3*2)/3) so 2^2 = 2^(6/3) so 2^2 = 2^2, long story short if you have common base being (a) in this example a^y = a^x, we can say y=x. if your still confused investigate exponents and logs. logs are this inverse of exponents, like division is the inverse of multiplication.(5 votes)

- according to this if x is either 3 0r 4 the given equation will always be = 24 and there would be no solution that will make it > 24 . how is that possible(3 votes)
- "according to this if x is either 3 0r 4 the given equation will always be = 24"

That is true, but...

"and there would be no solution that will make it > 24 ."

does not follow from that. The expression -x(2x-14) is greater than 24 for all values of x between 3 and 4.(2 votes)

- How would you solve x^2 + 4x - 7 >or equal to 5 considering there is already a -7 there?(2 votes)
- Original: x^2 + 4x - 7 >= 5 (Just copying the problem here)

Step 1: x^2 + 4x - 7 - 5 >= 0 (Move the 5 to the other side so you have 0 on the other side)

Step 2: x^2 + 4x - 12 >= 0 (Subtract the 5 from -7)

Step 3: (x+6)(x-2) >= 0 (Factor! Use FOIL or whatever method you prefer)

Step 4: Find the solutions to the equation, I'd advise you to sketch the graph out to solve it. You know that you want the part above the x axis so x<= -6 or x>= 2.

Hope this helped:)(3 votes)

- Is it correct if I factor the -2? Like:

-2(x^2-7x+12)>=0

-2(x-3)(x-4)>=0(3 votes)- You don't need to do it. Once you factor it by taking it out of the bracket, you can just divide both sides and flip the sign(1 vote)

## Video transcript

We've got the inequality negative x times
the expression 2x minus 14 is greater than or equal to
24. So I encar, encourage you to pause this video now and think about what the
solution set to this inequality would actually be,
and actually plot the solution set on a number
line. So I'm assuming you've given a go at it, so let's just try to simplify this a
little bit. So on the left hand side we could
distribute, we could distribute this negative x, and so if we
did that we would get negative 2, negative 2x squared,
negative times a negative is a positive, plus 14x is greater than or
equal to 24. Now I'm gonna put the, I'm gonna subtract
24 from both sides, just so that we just have a 0 here and then we can think about
factoring what we have here on the left. So we have negative 2x squared, plus 14x,
I'm gonna subtract 24 from both sides, so it's minus 24 is
greater than or equal to, I've subtracted 24 from the
right as well, so that's going to be greater than or
equal to 0. Now, I don't like, I don't like having
this negative 2 out front, so what I wanna do is I wanna divide this
left hand side by negative 2, but I can't just divide the left hand
side only by negative 2, I have to do divide the right hand side by
negative 2 as well. And anytime I multiply or divide both
sides of an inequality by a negative number, it's
going to flip the inequality. So if I divide both sides by negative 2,
I'm going to be left with x squared, positive x squared
minus, so I'm dividing by negative 2, so minus 7x, plus
12. And now since I divided by negative 2, I'm
gonna flip this inequality, is less than or equal to zero divided by
negative 2 is 0. So that simplified things a good bit, and
now let's see if we can factor this quadratic
expression. So two numbers whose product is positive
12, so that means they're gonna have the same sign, and whose, and
whose sum is negative 7. So if they have the same sign, and their
sum is negative 7, that tells us that they're
both going to be negative. And let's see, negative 3 and negative 4
seem to fit the bill. Their product is positive 12. Their sum is negative 7. So we could write this as x minus 3 times
x minus 4, is going to be less than or equal
to 0. So now this is the point that we're gonna
do a little bit of interesting logic. If the product of two things is less than,
is less than or equal to 0, what does that tell us, tell what do
we know about it? Well, that tells us that either, either
one or both of them is 0, or they have different
signs. The only way that you're gonna get less
than 0 is if one is positive and the other is negative, or one
is negative and the other is positive. So our 1 is negative, is that they have
different signs. So let's write that down, let's write that
down. So either, either x minus 3, x minus 3 is
less than or, less than, either x minus 3 is less
than or equal to 0 and, and x minus 4 is greater than or
equal to 0, x minus 4 is greater than or equal to 0, so
notice, note this one is non positive, this one is non
negative, they're either equal to 0 or they are have
different signs. So that's one situation, or, or the other
way around, or x minus 3 is non negative, it's greater than
or equal to 0, and x minus 4 is non positive, x minus 4 is
less than or equal to 0. Once again, they're either zero or
different signs, that's all I'm doing with this little with this little
logic work right over here. So what are these, what is this, what is
this simplified to? So x minus 3 less than or equal to 0, add
3 to both sides, you get x is less than or
equal to 3. And, and x minus 4 is greater than or
equal to 0. If you add 4 to both sides of this, you
get x is greater than or equal to 4. So what values of x are going to be less
than or equal to 3, and greater than or equal
to 4? Well anything that's less than or equal to
3 is not going to be greater than or equal to 4 and anything that's greater than or equal
to 4 is not going to be less than or equal to 3. So there's no x, there's no x value that
can satisfy this situation right over here, there's no
X value that will result in this one being negative, and
this one being, or this one being non positive and this one
being non negative. So let's go to this one right over here. So if we add 3 to both sides, we get x is
greater than or equal to 3 and, and we get adding 4 to both sides, x
is less than or equal to 4. Now does this make sense, that something
could be greater than or equal to 3 and less than or equal to 4? Sure, for example, well 3 is greater than
or equal to 3 and is less than or equal to 4, 4 is greater than or equal to 3 and it's less than or equal to 4 and anything in
between. So we can plot the solution set here. So this is actually all that matters. Cuz this one, there's no situation which
that would've been true. So this is the only thing, this is the only thing that's gonna make this or part
true. This part is always going to be false. So if we wanted to make the solution set,
it would look something like this. So if this is our possible values of x. So let's say that that is 0, so this is 1,
2, 3, and 4, 3 and 4, 3 and 4, x could be
greater than or equal to 3, so it's greater than or
equal to 3. But it's also let, it also has to be less
than or equal to 4 so we can't just go all past
4. Also, less than or equal, less is since
it's, less than or equal we can color in these dots, less
than or equal to 4. So anything in this range, including 3 and
4, that's why we circled in the dots, this would satisfy
this equation here. And if you wanted to think about it
visually, hey, you know we know that this type of thing,
we know that this type of thing, or this type of
thing, this type of thing right over here, these are,
these are parabola's. So how would that relate to this little
solution set that we just thought about right over
here? Well if you look at let's, let's just look
at one of these. Let's say we went to, let's say we went to
this form right over here. So all, everything we did, this is just
another way of thinking about negative 2x squared plus 14x minus 24 is
greater than or equal to 0. So this right over here, we have a
negative coefficient on the x squared term, that's going to be a
downward opening parabola. So when is that greater than or equal to
0? So if, if we though about the downward opening parabola, it could might
look something like this, it might look something like this if we're now thinking in 2
dimensions. And this is the, this is the, if you think
of this as the Y axis right over here. So when is that greater than or equal to
0? Well it's greater than or equal to 0, it's
above the X axis in this range for x right over
here. So that's one way of thinking about it. If we though about it from the point of view, not of that parabola, not of that
parabola, but this parabola right over here, when is x
squared minus 7x plus 12 less than or equal to 0? Well this is gonna be an up, upward
opening parabola. So if it has a positive coefficient here,
so this problem might look something like this, might look something like this, when
is it less than or equal to 0? Well once again, once again it's less than
or equal to 0 in that same range.