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## Algebra (all content)

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

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

## Want to join the conversation?

• What would the "one positive, one negative" rule from be like with an equation that goes up to x^3? Or x^4 or above? • 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.
• 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 at or so. One expression of the inequality is an upward opening parabola, but the other opens downward. • 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.
• 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? • 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. • 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.
• - 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 • 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!
• I didn´t understand why Sal changes the simbol of the inequality. Would be wrong if we don´t change the symbol? • 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? • 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.
• 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 • How would you solve x^2 + 4x - 7 >or equal to 5 considering there is already a -7 there? • 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:) 