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## Algebra (all content)

Sal solves the following challenge: Given that a>0, b<0, and a/b>a*b, what further can we learn about the variables? Created by Sal Khan.

## Want to join the conversation?

• at why is b squared
• Jonathan,
a/b < ab You multiplied each side by b to get
a*b/b < ab*b and the b/b is 1 and disappears so you get
a < ab * b which is
a < a *b * b and b*b is b² so
a < ab²
That is where the b² came from.
Then when you divide both sides by a you get
1 < b²
• When you divide A out at do you always replace the empty side with 1?
• When you divide any variable out of a equation, you always replace the number or variable with 1. 1 is like the benchmark in math, it always takes the place of the number if the initial number is divided out.
• At , can you take the square root of both sides? I tried that and it gives tells me that 1<b and -1<b. 1<b cannot be possible due to the given constraints. That leaves us with -1<b which is not what Sal concluded. What did I do wrong?
• Yes, we can take the square root of both sides, we just have to be very careful.

First of all, we know that the square root function is strictly growing over the interval [0, ∞).
This means that since 𝑏² > 1, then √(𝑏²) > √1

And because √1 = 1, we can write √(𝑏²) > 1

√(𝑏²) is the positive number whose square is equal to 𝑏².
Since 𝑏 is negative then −𝑏 is positive, and (−𝑏)² = 𝑏².
Thus, √(𝑏²) = −𝑏, which gives us the inequality
−𝑏 > 1

Multiplying both sides by −1 we get
𝑏 < −1
• At Sal changes the sign of the inequality saying that this is necessary, since we multiply both sides by a negative. Could anyone explain to me why this is the case? Like why do you need to change the sign and how do we know it, except from learning this as a rule?
• This is a often asked question in multiple videos.
So first way to think about it is finding values. If -x ≥ 3, put in values. If x=3, -3≥3 is false as would be any positive number. Try 0, 0≥3 still false. Try -3 and get -(-3)≥ 3 so 3≥3, finally true. The more negative we go, it would always stay true, so x ≤ -3.
Second way to show it, if -x ≥ 3, add x to both sides and subtract 3 on both sides to get -3 ≥ x, then flip the equation around to get x ≤ -3.
• Wouldn't the absolute value of 'b' also have to be greater than 'a' as the fraction -a/b is less than the multiplication product -(a*b)?
• No, you don't need the constraint that | b | > a.
Consider these examples.
a = 5 and b = -3. So "a" is larger than |-3|
5/(-3) is > 5(-3)
The key here is that dividing 2 integers will create a smaller negative number than multiplying them.
Hope this helps.
• I completely follow the tutorials, they make perfect sense when I watch him manipulate the expressions, but as soon as I try to do the practice problems my brain shuts off.
• at how can b>1 when at the top of the screen it says b<0
• It can't.
And in the 30 seconds after , Sal explains that b>1 cannot be part of the solution because b<0 is a restriction.
• Since we know b is negative, why don't we reverse the inequality when multiplying a/b * b?