Algebra (all content)
- Simplifying square-root expressions: no variables
- Simplifying square roots of fractions
- Simplifying rational exponent expressions: mixed exponents and radicals
- Simplifying square-root expressions: no variables (advanced)
- Intro to rationalizing the denominator
- Worked example: rationalizing the denominator
- Simplifying radical expressions (addition)
- Simplifying radical expressions (subtraction)
- Simplifying radical expressions: two variables
- Simplifying radical expressions: three variables
- Simplifying hairy expression with fractional exponents
Sal simplifies elaborate expressions with square roots. For example, he simplifies (4√20-3√45)/√35 as -√(1/7).
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- Did Sal make an error at2:25, where there was a negative sign before the -sqrt5/sqrt35, but at2:36he says the simplified version is simply a positive sqrt1/sqrt7?(47 votes)
- Yes he did! Good eye. He forgot to bring the negative over when we was simplifying! Oops! There should be a little bubble that pops up in the bottom corner that shows the correction though.
The correct answer should be -√(1/7)(41 votes)
- in the first example, at2:11you have simplified it to -sqrt5/sqrt35, the proceed to rewrite this as -sqrt(5/35) and end with the result of -sqrt(1/7).
When solving this on my own, after i reached the step you were at at2:11, I simplified it to:
-sqrt5 / (sqrt7)(sqrt5) and cancelled the sqrt5 to end with -1/sqrt(7).
I believe these are both correct, as your example of the whole fraction under the radical could simplify to my form, however which would be considered the most simple answer?(11 votes)
- They are both equivalent. It depends on how your instructor wants the answer to be really. Both are simple enough. Later on you'll learn to rationalize the denominator, so you'll most likely require to do that. You simply multiply - 1/√(7) by √(7) / √(7) which will give -√(7)/7.
P.S. He mas an error, forgetting to include his negative sign.(7 votes)
- Does anybody else find they can solve these and other problems but don't feel like they are fully understanding them?(5 votes)
- I think it's easy with maths to end up doing the manipulations, without really getting why the manipulations work. It's not satisfying that way (at least I find), as it feels like just going through the motions.
Something that helped me is when I realised that so much of the manipulations Sal is doing here are because of the the distributive property
( https://www.khanacademy.org/math/pre-algebra/pre-algebra-arith-prop/pre-algebra-ditributive-property/a/distributive-property-explained )
I have found that by thinking about that property and it's implications, and really understanding it, a lot of other stuff then fell into place.(5 votes)
- So, did Mr. Khan drop the negative sign when he simplified from -sqrt(5/35) to sqrt(1/7)?(5 votes)
- Yes, he dropped the minus sign. And, there was a correction box that pops up at about2:35in the video that said the answer should be - sqrt(1/7)(4 votes)
- Wait at2:24, the answer is actually -sqrt1/-sqrt7?(3 votes)
- The problem with that is a negative divided by a negative is a positive, so by doing this, you will improperly eliminate the negative sign(1 vote)
- when are you supposed to square root, simply divide, or simply multiply? I cant figure this out and keep getting my answers wrong. Iv'e been stuck on the practice problems for 2 hours now!(2 votes)
- When this happens to me, I use the hints in each problem. Walkthru the hints one by one. Compare what was done in the hint with what you did. If you did something different, chances are you have found your mistake. Try to learn the technique shown in the hint. Rework the problem from that point forward on your own. See if you can get it correct. If it is still wrong, go back to the hints. Step thru again. You may go thru more hints this time but will likely find another step that you didn't handle correctly. Use the hint as a tutorial to learn from your mistakes.(2 votes)
- isn't te awnser to the first question negative -sqrt(1/7) not sqrt(1/7)? at2:14-2:39(2 votes)
- Can any Genius? anyone rationalize this for me 1/√7 + 1/√2??(1 vote)
- Do each fraction individually.
1/√7 (√7/√7) = √7/7
1/√2 (√2/√2) = √2/2
So, your expression is now: √7/7 + √2/2
To add, convert to LCD of 14
√7/7 (2/2) + √2/2 (7/7) = 2√7/14 + 7√2/14 = (2√7+7√2)/14
Hope this helps.(2 votes)
- In what instance would the expression be incapable of being simplified? I don't see a video that explains the circumstances that lead up to this.... could someone explain this to me?(1 vote)
- One example would be the "empty set." Math does not allow any number to be divided by "zero."(1 vote)
- [Voiceover] We're asked to simplify the expression by removing all factors that are perfect squares from inside the radicals and combining the terms. So, let's see if we can do it and pause the video and give a-go at it before we do it together. Alright, so let's see how we can re-write these radicals. So, four times the square root of 20. Well, thats the same thing as four times the square root of four, times the square root of five, because 20 is the same thing as four times five. And 45, that's the same thing as nine times five. And the reason why I'm thinking about the four and I'm thinking about the nine is because those are perfect squares, so I could write this as, four times the square root of four times the square root of five. And then I could say, this part right over here is minus three, times the square root of nine, times the square root of five. The square root of 45 is the same thing as the square root of nine times five, which is the same thing as square root of nine times the square root of five, and then all of that is going to be over, all of that is over the square root of 35. Now, are there any perfect squares in 35? 35 is seven times five. No, neither of those are perfect squares. So, I could just leave that as square root of 35. And, let's see, the square root of four? Well, thats going to be two. This is the principal root, so we're thinking about the positive square root. The square root of nine is three. And so, this part right over is going to be four times two, times the square root of five, so it's going to be eight square roots of five. And then, this part over here is going to be minus three, times three, times the square root of five. So, minus nine square roots of five. And all of that is going to be over the square root of 35. Square root of 35. And, so let's see, if I have eight of something, and I subtract nine of that something, I'm gonna have negative one of that something. So, I could say negative one times the square root of five, or I could just say negative square root of five. Negative square root of five over the square root of 35. I actually think I could simplify this even more, because this is the same thing. This is equal to the negative of the square root of five over 35. Both the numerator and the denominator are divisible by five. So, we could divide them both by five, and we would get the square root of divide the numerator by five, you get one. Divide the denominator by five, you get seven. So, we can view this as the square root of 1/7th. Square root of 1/7th. And we are all done. Let's do another one of these. These are strangely, strangely fun. And once again, pause it, and see if you can work it out on your own. Perform the indicated operations. Alright, so let's first multiply. So, this essentially is doing the distributive property twice. And, actually let me just do it that way, so let's distribute the square root of five, plus the square root of six. Let's first multiply it times the square root of five. So, the square root of five times the square root of five is going to be five. Square root of five times the square root of six is the square root of 30. So, five plus the square root of 30. And then, when I take this expression, and I multiply it times the second term, times the negative square root of six. Well, negative square root of six times the square root of five is going to be the negative of the square root of 30. And then the negative of the square root of six times the square root of six is going to be, we're gonna subtract six. The square root of six times the square root of six is six, then we have the negative out there. And so, just like that, we are left with, well, let's see. Square root of 30 minus the square root of 30, well those cancel out, thats zero. And we're left with five minus six, which is going to be equal to negative one. And we are, we're, we're all done. Now, another way that you could've viewed this, is you could've seen a pattern here. You could've said, Well, this is the same thing as A plus B, times A minus B, or A is square root of five and B is square root of six, and we know that this will result in the difference of squares. This is will be A squared minus B squared. And, so this for this particular case it would be square root of five squared, minus square root of six squared. Which is of course equal to five minus six, which is equal to negative one. Either way, hopefully you found that vaguely entertaining.