Simplifying Expressions With Roots and Exponents
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- Опубліковано 20 лип 2024
- Now we have all these pesky terms with exponents and roots, and they come in all these complicated looking expressions. Square this, cube root that, whatever will we do? Don't worry, it's usually pretty easy to simplify these until they get into a more palatable form, and it's quite satisfying to get them there! Let's learn all the tricks.
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Damn, I feel like following the Mathematics (All of it) playlist, this one took a jump in complexity!
I started scratching my hair lol.
@@jerrytom8835 same
@@jerrytom8835 omg same holly f
he left out the in-between content regarding the concept and operation of exponents, so when he solves a mixed problem involved both exponents and roots, some of us will feel confused
Compare this 5-10 minute lesson to 1 hour lesson in school
This video took a curve with the my learning progress, but I will never leave this video until I fully grasp the concept.
I was doing fine up until this point...
same here I'm lost with this one
I'm also stuck. Weird to think this is 8th grade level mathematics and I'm STILL struggling with it at age 21. I only got half right on the comprehension check.
@@Canis_Fatalis Try to watch it again but each example analyze how he did it
I've been able to solve it all after an hour 😅
😂
🤣 You’ll get used to it.
The difficulty jump is wild. Yet, not unwanted. I know with time and repetition i'll be able to simplify with ease!
Damn, professor Dave aint fucking around this time....
Explanations of the answers:
1. (2x^2 y^4 z^5) ^3 -- As you recall, when cubing elements with exponent we simply multiply. So 2 cubed is 8, x^2 cubed is x^6, y^4 cubed is y^12, and z^5 cubed is z^15.
2. Remember that because the exponent is negative, we get rid of it by inverting the fraction. So now we have (x^2y over xy^2) squared. We multiply the exponents, yielding X^4 Y^2 over X^2 Y^4. That cancels to give us X^2 over Y^2.
3. Let's substitute Y=3, chosen at random. X^ (6+1) over X^(6-1) gives us X^7 over X^5. Cancel out, and we're left with X^2.
4. Good golly Dave, way to crank up the difficulty! This is just mean.
Okay, remember that the "2" in the exponent is also a 2/1. So now we can multiply the exponents. X^1/3 by 2/1 gives X^2/3. The other term, X^2/3s multiplied by 2/1 gives us X^4/3s. So now we have X^2/3 times X^4/3 divided by X^3/2.
Now we go ahead and multiply X^2/3 by X^4/3, which means we add the numerators, giving us X^6/3. (which is the same as X-squared, but in this case we'll convert it to X^4/2, a common denominator.) Now when we divide X^4/2 by X^3/2, we subtract the numerators, yielding X^1/2. Whew!
(At least I think that's how it works. If I've come to the right answers by erroneous methods, by all means show us your method.)
Good detailed explanations!
Another way for option 4 would to do this (the way I did it):
(x^1/3 * x^2/3)^2 = (x^1/3 * x^2/3) (x^1/3 * x^2/3)
Since we are multiplying exponents, we add them. So in that case, it's just adding fractions:
(x^1)(x^1)
Anything to the power of 1 is itself:
(x)(x)
This is the same as x to the power of 2:
x^2
That's the same as x^2/1:
x^2/1 divided by x^3/2
When dividing exponents, you have to subtract. So we are substracting fractions, and to do this, we have to change the equation to where both have the same denominator (LCM):
x^4/2 divided by x^3/2
Subtract fractions:
x^1/2
Also note that your method is not bad, and actually I like yours better. I simply stated this as an alternative method to anybody who may come across your comment.
how to know if we need to convert and how to convert on situations like this please explain and also the conversion of the 4th question x raised to 6/3
Multiplying Two Different Fractions Means:
Adding Those Fraction's Numerators , For Addition , we must first we assured that their denominators are common (or same) if they are not we'll have to find out Least common multiple (LCM) of denominators .
After finding LCM we'll have to mutiply both denominators with that factor that will make the value of denominator equal to LCM
From then on we can continue the addition of numerators, 'cause now the denominators are equal.
But in this case their denominators are already common so we just have to add their numerators which gives us : x^6/3
I hope that clear up your confusion!
for the second one, how come we have to multiply the exponents? Is there a symbol or exponent? I thought we got rid of the -2 so we can invert the fraction. What is multiplying it now?
Nevermind, just rewatched the video. You don't get rid of the -2. You actually just invert it's operation. Thanks for the help anyways
man, i really love your videos! i used to hate math but grew to love the subject because your videos made me realize that math isn't that bad at all. thanks a lot, professor dave!
This was a tuff vid, but just a few more replaying this video. I got it now thanks!!
On 3x^2√32y^2 - 4y √18x^4, they cut the 32 in half to take its root, and they leave the √2 since you cant do anything with it. They combine 3x • 4y because mutliplication is associative (doesnt matter what order you multiply), so it is valid, and left us with 12x^2y√2
Same thing on the other side.
It took me a few days to understand this, and I don't want other people to face the same issue so hopefully this helps.
Ty
Bro it's so complex but u made it easy asf!! I really love your videos and still go on w learning mathematics with you!!
I'm 26years old, looking forward to completing the playlist to get prepared for engineering education that I'm willing to start soon. This playlist been perfect so far. I'd only wish if there were more comprehension questions in the video or linked to it. Thanks a lot prof!
28 y/o here, about to start undergraduate studies in pharmaceutical sciences. It's never too late to start! All the best for your engineering courses.
I am 14
@@letzte_maahsname how's it going
@@Danny-df4pokeep up the good work 💪
Literally me
hello, question for you while I am understanding for the most part how you solve for example the 64/x6 2/3 square. but i never really understood what it would be used for or the many variations for solving equations?
This can be used for bulky equations with tons of stuff on both sides to simplify it to something like x^4=81
For practice question #4 is my method right? Pls correct me if not.
x^6/3 divided by x^3/2 but we should make it have a common denominator first...
so: 6/3 x 1/2 and 3/2 x 1/3 which will give us 6/6 - 3/6 or simply 1-1/2
so: x^1/2 or square root is the answer.
Explanation of the third exercise:
So we have x^(2y+1)/x^(2y-1)
The first term can also be written as x^2y*x^1
The second one as : x^2y*x-1
So we have : x^(2y+1)-(2y-1) taking into consideration that when we have minus before a parathesis we multiply the terms by -1 therefore the +2y cancels with -2y and because -1 became +1 we remain with +1+1 => x^2
The last one it s even essier if you dont do the mistake i did and start as you should by calculating the parathesis, by doing so you will have x^2/x^3/2 which equals x^1/2
i had problem only with the third and this answer explanation to me everything, ty
Hi again Dave, sorry about my previous comment. I strugled a little bit more and found what I had made wrong. Now I get the same result as you. So, it must be right to flipp the expressin inside the brackets. And thanks again for your exelent lessons. / Ingvar
I'm confused, in the last example you took the square root of 16y² and got 4y. Couldn't it be -4y as well since an even root has two possible answers? Same goes for the square root of 9x^4, in the same example, which became 3x²; couldn't it also be -3x²?
When taking the square root, we only consider the positive case, and when solving equations (like x^2=16) we consider both positive and negative. If you would like an explanation for why we do this, I made a comment about it in the previous video in the playlist :)
I saw Leonardo Da Vinci's biography and came to know how a person can be jack of all trades and here I came to learn some maths after getting inspired by Da Vinci!
Professor Dave is making my life much easier. Thank u very much ❤️
Can you please explain the second problem? I don't get it.
@profesor dave explains could you help me out on question 3? I can either get (2y+1)/(2y-1) or 0 not x²
it's because you subtract them, (2y+1)-(2y-1) and then the - in the parrenthesies flip into + because - times - equals +, so it's 2y+1-2y+1, 2y cancels and you're left with 1+1 equals 2
In no.4, where did you get the 4/2?
If you flip the fraction then the exponents change symbol like from negative to positive right?
Yeah
In 6:53, you can turn the root (9•2)x⁴ into 3x² because when you combine the two, you need to take the square root of them to get them outside the square root
If that makes sense
Short answer for last question. it might be wrong
top side we sum the powers in paranthesis 1/3+2/3=1
x^1=x
so we have x in paranthesis
x^2 we have at top
down side has 3/2 divider is 2 so we have to make the upside same like x^2=x^4/2
now we drag the down side 3/2 to upside as negative
4/2-3/2 = 1/2
so we are left with x^1/2
when i try to explain it takes some time but its short tbh
@ 4:33 Why do you flip the fraction to make the exponent positive?
4:15
in check comprehension, in the second in the first line i got x^2y^-2, but if you change the y^-2 in 1/y^2, and then you do x^2/1 times 1/y^2, the result is the same. in the last, i just wrote x^1/2 like square root of x, so is the same
Hi I'm new and I'm learning I'm trying to catch up of my school gade ty Dave for all of ur mathematics videos ^v^
In the last example problem, what happened to the ys?
U teach better than my school teacher.
U are the beat professor Dave!
Hello Prof! I'm afraid I didn't get the (second?) question (xy / x2y)-2 . The answer I keep getting is x2.
I flipped it since it's an inverse squared operation. (x2y/xy)2
I squared the nominator and denominator (x2y)2 / (xy)2 to get x4y2 / x2y2.
Since Y2 is common, it's taken off to leave x2
Please how did you arrive at x2/y2 ?
How did I miss the squared on letter y of the numerator of the question. Lol
I got them all!
((xy2)÷(x2y)) -2 = (y÷x) 2 ( we get rid of the x above and the y in the base then we can flip the nominator and the denominator to get (x÷y) 2 which gives us x2÷y2 as the final result
y^2 is not common, you square the y^2 to get y^4, after which you divide to get (x^2*y^-2). inverse square to get x^2*((1/y)^2) then replace the 1 with x^2 to get x^2 / y^2
Thanks Dave
Great explanations.. I am just really wishing you had a link to how you got the answers to the comprehension questions at the end of all of your videos..
most of them show the work
@@ProfessorDaveExplains I'm up to lesson 35 and didn't see any workings for any of the comprehension questions yet.
@@ProfessorDaveExplains Hello, David. Thank you for the lesson.
I would like to clarify one point:
in the part (9x^5•y^5) / x^6, we divide x^5 by x^6 and get x^-1 and this is the same as 1\x. But then why do we still have 9 there, and not 10?
After all, we have 9 + 1 remaining from the reduction operation, don't we?
@@user-ox3bp4rz3n well no because that 1 is part of the exponent
@@user-ox3bp4rz3n It's because 1 is multiplying, it is not adding. When it was x^-1 it was multiplying 9, so when it became 1/x, 1 was still multiplying, NOT adding, so 9 times 1 is still 9, not 10 :)
This is extremely hard. especially because im 13 years old. but i went over this video with my dad and i then watched the upcoming videos and im starting to get the hang of it. it was tough but it payed of. best of luck.
because this is high school tier math :) you will get there one day bud. take it easy and good luck with your studies!
@@teos4664 Thank. It got way easier. Good luck to you too.
can some one please explain the third practice question?
In the compression x ^ 1/2 can be simplified even further to square root of x which is which is just X is the simplest form
Yes that is true
wow thank you a lot, how can i give my thanks to appreciate your guidance
Hey how come in the previous example 5a[(2x+y)-(x+y)] you got 5a^x by elimination.. but in the comprehension exercise 2 x^(2y+1)/x^(2y-1) which is x[(2y+1)-(2y-1)] and you got X^2 . .. should the 2y and 1 have been eliminated resulting in x^0? I keep getting x^0. Not sure if this was answered i didnt check the comment thoroughly. Thanks
I got it... when u distribute the negative sign it became 2y+1 . Making it x^2 .
Hi David . Thanks for the videos. Can I just ask, at 3:31 you multiply the numerators x^3 and x^2 and you write x^5 and not x^6.
Why would this be?
Thanks once again
Well x^3 is x x x, and x^2 is x x, so that's five all together, or x^5. You only multiply the exponents when raising one to the other.
@@ProfessorDaveExplains Thanks :)
6:19
but how did it work if by the denominator just can not be 0?
At 6:38 why is the left over root 2 not root 2y? Isn't 16x2 in the brackets equal to 16y x 2y? So when you remove the 4y should there still be a y left
but the square root of y^2 is y, so you take y^2 completely out of the root
@@ProfessorDaveExplains ohhhhhhhhhhhh of course. Thanks so much for replying by the way, you're amazing. I'm currently studying for the GAMSAT (MCAT for Australia and the UK) and your channel has helped me so much (ive been watching heaps of your chem videos too). I'm currently doing my masters in medical research, and so I have a pretty good grasp of science as a whole, but my mathematics is so lacking it's embarrassing. There's so many little rules like this that accumulate into a solid understanding of mathematics (regretting not trying harder in highschool every day) and unfortunately it doesn't seem like there are an shortcuts. Would you have any recommendations/sources, apart from going topic by topic, where someone a little further on in their careers can go and practice the basics? (Really fishing for shortcuts here -_-)
well i really tried my best to provide an ultra-comprehensive math course that covers all the basics in addition to the advanced topics, so i would just head into my mathematics playlist and everything you need should be in there! good luck :)
@@ProfessorDaveExplains Thanks so much for your reply again. And I think this is the first maths one I've watched so I will definitely continue.
I've been using tutorial videos to study for many years now and honestly I've only found your channel recently. At the moment I actually prefer your stuff over Khan academy or others because you start simple (for people like me) but still extend the concepts to their complexity which is where you actually start realistically applying them, and your comprehension section at the end is great. Also your intro jingle is dope.
Thanks again dave, i honestly can't express how much people including myself appreciate the work people like you do. Science education is so important!
why does the negative sign distribute at 6:06 ?
I finally get it thanks!
Explanation of 2nd and 4th one:
2nd ques-
(xy2/x2y)^-2
= (x2y/xy2)^2 (becoz there is -2 whole power ,so to make it positive we have to reverse it , as x-1=1/x)
=x4y2/x2y4 (now cut x2/x4 and y2/y4)
=>x2/y2 final ans.
4th ques:
(X⅓ . X⅔)^2\X³/²
=(X⅓×2 . X⅔×2)/X³/² (solve power individually )
=(X⅔. X⁴/³)/X³/² (now add numerator powers 2/3+4/3)
=X⁶/³ divided by X³/² ( now cut 6/3)
=X2-³/² =X²/¹ --- ³/²
=X4-3/2 => X1/2 (final ans)
Thank you for listening my Explanation 💜
Can you explain #3 for me. I feel like #2 was easier to understand for my brain lol
Where can i find exercise to practice?
Problem at mark of 4:07 the flipping over i don't get it
The answer of question 4 is 1/x? right?
6:36 that's is y² how you put only y. The y is common for both 4 and √2.
the square root of y^2 is y. since y * y = y^2. So he basically took the two factors 16 and y^2 out of the square root which became 4y
y is also inside the square root
Edit: read my reply since that's confused
Yeah, this hard to do, but it'll be easy to forget since a lot of this stuff I'd be doing without no explanation. Like the flipping the fraction to flip the negative exponent, then having to square both the top and bottom. I was hoping to do the general chemistry playlist just from the information in these videos but I can tell that won't be an option
For anyone else that's confused, there are a few videos and websites online that explains this very nicely. The best thing in my opinion is just to revisit fractions and exponents, since there's no harm in that. But if you don't want to, look up reciprocals. Also look the rules of exponents, a really good site for it is called mathematics monsters. Thinking about it now, I'm sure Dave covered all of this in his arithmetic playlist so check if he did first.
Hi Dave, your lessons are realy interesting. But, now I have got a problem. In the comprehension at the end of the lesson I solved the problem your way and came to the same result as you. However I was a little questining about the flippening of the number inside the brackets and change of the exponent. So, I put in real number in the expression and solved the problem the simpiest way. But that did not give me the same answer as I got when I used your "flipping". I put in X = 3 and Y = 9 and got 15.52 / 9 or 1.732 with mine way. Your way I got 9/81 or 0.111. Quite a differens. Are you sure your way of doing it is correct?
I just put it in the calculator, and both ways should be 1/9 or 0.1 repeated. So don't worry, his way is correct.
Algebra is about using variables (letters), so when we change those variables to random numbers with no justification at all, it won't always be right, that's why it didn't gave you the correct answer. I recommend you only using letters in these scenarios :)
Why no PEMDAS at 2:48? I still got the right result using PEMDAS, but what's the point of doing it the way it's done in the video?
In one of his videos he explains how distributive property works. It won't matter if you distribute first and then do inside the parentheses, or do the partentheses and then distribute.
For example:
5(2+3)
You can do PEMDAS:
5(5)
25
Or you can distribute:
10+15
25
Both ways are valid methods, it's simply up to one's preferences.
can someone explain why x^(2y+1) / x^(2y-1)= x^2 ?
because since they both have the base as X so we will minus the exponents like this
[ ( 2y+1) - (2y-1) ] = 2y + 1 - 2y + 1 = 2 then it would be x^2
Sure, let's break it down.
So, you have something that looks like this: ((x^(1 part) * x^(2 parts))^2)/x^(3 parts).
Let's start with (x^(1 part) * x^(2 parts)). That means x has 1 part plus 2 parts, which equals 3 parts or just x.
Now we square it (meaning we times it by itself) so now it looks like this: (x * x) or x^2.
In the denominator, we have x^(3 parts). That's like x, but times itself again, so we get 1.5 parts (x^1.5).
Finally, we subtract the bottom part from the top part. So, 2 parts (from x^2) - 1.5 parts (from x^1.5) equals 0.5 parts or x^0.5.
So at the end, you get x^0.5. And that's the answer! You can also say "x to the power of half" or "square root of x".
how did you get (x^6)^1/3?
(x⁶)^ (1/3)
=Cube root(x⁶)
=x^(6/3)
At 6:35 why we only took 4y why not 4y^2 ??
6:37 I'm a little confused by this part. I thought 3xsquared and 4y wouldn't be like terms since they're attached to different variables, like 9 apples and 4 oranges.
same here!
When you are doing multiplication (or addition), the order of operations doesn't matter. Whichever order you choose, you'll always get the same result. 2x3x4x5 is the same as 3x4x5x2 (both equal to 120). AxBxCxD is the same as BxCxDxA. And likewise 3X4Y is the same as 12XY, only the order changed.
@@ristaglista3378 where is another y there is y²
ty
Great series! I’ve always said « all squared » rather than « quantity squared » , but I am Australian...
We say "whole squared" here in South Asia.
You know its getting serious when he starts sounding like a quarterback changing the play
Mh, as others have said, this is becoming really difficult all of a sudden. I don't even think the exercises are that hard. I mean, I almost got them right, it's just that there's so many things I have to keep track of. I really don't think I'll ever be able to memorize those. Maybe I'll skip for now and hopefully come back and understand better once I'm further into the playlist. I'm in the last year of HS, all I want is to pass the final exams. Hopefully I can do it.
how is it now?
I GOT IT ALL RIGHT!
could anyone explain how did we get X1/2 in the comprehension?
Just take notes of all the techniques he uses and use them yourself one step at a time. I'll try to explain with words, but it's gonna look a bit messy. Anyway, this is how I did it and got X to the power of 1/2:
1) distribute the exponent in the upper fraction (leave bottom fraction as it is for the time being)
2) multiply the X's (just add their exponents together)
3) then you get ( *X* to the power of 6/3) / ( *X* to the power of 3/2)
4) since 6/3 = 2, you end up with *X* squared / ( *X* to the power of 3/2)
5) subtract the exponents from each other
You end up with *X* to the power of 1/2, which is the same as the root of *X* .
@@ristaglista3378 Thanks alot! but my problem is with distributing the first step , i can't really find an example in this video or the exponents one about squaring variables with fractioned exponents.
(x⅓)² =? and (x⅔)²= ? do we deal with fractioned exponents as normal fractions ? because when we multiply fractions we multiply the denominator and the numerator and it doesn't seem that you've done that!
the square of a number 2² is 2x2 but what about the square in this case ? ⅓ x ⅓ and ⅔ x ⅔ ?
ua-cam.com/video/lWAsS82pyYA/v-deo.html Got it!
thanks again.
@@ristaglista3378 , thanks a lot dude:^)(◕ᴗ◕✿)(◕ᴗ◕✿)w(°ｏ°)w♪┌|∵|┘♪
@@ristaglista3378 thanks to remember the addition of exponents
(X⅓ • X⅔)²/X³/²
= (X²/³ • X⁴/³)/X³/²
= (³√(X² • X⁴)/X³/²
= (³√X^6)/X³/²
= (³√X³ • X³)/X³/²
[the cubes cancel out the cube root]
= X • X/(√X³)
= X²/(√X² • X)
[the square cancels out the square root]
= X²/X • √X
[one X from X² cancels out with the X in the denominator]
= X/√X
= X/X½
[there are two halves in one whole]
= X½ • X½ / X½
[one term from the numerator cancels out with the denominator]
= X½
Thank.
sorry but why isnt the second answer x2 over y-2
Thanks for the lessons. I don't understand why at 3:36 x cube times x square is x to the 5th power not the 6th power. Please help.
☺️ I um found your video explaining. Thanks for the great videos!😁👍
Yeah Idk why this happened too
Question 4 in the comprehension was tricky but atleast i now know how to solve it
6:33 Really need a more in-depth explanation on this one because you went so fast, I just can't comprehend how a square root of (16•2)y² suddenly became a 4y square root 2, and somehow the y² became a single y?? Idk at all...
If you watch the Playlist in order I think the video before this one explains how to simplify square roots and cubes roots.
Basically the inverse operation of y^2 or y squared would be to get the squared root so instead of (16•2)y^2 you can do the square root of 16 and 2. This looks like /16 • /2... 4 is the square root of 16 so that's where the 4 comes from and /2 or square root of 2 comes from.
You leave the square root of 2 looking as is because if you solved for that number you will get a long repeating number that'll make the while equation look ugly. You want your equation to be simple aka simplified so that's why you dont solve for it. Anyways, now your equation should look something like 4y•/2. The reason the 4 gets a y and not the 2 is because you solved for y by using the square root! Because you can't get a whole number by solving for the square root of 2 aka /2 you leave it there alone, no y because that was the whole point of doing the square root. Remember the inverse operation of a variable squared is to find the square root and vice versa. I hope this helped but if it didn't, I'd recommend going back a video in the Playlist in understanding those basics of simplifying square roots.
@@itsdemonz9814 You solved a huge headache for me!
I couldn't figure out why the y^2 became a y and you reminded me that you are finding the square root of y^2 which is of course y.
Thank you so much!
@@MichaelLeightonsKarlyPilkboys no problem. These equations can get tricky. Glad I could help
Nice👍
I need to learn but with you I can’t
So you mean that u can't learn it with that teacher?!
can someone explain to me the ;ast problem i couldnt do it
I'd like to see the answer worked out too, please!
(X⅓ • X⅔)²/X³/²
= (X²/³ • X⁴/³)/X³/²
= (³√(X² • X⁴)/X³/²
= (³√X^6)/X³/²
= (³√X³ • X³)/X³/²
[the cubes cancel out the cube root]
= X • X/(√X³)
= X²/(√X² • X)
[the square cancels out the square root]
= X²/X • √X
[one X from X² cancels out with the X in the denominator]
= X/√X
= X/X½
[there are two halves in one whole]
= X½ • X½ / X½
[one term from the numerator cancels out with the denominator]
= X½
Im having trouble figuring out how CQ#3 is x²..... I just got x.
Where did i go wrong?
same here. It contradicts the previous example he had. Which is subtracting the elements.
How did (3√x^6)^2 become (x6)^1/3?
Cube root (^3√) is the same as saying ^1/3
^2√ = ^1/2
^4√ = ^1/4
^3√2^2 = 2^2/3
^5√3^6 = 3^6/5
The exponent outside is the denominator, and the exponent inside is the numerator.
Do you see the pattern?
One thing to notice if some people didn't get it: the whole expression in parenthesis changes sign if a minus is in front of it.
it took me 60 min to complete this video, but now I know Kong-fu :)
Can anyone explain me the last comprehension
yeah this one is like a lighting jump compared to the others lol
Hi Professor Dave Everything was going fine, but I am not able to undertand any single questions here, kindly make it easier
It's difficult for you because you didn't watch his other videos or you don't know the basics. It makes sense, and I commented explaining how it works.
Many of your videos say that they're private for some reason 🤔
patience, friend! i release one clip per day.
Now I remember why I hated math🥲.
I was starting to love it until I saw this, Then I remembered when I used to flip things all over the place and memories the steps and forget on the exam
Edit: after repeating I now understand but I'm sure I'll forget and do it in another way next time and mess everything up
Friend, your skills won't grow if you're preparing yourself to mess up. Stay diligent, yet easy on yourself. Mistakes are chances to learn, so appreciate them!
the videos were very clear before this on, but this one took a huge elevation in compleixty
I couldn't keep up with this one.
7:41 x square, please explain this.?
3rd problem
You take the denominator multiplied by numerator so "x ^ - (2y-1)" then the sum of exponents will be "2y + 1 - 2y + 1" the 2y goes out and only "1 + 1" is left.
A question: In the example 4x^3y (3xy^2 / 2x^3)^2, I would simplify what’s inside the parenthesis first, and then square it. The result is the same but I wonder whether that would always be the case. Anyone?
So long as there is only multiplication and division inside the parenthesis, it will make no difference. This is because exponentiation is nothing but repeated multiplication, and you can multiply and divide in any order you like. Watch out if there is addition and/or subraction inside the parenthesis though.
Possible spoiler!
Can we simplify question 4 even further by writing x^1/2=root of x?
Edit: Also, when simplifying algebraic expressions, do we strictly abide by the same rules of PEMDAS?
when evaluating algebraic equalities/inequalities, do we perform PEMDAS backwards?
It is a matter of choice. They are both equivalent statements, and the word "simplify" is a bit of a misnomer in this situation. Neither is form is simpler than the other, they are merely usefull for different situations.
Yes
This is the point in 7th grade where I realized math is getting harder
Brain-storming! btw loved it, got 3 right, the last one had a silly mistake.
I feel like a couple of rules could've been explained before this video like inverting the fraction when you're applying a negative exponent to it
All explained earlier in the series.
super....
What grade is this taught in?
Oh I dunno, somewhere between 8th and 10th I think.
@@ProfessorDaveExplains Kk
@@Samuelunar more like between 10th and 11th.
@@teos4664 I learned this in 7th, although I didn't understand it at the time
(a+b)^2 binom de newton
I'm not understanding how you got to the answer for the last one.
@@balaganeshkumar3318
taken from the comment section:
Rista Glista
for 5 måneder siden (redigeret)
Just take notes of all the techniques he uses and use them yourself one step at a time. I'll try to explain with words, but it's gonna look a bit messy. Anyway, this is how I did it and got X to the power of 1/2:
1) distribute the exponent in the upper fraction (leave bottom fraction as it is for the time being)
2) multiply the X's (just add their exponents together)
3) then you get ( X to the power of 6/3) / ( X to the power of 3/2)
4) since 6/3 = 2, you end up with X squared / ( X to the power of 3/2)
5) subtract the exponents from each other
You end up with X to the power of 1/2, which is the same as the root of X .
(X⅓ • X⅔)²/X³/²
= (X²/³ • X⁴/³)/X³/²
= (³√(X² • X⁴)/X³/²
= (³√X^6)/X³/²
= (³√X³ • X³)/X³/²
[the cubes cancel out the cube root]
= X • X/(√X³)
= X²/(√X² • X)
[the square cancels out the square root]
= X²/X • √X
[one X from X² cancels out with the X in the denominator]
= X/√X
= X/X½
[there are two halves in one whole]
= X½ • X½ / X½
[one term from the numerator cancels out with the denominator]
= X½
@@zunaslogic4128 you are overcomplicating it, this source explains it well quickmath.com/webMathematica3/quickmath/algebra/simplify/basic.jsp#c=solve&v1=%255Cfrac%257B%255Cleft%2528%255Cleft%2528x%255E%257B%255Cfrac%257B1%257D%257B3%257D%257D%255C%2520%255Ccdot%255C%2520x%255E%257B%255Cfrac%257B2%257D%257B3%257D%257D%255Cright%2529%255E2%255Cright%2529%257D%257B%255Cleft%2528x%255E%257B%255Cfrac%257B3%257D%257B2%257D%257D%255Cright%2529%257D
@@lynxica2133 eh. I'm slow at math haha
@@zunaslogic4128 nono its alright lol ur method works fine (in some circumstances your method works better) so good job on that! :D
(x^1/3 . x^2/3 ) 2 / x^3/2 = {x ^(1/3+2/3) }^2 / x^3/2 , {x^(3/3) }^2/x^3/2 , (x) ^2 / x^3/2 , x^(2 - 3/2) , x^ (4 -3/2) = x^1/2
Can someone explain me the last one please
(X⅓ • X⅔)²/X³/²
= (X²/³ • X⁴/³)/X³/²
= (³√(X² • X⁴)/X³/²
= (³√X^6)/X³/²
= (³√X³ • X³)/X³/²
[the cubes cancel out the cube root]
= X • X/(√X³)
= X²/(√X² • X)
[the square cancels out the square root]
= X²/X • √X
[one X from X² cancels out with the X in the denominator]
= X/√X
= X/X½
[there are two halves in one whole]
= X½ • X½ / X½
[one term from the numerator cancels out with the denominator]
= X½
@@zunaslogic4128 thank you very much
Oof, this was a meat grinder of a lesson
Someone explain 6:50 to me please.
Which part specifically don't you understand from that process? In √(9 • 2) x⁴, the reason why Prof. Dave broke up 18 like that is so that we can get the perfect square 9 and get the root of it which is 3, but we can also do the same thing to x⁴, its root being x². By doing this we leave only 2 inside the square root and we can combine like terms, leaving us with 12x²y √2.
Sorry for being late
1st. Let's focus on the Root (32y²). We can convert that to Root (16 • 2 • y²), that would equal to 4x y • Root(2), and now we just multiply 3x² by 4y Root(2). which is just 12x²y Root(2). Now we do 4y • Root(18x⁴) which is converted to Root(9 •2 •x⁴) which is 3x² • Root(2) now We multiply 4y by 3x² • Root(2) which is 12x²y • Root(2) now we subtract 12x²y • Root(2) - 12x²y • Root(2) which is equal to 0
@@James-hs1eqbro why he wrote y from y2 at 3rd step and x2 but it was x4??
Can someone kindly explain to me why I lack the braincells
cos
At this point I’m confused.. 5:55
I officially hate math forever now. Couldn't ask for a better teacher, and I still can't get this crap!
Goodbye forever, math. My sweet, sweet math. I'll miss you. NOT!
I've got a bit of an ego after acing the practice problems 😎
Me too haha ifeel like I’m nicola Tesla
in india quantity = whole square
❤❤❤🎉🎉😊😊😊