You Sir, are like the Bob Ross of mathematics -- soothing away confusions in an amiable manner...wish I had you as a maths tutor in my schooling years! 😌
Me too. just turned 66 a few months ago. Math has been the bain of my life and it makes me mad. I love science but when it runs into math, I collapse. Being so poor in math has held me back.
(2√6)/3 What's under the radical simplifies quite easily to √(8/3) or √8/√3. This simplifies to (2√2)/√3. Can't have a radical in the denominator, though, so we multiply by √3/√3 and get (2√6)/3.
Math problems that ask you to 'simplify' always remind me of a math teacher I had who ruled the class by sarcasm. I was sent to the board to solve a 'simplify' problem in front of the class, but instead of the thing getting smaller the way it was supposed to, it just got bigger and bigger. After 5 minutes of struggle, the teacher's voice resounded from the back of the class: "READ the question, Carvill. It says SIMPLIFY, not COMPLICATE!". By mutual agreement I was demoted to 'pass' math by the end of the week, so in a way I had managed to 'simplify' two people's problems: mine and the teacher's.
My friend was making Ds in post graduate classes. I imagined shooting arrows into an A target. My friend asked me to help her learn how to choose multiple choice answers. We talked 3 hours over the phone every night. I taught her that you have to learn that three of the answers may be partially correct. One will be completely correct. Finally she caught on. She made a B. I made an A. Of 1500 students who started the two year program. 500 made it into the second year. 30 got jobs. We both were hired. She was from Pakistan. She was taught how to give oral, and written essays in British schools. I knew the books orally, backwards and forwards by explaining all the answers, jumping back and forth through the pages to answer every one of her questions. So few made it. We saved each other. I had four new teaching certificates, which raised my salary, and earned yearly stipends, but it was like going through boot camp, working, going to school, taking care of two kids, and studying three hours a night. I always taught my own kids what they did not understand, no matter what time of night.
I am a math and a physics teacher, and I don't see why you can't have an irrational number in the denominator (other than older math teachers hate it). You can divide by an irrational number just as well as any other number. I wonder if this is a "rule" that was invented by math teachers from a time when calculators weren't so ubiquitous.
a little different then what I did. I recognized that 8 is the same as 4 * 2. So I factored the numerator to 2(sqroot(2)) divided by sqroot(of 3). Then I multiplied the unfactored 2 by 3 and reduced the denominator to 3. (sqroot of (n) x sqroot of (n) is just n). this of course left me with (2 x sqroot(6) / 3) which is the answer.
A lot of borderline students will thank you because you explained that 1 + 2 = 3 sixteen different ways and they finally got it! That makes you feel good! But your more advanced students got bored and started acting up in class and got kicked out of school! Does that make you feel good, too??
2 square root (2/3). Factor out a square root of 4 to get 2 outside the radical, which is now to be multiplied by the square root of ((1*14)/(3*7)-->square root (14/21 = 2/3).
I did it converting the fractions to decimals because I think in decimals and not complex fractions. You have the sq root of (4/7 X 14/3) or sq root of (.571 X 4.666) or sq root of 2.665 which is 1.63. 2 times the sq root of 6 divided by 3, to me, the problem is not completely solved yet.
Sadly, you wouldn't have gotten any points for your answer. The question is asking you to write it in a simpler form. The answer involves an irrational number that can never be written completely in decimal form because the decimals go on forever.
I found that the sq rt of 2/3 × 2 has the same value on my calculator as 2 × sq rt of 6 ÷ 3, so, apparently, the calculator allows irrational numbers in the denominator of a radical.
I factored out the 4 after reducing and got 2√2/3. What's wrong with that? In addition to not allowing a radical in the denominator, is there another rule that you can't have a fraction in the radical?
@@skellington2000 "dividing by irrationals is well defined." Yes, I didn't understand John's reference to a rule that an irrational can't be in the denominator. He's said that in other videos, too. But then, there's a lot about math that I don't know. It would seem to me that multiplying by, dividing into, adding, or subtracting an irrational would be no better than dividing by one. Anyway, my question wasn't about having whole numbers in the denominator. It was about having fractions in a radical. As for having a fraction in the denominator, I can see why that would be unacceptable. That implies another operation that needs to be carried out (multiply the numerator by the inverse of the denominator). That actually would be more simplified. For example, if you get 7/(1/2), that should be simplified to 14.
@@skellington2000 As a "re-student" I'm looking for rules, principles, and methods for solving certain types of problems. One of the things I've gained here is a better understanding of logs. They were always a problem for me when I was in school. I'm not really concerned with conventions, but when they come up I'm curious about them. Thanks for the feedback.
@@skellington2000 I'm hoping to get to the point where I can recognize a word problem that will require the use of logs -- and when to use inverse logs. I'm still not nearly as comfortable with logs as I am with exponents; it's a work in progress. Never too old to learn, and thankfully for me, never too old to get excited when I discover something. (74) I have no reason other than a desire to learn and the appreciation of mathematical beauty when I see it. There is so much beauty in math, and dammit, I never had a teacher who mentioned that until I took a calculus course 17 years after graduating from college. (That was kind of a mistake, since it had been so long since I had college algebra.)
@@skellington2000 If this started with 1 bacterium, I could do it. But I have no idea how to start it at 100. Can you give me a hint about what to do with that 100? At least I do realize that the final count should be 9900. And I did count on my fingers that it would be 6 and a fraction hours, but that doesn't help at all. I feel that there will be a 2^x in here, somewhere. Is that right?
@@skellington2000 I didn't think you'd answer so quickly. I came back to say I figured out that I could calculate from 1 to 10,000 and then for 1 to 100, and then subtract the second from the first. I tried it. It matches yours. Final answer: 6.64385 hrs. But doing this I noticed something amazing. Then I face-palmed. The amount of time from 1 to 100 will be the same as the amount of time from 100 to 10,000! Well DUH! Of COURSE! 1 to 100 is the same ratio as 100 to 10,000. No matter what method you use, you end up with 2/log2! Thanks so much for that, and for your help.
Four minutes in and you still haven't started to simplify. Get to the point already! Correction: five minutes in and you still haven't started to simplify...
Tell you what spudlo! I think the identical thing when I do these videos. This guy can drive me nuts but doggone it! He is correct. I've learned to skip certain parts. I go about three quarters in now and work my way backwards. You did hit the nail on the head! Thanks!!!
I got 2 times radical 2/3 as my answer before watching the explanation. And now I know why it's wrong because you can't have an irrational number as the denominator. As 2 times radical 2/3 = 2 times radical 2/radical 3. So I guess that we can't have the radical of a fraction as an adequate solution, right?
What you did isn't wrong. You absolutely can have an irrational number in the denominator. Stone people just have a preference not to. 2√(2/3) and (2√6)/3 are just different ways of expressing the same value.
Once again I think I know everything but Mr. Mathman explains things in a perfectly sequential fashion and I know I must listen. I do find this type of equation vexing because I can't follow a set of procedures. Some of these processes are so vague. These processes have no logical procession. There is some random memorization. I understand perfect squares. Understanding where to intercede when the number cannot be perfectly squared could be quandry city. I will say, once this procedure is learned it can up your I.Q. It is hard seeing where this has an application in anything I know of in life. Maybe somehow it does. Only the subconcious knows for sure. Regardless this is still vexing when you can't memorize everything.
Hello he is dedicated to teaching. And the fact that someone thinks his videos are too long just supports his teaching theory. Comprehensive Approach... And you all have FF capabilities 😂
Denying the value of rote memorization is a real fallacy of education in today's world. If you don't like this opinion, well, you don't. This is why I have that opinion: 27 x 2 = 54. No brainer, right? Except, today's high school graduates can't solve this without a calculator. They don't know their multiplication tables, long division is beyond comprehension, and squares and roots are from another universe. I have seen it, first hand, in Honors Graduates. Gimme a break. I am well aware that the students that excel in Math/Science can do this stuff, but the ordinary student: No they cannot, not without their calculator. In my day, starting in 3rd or 4th grade, memorization started, beginning with addition, then subtraction, then multiplication and division. These are very useful skills in the everyday lives of everyday people. The idea that All Students must "Comprehend" theory is absurd. Most have no use for math beyond those four basic arithmetical skills. Ask yourself, how often do you use squares, roots, algebra, or trigonometry in your everyday lives. Leave analytical geometry and calculus out, as it's a totally foreign language to most of those who do use algebra and trig on a daily basis. Machinists and mechanical inspectors, etc. let alone ordinary people that do not. Please understand I am NOT saying this channel has no value, It most certainly does. I am saying that some of the underlying subsumptions of educators of today are just plain bogus.
Doesn't everybody recognize that "simplicity" is in the eye of the beholder. Somebody says that you can't have a square root n the denominator when a square root in the numerator is fine! Somehow, all the intermediate steps are NOT the right answer, but nothing at all would keep someone from taking the simple answer here and doing further manipulations. Then who decides which is "simpler". The real math comes in the entirely unproved statement at the 8:26 minute point that Root(8/3) = Root(8) divided by Root(3). How can a teacher convince a ninth grader of that?
It does not take 18 minutes to simplify this - you are making it unnecessarily complicated. You can immediately pull out 4 from the sqrt and divide 14 by 7, getting 2 sqrt(2/3). If you do not want a sqrt in the denominator, you can multiply by sqrt(3)/sqrt(3), getting 2/3 sqrt(6).
I'm 68 from UK..Got confused initially because of a dot between the fractions..Why wasn't there a multiplication symbol then? or does dot mean that already?
Interesting but so much irritating waffle that spoils the overall presentation. At 75 years of age it’s good to exercise my mind. I can still do most of them in my head.
In the description of the video it says 'and rationalize the denominator'. This would almost certainly be in the actual problem and just got skipped in the video. If it is there, you know you can't have that √3 in the denominator. If it isn't, well, strictly speaking it's not wrong and the problem setter got lazy. Mind you, in a school setting, better do it anyway in case the person correcting you is just as lazy.
You approach is way over complicated. SQR(4/7 * 14/3) can be immediately simplified to SQR(4 * 2/3) SQR(4) * SQR(2/3) then remove 2 squared 2 * SQR(2/3) this can be left as written unless you hold the belief that you can't have an irrational denominator 2 * SQR[(2/3) ^ 3/3] then remove the irrational number 2 * SQR(6/9) 2 * SQR(6)/SQR(9) 2 * SQR(6)/3 (2/3) * SQR(6)
I don't agree with how you did the final stage of the simplification because the aim in mathematics is to solve the problem with as few steps as possible thus eliminating the chances of making errors. After you got √8/√3 why did you got to √24 instead of simplifying √8 to √2*√4 then proceeding to multiplying by √3/√3. and getting the final answer?
It is the same thing. ✓24 then has to be reduced. So you have to convert it to 2✓6. Factor out the 2 forst or last.. same answer, same number of steps.
@@JohnnyHughes1 These problems are meant for kids who are learning maths for the 1st time. So working with smaller numbers makes it easier for them!! ✓24 is more intimidating to them than ✓8.
You should always rationalise the denominator as it's considered bad practice in maths to have a surd (or radicals) as the denominator. Far easier to work with fractions that have a whole number as a denominator.
Interesting, if {A/SQRT(x) }, where (SQRT(x))-1 is defined as an irrational number is irrational, then you are saying for this example that an irrational number multiplied by another irrational number (SQRT(x)/SQRT(x) is okay. That is literally like saying SQRT(x)/irrational equals one. It seems like the order of operations (once (SQRT(x))-1 is defined as irrational, would preclude this solution. The answer of SQRT(8/3) should be acceptable because it avoids a special case where an irrational number divided by another irrational number is equal to one. If you just simplify the fraction down to 24/9 by multiplying (8/3) by 3 inside the square root bracket in the first place and take the square root of that, you get the same simplified answer you show directly without introducing irrational numbers. Or maybe, defining a square root in the denominator as irrational, has no real place in the age of computers. Thanks for another thought provoking "simple" math problem:)
i do not know if it is a UK thing or not once i got to √24/3 did 24 = 2*2*2*3 giving me √2*2*2*3 and always taught any pair of numbers move outside = 2√2*3 = 2√6
unnecessarily complicated. As with all such thngs, start with the inside, in this case, simplify 4/7 x 14/3 cancel the 7 to get 4x2/3 now take the square root to get 2√(2/3) rationalise the denominator by ensuring that there are only perfect squares in the denominator in side the square root. then bring those squares out of the square root. in this case replacing 1 /√9 with 1/3 2√(6/9) = 2√6/√9= (2/3)√6 Answer
I got "squareroot of 24 over 3. " (Which I guess is the same thing as twotimes squareoot of six, divided by three.) So close. Oh, so close. This is why math can be so frustrating...
Maths is one of the simplest subjects! You must 1st read and understand the question/problem, what's required, the rules, formulas,properties, mathematical operations, etc. In this particular case you are required to SIMPLIFY! In maths one must try to work with the smallest numbers possible. So you should have simplified what's under the square root 1st. The rest would have been easier using what I said in my 2nd sentence.
You Sir, are like the Bob Ross of mathematics -- soothing away confusions in an amiable manner...wish I had you as a maths tutor in my schooling years! 😌
This year has been 50 years out of school. Thank you alot of refresh
INDEED! I commonly get these right, but only because I've been watching your videos.
Thank you I am 66 this year and still am up for learning 👍
Me too. just turned 66 a few months ago. Math has been the bain of my life and it makes me mad. I love science but when it runs into math, I collapse. Being so poor in math has held me back.
Greetings. Hello Julie. Never stop learning.
Same here. More interesting to learn now.
67 and love to learn, school is never OUT❤❤❤
(2√6)/3
What's under the radical simplifies quite easily to √(8/3) or √8/√3. This simplifies to (2√2)/√3. Can't have a radical in the denominator, though, so we multiply by √3/√3 and get (2√6)/3.
18 min video explained in 30 sec 😊👍👍👍
Sorry, why can't we have a sqrt in the denominator? this can be written as 2√(2/3)
@@jonjohns8145 A square root in the denominator is not considered the simplest form, and the whole point of the question is to simplify.
@@jonjohns8145You can have a square root in the denominator. This guy just prefers not to.
Did it the same way in my head. Why complicate it.
Math problems that ask you to 'simplify' always remind me of a math teacher I had who ruled the class by sarcasm. I was sent to the board to solve a 'simplify' problem in front of the class, but instead of the thing getting smaller the way it was supposed to, it just got bigger and bigger. After 5 minutes of struggle, the teacher's voice resounded from the back of the class: "READ the question, Carvill. It says SIMPLIFY, not COMPLICATE!". By mutual agreement I was demoted to 'pass' math by the end of the week, so in a way I had managed to 'simplify' two people's problems: mine and the teacher's.
😮😅😂
Aww :( But tbh I would wanna be in ha class for a day
This is good exercise for the brain. 🎉
My friend was making Ds in post graduate classes. I imagined shooting arrows into an A target. My friend asked me to help her learn how to choose multiple choice answers. We talked 3 hours over the phone every night. I taught her that you have to learn that three of the answers may be partially correct. One will be completely correct. Finally she caught on. She made a B. I made an A. Of 1500 students who started the two year program. 500 made it into the second year. 30 got jobs. We both were hired. She was from Pakistan. She was taught how to give oral, and written essays in British schools. I knew the books orally, backwards and forwards by explaining all the answers, jumping back and forth through the pages to answer every one of her questions. So few made it. We saved each other. I had four new teaching certificates, which raised my salary, and earned yearly stipends, but it was like going through boot camp, working, going to school, taking care of two kids, and studying three hours a night. I always taught my own kids what they did not understand, no matter what time of night.
Thank you! Appreciate your teaching. Blessings
thanks for another great lesson.
I am a math and a physics teacher, and I don't see why you can't have an irrational number in the denominator (other than older math teachers hate it). You can divide by an irrational number just as well as any other number. I wonder if this is a "rule" that was invented by math teachers from a time when calculators weren't so ubiquitous.
a little different then what I did. I recognized that 8 is the same as 4 * 2. So I factored the numerator to 2(sqroot(2)) divided by sqroot(of 3). Then I multiplied the unfactored 2 by 3 and reduced the denominator to 3. (sqroot of (n) x sqroot of (n) is just n). this of course left me with (2 x sqroot(6) / 3) which is the answer.
A lot of borderline students will thank you because you explained that 1 + 2 = 3 sixteen different ways and they finally got it! That makes you feel good! But your more advanced students got bored and started acting up in class and got kicked out of school! Does that make you feel good, too??
Love your method of teaching, I'm also here to just ask what is that awesome chalkboard you're using :-) ?
2√6/3
You r an outstanding teacher.
(4/7=0,57) (14/3=4,66) -> 0,57*4,66=2,66 -> SQ 2,66= 1,63
2 square root (2/3). Factor out a square root of 4 to get 2 outside the radical, which is now to be multiplied by the square root of ((1*14)/(3*7)-->square root (14/21 = 2/3).
drop the 'go ahead and' expression
I did it converting the fractions to decimals because I think in decimals and not complex fractions. You have the sq root of (4/7 X 14/3) or sq root of (.571 X 4.666) or sq root of 2.665 which is 1.63. 2 times the sq root of 6 divided by 3, to me, the problem is not completely solved yet.
It was not a calculator question. You were asked to simplify the expression, not solve it.
Sadly, you wouldn't have gotten any points for your answer. The question is asking you to write it in a simpler form. The answer involves an irrational number that can never be written completely in decimal form because the decimals go on forever.
I agree with your philosophy. I did the same and i my head. In the real world we need precise answers not just mental gymnastics.
I found that the sq rt of 2/3 × 2 has the same value on my calculator as 2 × sq rt of 6 ÷ 3, so, apparently, the calculator allows irrational numbers in the denominator of a radical.
I did it mentally, and I got the same answer as yours. Doubled check using a calculator, and the results are the same.
2radical6/3
Wonderful!
(8/3)^.5= +/-2/3•√6
Thank you
Oh, Simplification is so complicated!
This is a comedy line!
I factored out the 4 after reducing and got 2√2/3. What's wrong with that? In addition to not allowing a radical in the denominator, is there another rule that you can't have a fraction in the radical?
@@skellington2000 "dividing by irrationals is well defined."
Yes, I didn't understand John's reference to a rule that an irrational can't be in the denominator. He's said that in other videos, too. But then, there's a lot about math that I don't know. It would seem to me that multiplying by, dividing into, adding, or subtracting an irrational would be no better than dividing by one.
Anyway, my question wasn't about having whole numbers in the denominator. It was about having fractions in a radical. As for having a fraction in the denominator, I can see why that would be unacceptable. That implies another operation that needs to be carried out (multiply the numerator by the inverse of the denominator). That actually would be more simplified. For example, if you get 7/(1/2), that should be simplified to 14.
@@skellington2000 As a "re-student" I'm looking for rules, principles, and methods for solving certain types of problems. One of the things I've gained here is a better understanding of logs. They were always a problem for me when I was in school.
I'm not really concerned with conventions, but when they come up I'm curious about them.
Thanks for the feedback.
@@skellington2000 I'm hoping to get to the point where I can recognize a word problem that will require the use of logs -- and when to use inverse logs. I'm still not nearly as comfortable with logs as I am with exponents; it's a work in progress. Never too old to learn, and thankfully for me, never too old to get excited when I discover something. (74) I have no reason other than a desire to learn and the appreciation of mathematical beauty when I see it. There is so much beauty in math, and dammit, I never had a teacher who mentioned that until I took a calculus course 17 years after graduating from college. (That was kind of a mistake, since it had been so long since I had college algebra.)
@@skellington2000 If this started with 1 bacterium, I could do it. But I have no idea how to start it at 100. Can you give me a hint about what to do with that 100? At least I do realize that the final count should be 9900. And I did count on my fingers that it would be 6 and a fraction hours, but that doesn't help at all. I feel that there will be a 2^x in here, somewhere. Is that right?
@@skellington2000 I didn't think you'd answer so quickly. I came back to say I figured out that I could calculate from 1 to 10,000 and then for 1 to 100, and then subtract the second from the first.
I tried it. It matches yours. Final answer: 6.64385 hrs. But doing this I noticed something amazing. Then I face-palmed. The amount of time from 1 to 100 will be the same as the amount of time from 100 to 10,000! Well DUH! Of COURSE! 1 to 100 is the same ratio as 100 to 10,000. No matter what method you use, you end up with 2/log2!
Thanks so much for that, and for your help.
Thanks!
Four minutes in and you still haven't started to simplify. Get to the point already!
Correction: five minutes in and you still haven't started to simplify...
Tell you what spudlo! I think the identical thing when I do these videos. This guy can drive me nuts but doggone it! He is correct. I've learned to skip certain parts. I go about three quarters in now and work my way backwards. You did hit the nail on the head! Thanks!!!
You're the best🙌🏿
I got as far as √8/3. So, I 'm going back in to learn the rest.🎉
teacher does not like sqrt(3) in denominator... so that the more proper answer is
sqrt(24)/3
sqrt(6×4)/3
2sqrt(6)/3
2/3 sqrt (6)
Ah, a square root of fractions...
First simplify the fractions: 4/7 . 14/3 = 4/1 . 2/3 = 8/3
Then take the square root: V(8/3) = V8 / V3 = V8 . V3 / 3 = V24 / 3 = 2V6/3
The answer is 2 to the square of 6 over 3
(2sqrt(6))/3
Wow so complicated for a simple problem.
2 and 2/3
2√(2/3)
2(2/3)^1/2 is a solution also
I got 2 times radical 2/3 as my answer before watching the explanation. And now I know why it's wrong because you can't have an irrational number as the denominator.
As 2 times radical 2/3 = 2 times radical 2/radical 3.
So I guess that we can't have the radical of a fraction as an adequate solution, right?
What you did isn't wrong. You absolutely can have an irrational number in the denominator.
Stone people just have a preference not to.
2√(2/3) and (2√6)/3 are just different ways of expressing the same value.
My answer is 9😊😊
Once again I think I know everything but Mr. Mathman explains things in a perfectly sequential fashion and I know I must listen. I do find this type of equation vexing because I can't follow a set of procedures. Some of these processes are so vague. These processes have no logical procession. There is some random memorization. I understand perfect squares. Understanding where to intercede when the number cannot be perfectly squared could be quandry city. I will say, once this procedure is learned it can up your I.Q. It is hard seeing where this has an application in anything I know of in life. Maybe somehow it does. Only the subconcious knows for sure. Regardless this is still vexing when you can't memorize everything.
8/3
Hello he is dedicated to teaching. And the fact that someone thinks his videos are too long just supports his teaching theory. Comprehensive Approach... And you all have FF capabilities 😂
I came up with the dog ate my homework.
Holy smoly! I'm lost. I have to go back to the square part.
Denying the value of rote memorization is a real fallacy of education in today's world. If you don't like this opinion, well, you don't. This is why I have that opinion: 27 x 2 = 54. No brainer, right? Except, today's high school graduates can't solve this without a calculator. They don't know their multiplication tables, long division is beyond comprehension, and squares and roots are from another universe. I have seen it, first hand, in Honors Graduates. Gimme a break. I am well aware that the students that excel in Math/Science can do this stuff, but the ordinary student: No they cannot, not without their calculator. In my day, starting in 3rd or 4th grade, memorization started, beginning with addition, then subtraction, then multiplication and division. These are very useful skills in the everyday lives of everyday people. The idea that All Students must "Comprehend" theory is absurd. Most have no use for math beyond those four basic arithmetical skills. Ask yourself, how often do you use squares, roots, algebra, or trigonometry in your everyday lives. Leave analytical geometry and calculus out, as it's a totally foreign language to most of those who do use algebra and trig on a daily basis. Machinists and mechanical inspectors, etc. let alone ordinary people that do not.
Please understand I am NOT saying this channel has no value, It most certainly does. I am saying that some of the underlying subsumptions of educators of today are just plain bogus.
2.6
New here and subscribed.
Doesn't everybody recognize that "simplicity" is in the eye of the beholder. Somebody says that you can't have a square root n the denominator when a square root in the numerator is fine! Somehow, all the intermediate steps are NOT the right answer, but nothing at all would keep someone from taking the simple answer here and doing further manipulations. Then who decides which is "simpler".
The real math comes in the entirely unproved statement at the 8:26 minute point that Root(8/3) = Root(8) divided by Root(3). How can a teacher convince a ninth grader of that?
It does not take 18 minutes to simplify this - you are making it unnecessarily complicated. You can immediately pull out 4 from the sqrt and divide 14 by 7, getting 2 sqrt(2/3). If you do not want a sqrt in the denominator, you can multiply by sqrt(3)/sqrt(3), getting 2/3 sqrt(6).
You are correct. I would loose my mind sitting in his class listening to him over explain EVERYTHING!
56/21
(8/3)^½
Like for example sq.rt of 9=3 or sq rt of 25=5 am i correct sir
2/3 x sqrt 3😊
I would stop at 2*sqr(2/3)
9
wow, you make math's look hard.. why not just use a base denominator of 21, then get the the square root of the sum. easy.
4
I'm 68 from UK..Got confused initially because of a dot between the fractions..Why wasn't there a multiplication symbol then? or does dot mean that already?
2√2/√3
There is a perfect square at the start with the 4. Pull the 2 out and it'll be way simpler.
Yes it is. Then resolve the SQRT(2/3)
2 is the correct answer.
√(8/3)
= 2√2 / √3
= 2√6 / 3
2*SQR2/3
2 2/3
1.633
Falling asleap! Dude!
what is the final answer
Interesting but so much irritating waffle that spoils the overall presentation. At 75 years of age it’s good to exercise my mind. I can still do most of them in my head.
Why it took 18 minutes to explain this!
2 root 2 over 3
2/3square 6
Answer is 2
Too much going in circles for me to follow.
Please go back and watch it step by step ! He is giving a very good explanation ! Good luck !
28 / 3√7
Simplify
what's wrong with the answer:
2(2/3)^1/2 ... I know it can be simplified to 2/3(6)^1/2 ... but still wondering if the former is acceptable??
Thinking the same thing. My first solution was 2 times the root of 2/3. Does he not like any denominators in a root? Calculates to the same value.
You have to simplify the radical expression.
In the description of the video it says 'and rationalize the denominator'. This would almost certainly be in the actual problem and just got skipped in the video.
If it is there, you know you can't have that √3 in the denominator. If it isn't, well, strictly speaking it's not wrong and the problem setter got lazy.
Mind you, in a school setting, better do it anyway in case the person correcting you is just as lazy.
@@betchayarazaruiz5817They did that. The question is why do you then have to also rationalise the denominator.
You approach is way over complicated.
SQR(4/7 * 14/3) can be immediately simplified to SQR(4 * 2/3)
SQR(4) * SQR(2/3) then remove 2 squared
2 * SQR(2/3) this can be left as written unless you hold the belief that you can't have an irrational denominator
2 * SQR[(2/3) ^ 3/3] then remove the irrational number
2 * SQR(6/9)
2 * SQR(6)/SQR(9)
2 * SQR(6)/3
(2/3) * SQR(6)
I don't agree with how you did the final stage of the simplification because the aim in mathematics is to solve the problem with as few steps as possible thus eliminating the chances of making errors. After you got √8/√3 why did you got to √24 instead of simplifying √8 to √2*√4 then proceeding to multiplying by √3/√3. and getting the final answer?
He lets it be difficult
It is the same thing. ✓24 then has to be reduced. So you have to convert it to 2✓6. Factor out the 2 forst or last.. same answer, same number of steps.
@@JohnnyHughes1 These problems are meant for kids who are learning maths for the 1st time. So working with smaller numbers makes it easier for them!! ✓24 is more intimidating to them than ✓8.
OK So I got Square root 8/3
but don't really understand why 2(square root 6)/3 is considered simpler
You should always rationalise the denominator as it's considered bad practice in maths to have a surd (or radicals) as the denominator. Far easier to work with fractions that have a whole number as a denominator.
It's not simpler. It's exactly the same.
1/6
2
Just a little less unnecessary narrative, and direct approach to the subject matter.
The convolutions are arbitrary
Did you get this right? Sorry, genius.
Interesting, if {A/SQRT(x) }, where (SQRT(x))-1 is defined as an irrational number is irrational, then you are saying for this example that an irrational number multiplied by another irrational number (SQRT(x)/SQRT(x) is okay. That is literally like saying SQRT(x)/irrational equals one. It seems like the order of operations (once (SQRT(x))-1 is defined as irrational, would preclude this solution. The answer of SQRT(8/3) should be acceptable because it avoids a special case where an irrational number divided by another irrational number is equal to one. If you just simplify the fraction down to 24/9 by multiplying (8/3) by 3 inside the square root bracket in the first place and take the square root of that, you get the same simplified answer you show directly without introducing irrational numbers. Or maybe, defining a square root in the denominator as irrational, has no real place in the age of computers. Thanks for another thought provoking "simple" math problem:)
It is only wrong if the teacher specifically says to simplify all fractions to their simplest form.
9 16 25 36 49 64 81 100 121 144
Wait a minute, the problem in the title and the problem in the video are not the same!
i do not know if it is a UK thing or not
once i got to √24/3 did 24 = 2*2*2*3 giving me √2*2*2*3 and always taught any pair of numbers move outside = 2√2*3 = 2√6
How can you solve this
In my head 2/3 sqrt(6)
I got 2 x sqrt of (2/3). I guess it’s not simplified enough
@Teddy Same here! We were close though. ;)
It is. This guy just likes to end up with no √ in the denominator.
That's his preference. It's not simplification.
unnecessarily complicated.
As with all such thngs, start with the inside, in this case,
simplify 4/7 x 14/3
cancel the 7 to get 4x2/3
now take the square root to get
2√(2/3)
rationalise the denominator by ensuring that there are only perfect squares in the denominator in side the square root. then bring those squares out of the square root.
in this case replacing 1 /√9 with 1/3
2√(6/9) = 2√6/√9= (2/3)√6 Answer
Nice I agree
4/1 x 2/3 = the square root of 8/3?
Ok SqR 24= 2•2•2•3 over
SqR. 9= 3•3
=. 2 SqR 6
Over3
√((4/7)*(14/3))
√((4/1)*(2/3))
√4 * √(2/3)
2√(2/3)
2√(6/9)
2√6 * √(1/9)
2√6 * (1/3)
(2√6)/3
I got "squareroot of 24 over 3. " (Which I guess is the same thing as twotimes squareoot of six, divided by three.)
So close. Oh, so close. This is why math can be so frustrating...
Maths is one of the simplest subjects! You must 1st read and understand the question/problem, what's required, the rules, formulas,properties, mathematical operations, etc. In this particular case you are required to SIMPLIFY! In maths one must try to work with the smallest numbers possible. So you should have simplified what's under the square root 1st. The rest would have been easier using what I said in my 2nd sentence.
Two ways to solve arithmetically or algebraically....
Does not take so much time.
2√(2/3) ?
i tried to do it in my head, but i had to use paper.
Why so complicated? You multiply numerator by numerator and denominator by denominator and then simplify 56/21 to 8/3, so it's square root of ✓8/3
so the square root of 24 is 4.6?