Help with this rational! Rationalize the denominator with radicals, Reddit r/askmaths
HTML-код
- Опубликовано: 6 июн 2024
- This tutorial not only answers an algebra question from Reddit but also provides a detailed overview of how to rationalize the denominator with different radicals.
/ g3wguc2hwt
im the one who asked the question on reddit! thanks so much!
why does it say 3 months ago??? the video came out half an hour ago
Same question
@@ibrahimali3192comments like this can appear when the video is uploaded unlisted before being public. im not sure how this comment got 3 months ago with the context not fitting with the date
@@ibrahimali3192if im not mistaken, he unlisted the video, so only people with the video link can watch it, which he sent to the op
Brbp unlisted this video and put the link in reddit. 1 hour ago, he changed it to be published.
I love how in 5:48 you tap the board and you look so confused as to why the magic didn't activate and erase the board.
8:15 ta-da!
Write ⁵√12 as (⁵√-3)(⁵√-4) and then ⁵√-4 will cancel each other out and you'll be left with (⁵√-3)/4 which can be simplified to -⁵√3/4
That's the way I thought too
You can’t expand roots as negatives since you need to resolve the sign of a number internally before you expand a number
@@aperson6081 nth root of x will be positive if n is even, else it will be the same sign as x 🙄 simple maths
@@alexandermcclure6185 That's rude man
@alexandermcclure True, but I see what they were tryna say. They probably confused it for √mn ≠ √(-m)(-n), which is a rule ONLY for roots with positive even orders.
I started watching your videos a few years back when I went back to school in my 40s for a biology degree. I needed a LOT of help learning calculus and you were one of the reasons I passed the class.
Now I'm graduated, don't 'need' to know how it all works anymore since I can just plug things into programs to do the math for me, but I continue to watch your videos just because they're fun to watch. Hope you keep it going for many more years.
I love your videos. Not because I need the help. I am old enough to have seen and done most of the exercises a hundred times already.
First it is a good relaxing way to see math I don't any more. But the most important for me is that I like to see how you teach it so I can improve on teaching. Basic your videos have become meta for me
Thank you for your hard work
You can rewrite the problem like this: 1/4 * 12^(1/5)/(-4) ^(1/5). Then according to power rules you can do that: 1/4 * (12/(-4))^(1/5) (same power, different base). Which leads to: 1/4 * (-3)^(1/5) or (-3)^(1/5)/4
With a cube root, multiply it 3 times, a 4th root, multiply it 4 times, etc. It's not always the shortest way, but it will work.
Yeah, I did it similarly.
For clarity, let us call the fifth root function as fifth(x).
fifth(12)/[4 * fifth(-4)] = fifth(12) / [-4 * fifth(4)].
Now, I can multiply both the numerator and the denominator by fifth(4^4). In fact, by doing the multiplication, the denominator becomes -4*fifth(4*4^4) = -4*fifth(4^5) = -4*4.
Likewise, the numerator becomes fifth(12 * 4^4). 12 = 3 * 4, thus the numerator becomes fifth(3*4*4^4) = fifth(3*4^5) = 4 * fifth(3).
Now, we have: [4 * fifth(3)] / [4 * (-4)]. We can simplify the 4s and get:
fifth(3) / (- 4) = - fifth(3) / 4.
That speedrun strat at the end was impressive.
I love that quick method!!
I mean, I'm somewhat good at math, but when I solved this in memory in like half a minute and saw a 9 minute video I got interested if I got it right. And I did, because I've done it in a way that's, like, 5 times easier? Just split 12^-5 and -4^-5 into (3*4)^-5 and (-1*4)^-5 and then the roots of 4 reduce, leaving you -1^-5 which is just -1 at the bottom and 3^-5 at the top, which is already the answer. I don't see why you wouldn't do that, multiplying things instead (which has more potential to simply make an error in the numbers).
Remember that x^-5 is actually not a root but a fraction like this -- 1/x^5.
Roots are written as x^(1/5)
And then u watch the whole video till the end XD
@@DamianBR Right, as usual, I've mixed them up :D
I think using rational exponents makes this question 1000x easier. I was also confused on how to even do this with the radicals, but changing to rational exponents made it trivial
I'm happy i was able to think of the quick method by myself
At 1:55, you can use two 2^(1/3)'s in both the numerator and denominator. The denominator then simplifies to 2 while you just multiply the two numerator 2^(1/3)'s together to get 4^(1/3). Then you have 6*4^(1/3)/2 = 3*4^(1/3).
Also, in the current problem, you can put four (-4)^(1/5)'s in both the numerator and denominator. This would lead to 12^(1/5) * (-4)^(1/5) * (-4)^(1/5) * (-4)^(1/5) * (-4)^(1/5)/ (-4)^(1/5) / (-4)^(1/5) / (-4)^(1/5) / (-4)^(1/5) / (-4)^(1/5)/ 4 = (12 * (-4)^4)^(1/5)/ (-4)/4 = ((-3) * (-4) * (-4)^4)^(1/5)/ (-4)/4 = ((-3) * (-4)^5)^(1/5)/ (-4)/4 = (-4) * (-3)^(1/5)/ (-4)/4 = (-3)^(1/5)/4 = -3^(1/5)/4.
hey hii, thank u for this explanation...it helped a lot!!
"Is he lying to me?"
"NO, nonono"
"Ok, that'd be great"
Wonderful 👍👍👏👏
How come at the last method you don't put the 5th root of 4 on the bottom denominator?
I understand easily with powers like 2^0.2 or 2^1/5
you can see his stash of markers at the bottom right LOL
Was wondering why you didn't factor the top (fifth root of 12) into (fifth root of -3) × (fifth root of -4) and cancel the top and bottom at the beginning?
Just watched last 20 seconds... answered my own question... 😅
@@mr9512 It also gets rid of the problem of why the square root of 1 isn't the square root of -1 times the square root of -1. He explained why what doesn't work for even roots did work in this situation.
Or you could just multiply the entire thing by the fith root of negative four to the power of four both up and down. That would create the same result.
Is this correct? We're not raising to the fifth power, but to 1/5 so "putting the minus sign outside" shouldn't apply, should it?
Minus sign is because of the odd number of the root. -1³ is -1. Cubic root of -1 will still be applied.
simply said: if i want the fifth root of -32, i can say -2*-2*-2*-2*-2, and in the end, one minus-sign will remain, giving -32, therefore its -2 which is equivalent to -1 * fifth root 32
another reason:
when you take y = x^5 it is defined for all x and y as real numbers with no restrictions.
when taking the inverse of it, y_ = fifth root x, we will switch the range and domain of the original function
since the original function has no restrictions on domain and range, its inverse also doesnt have any restrictions, therefore defining it for negative x too.
the same logic also says why y = square root x isnt defined for negative x.
y = x^2 has the restriction y >= 0, so the inverse y_ = square root x has the restriction x >= 0, making it undefined for negative x
-1^x, where x is an odd integer = -1. The reverse fact is also true, and therefore, -1 can always be written to the odd power of -1.
@@Sg190th-- You are missing grouping symbols: (-1)^3 = -1.
@@ronaldking1054-- You are missing grouping symbols: (-1)^x, where x is an odd integer, equals -1.
3:30 this seems a really confusing way to explain this instead of just saying you need the same power inside as the "term" of the radical. If you have cube root of 2^1 you need to get that 1 to be a three, so you multiply by cube root of 2^2. if it's 21st root \you just multiply by 2^20 on the inside.
Do you not need to be careful of expanding roots to negatives - I works with odd powers but I thought with even powers it breaks stuff? Amazing video nevertheless.
Bro erased the board with just a tap.
available at Amazon
But the tap only works if the answer is correct 🤫.
I solved it correctly, using the laws of exponentiation.
Can you teach me how can I sove this problem, please?
sqr(a)+sqr(ab)+sqr(abc)=12
sqr(b)+sqr(bc)+sqr(abc)=21
sqr(c)+sqr(ac)+sqr(abc)=30
Find: (a^2 + b^2 + c^2)
Did it the extra long way but hey I guess I still got it right
i asked myself, "why didnt he js combing the radicals" and then i was like oh
Kindly upload video in hinglish
8:17 I knew I wasn't tripping lmfao there is a fast method but I guess you wanted showcase on how to rationalize fractions with nth roots, which is respectable.
Mistake at 5:46 . You cant not simply put out the minus out of the 5th-root in front. By this logic the 3rd root of -8 is equal to -2, which is wrong. The 3rd root of -8 is equal to 1+sqrt(3)*i . the 5th-root of negativ 4 is not equal to the negativ of the 5th-root of 4. You have to work with the absolut smallest argument
x^3 + 8 = (x + 2)(x^2 - 2x + 4). You picked to deal with only the complex root while ignoring the real root, which is easiest to find. The real odd root of a negative number is always negative, and while there are complex roots, there is always a real one, and the real one should be the first to attempt to show a basic math person.
I forgot that you also ignored that another third root of -8 is 1-sqrt(3)*i.
Incorrect. While the equation x^3=-8 has 3 answers, the the n-th root of ANY number has only ONE answer for any n. Same as the sqrt(4) is equal to 2 and Not to plus and minus 2. Dont confuse polynoms with roots. For roots you only take one answers, and its the one with the smallest absolut argument. (Checking the wikipedia page on my language (german) comfirms my answer.)
@@jurineu2401 Wrong. There are as many roots as there are powers. Most of them are complex.
@@jurineu2401 -2 * -2 * -2 = -8. Basic definition of a cube, but you stated this isn't true. You are wrong. This is the real root. It is the only real root, which you claimed a complex number was the simplest argument while a real number with no complex component is the actually the simplest answer. No branch of mathematics had to be created in order to find this root. Real analysis works.
Next (1 + sqrt(3)*i)(1 + sqrt(3)*i) = 1 + 2sqrt(3)*i - 3 = -2 + 2 sqrt(3)i
(-2 + 2sqrt(3)i)(1 + sqrt(3)*i) = -2 - 2 sqrt(3)i + 2sqrt(3)i - -6 = -8 by the basic definition of a cube. This is your root.
Next (1 - sqrt(3)*i)(1 - sqrt(3)*i) = 1 -2 sqrt(3)*i - 3 = -2 - 2*sqrt(3)*i
-1 (2 + 2*sqrt(3)*i)(1 - sqrt(3)*i) = -1 (2 - 2*sqrt(3)*i +2 * sqrt(3) * i + 6)
= -1 * 8 = -8. Basic definition of a cube.
They all have different values.
Isn't there a negative in the original question?
that's like a misprinted dot or smth if it's a negative sign it would be longer(as long as the negative sign in the -4)
@@mosescheung5794 Possibly
pls do the question i gave u on your calculus channel
Just divide 12 by 4 and get to the answer quicker