Simplify this Radical Expression | Step-by-Step Explanation

[@mshanmukhavalli4567]

Nice video.very helpful for me

[@PreMath]

Glad to hear that! You are awesome Shan👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃 Love and prayers from the USA! 😃

[@ramtirthparashar5988]

Pl. solve more and more problems. Thank you.

[@ramtirthparashar5988]

Nice problem.pl. send more difficult problems with solutions.

[@shivanandankolekar9244]

*it I don't want it

[@gmutubeacct]

Can be solved in 3 steps by multiplying numerator and denominator by ((sqrt(3) + sqrt(5) + sqrt(7)) - sqrt(5)) Treat the denominator as ((sqrt(3) + sqrt(5) + sqrt(7)) + sqrt(5))

[@PreMath]

Thanks Ram for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@pranavk.p.6333]

@@PreMath you should reload the same otherwise people would get confused watching this video.

[@SDon-wn8xb]

@Ram Narayan Can you illustrate fully your suggested method, for it seems to me your method may be more involved than that illustrated here, no?

[@hyperloid2143]

Illustrate please

[@teambellavsteamalice]

Lets see if √3, √5 and √7 are a, b and c you get this: n = (ab+bc+ac) + 5 d = (a+b+c) + a Then you'd multiply both by a+c, which can be written as (a+b+c) - a, note (a+b+c)^2 = 2*(ab+bc+ac ) + 15 (the sum of squares 3+5+7) so the denominator becomes this square minus a^2 (which is 5). That makes it equal to twice the numerator! In the numerator only remains the added factor (a+b+c) - b which is of course a+c so you quickly get the answer (√3+√7)/2 Well done!

[@phoenixslender6541]

Incredible solution! Thank you for uploading amazing video everyday!

[@HassanLakiss]

Thank you for an interesting problem and you very good explanation of how to get to the answer. I would respectfully say that it could also be done by rationalization of the denominator once it has been grouped as you did in your first line. Thank you

[@billmorrigan386]

Very nice problem! After factoring the numerator I immediately saw the structure of ab/(a+b); so naturally, I went 2ab = (a+b)^2 - (a^2 + b^2) but it looked like a circular dead end. So I also flipped the whole fraction to (a+b)/ab = 1/a + 1/b and momentarily stumbled upon the exact same solution: 1/(√5+√7) + 1/(√3+√5) = (√7-√5)/2 + (√5-√3)/2 = (√7-√3)/2. Flipping it back gives 2/(√7-√3) = (√7+√3)/2. That's a beautiful problem. I didn't expect that flipping the fraction would work. I was actually surprised to find it out. Although this problem is extremely simple and I stumbled upon the solution in a few minutes, I really like it. A huge like to this video too!

[@fmstudio085]

You're genius, many thanks 👍🏻 I already suggested this perfect channel to all my friends which intended to obtain good math skills

[@imonkalyanbarua]

So smooth! Very nice! 😇👏👏👏

[@sangpham2931]

Tuyệt Tuyệt vời, giải rất hay (tốt )rễ hiểu, xin cảm ơn chương trình đã cho ra nhiều đề (bài toán )rất khó nhưng giảng viên giải rất hay, rễ hiểu ,một lần nữa xin cảm ơn

[@JLvatron]

amazing! I almost got it, I just gave up too early but I was on the right track.

[@pratimadevi2214]

Me tooooo

[@Rudxain]

I only went as far as doing the prime factorization lol

[@adityakumarsingh8012]

easiest method can be by multiplying numerator and denominator by 2 then making (root(5)+root(7)+root(3))^2 - (root(5))^2 at numerator and applying diffrence of squares formula

[@duongduong2939]

R u sure 😂😂😂

[@rafaelivan1830]

this has become a new fave channel for me :D its fun trying to solve the problem you shared and its mindblowing to see how you solved those !

[@PreMath]

Awesome, thank you Rafael! Glad to hear that! You are awesome👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃 Love and prayers from the USA! 😃

[@Thathugheschap]

Love these for keeping my brain working. Thanks 👍

[@Leonex52]

If one knows about field theory then the solution can be quite simple. Note that this is a element belongs to the field Q(sqrt(3),sqrt(5),sqrt(7)) , and the elements of the field can take the form of a*sqrt(3)+b*sqrt(5)+c*sqrt(7). Then solve for a, b and c.

[@soloduhaelena3412]

Thаnk you for your interesting and sometimes unexpected explanation . It's a pleasure to follow you step by step.

[@Ramkabharosa]

Suppose (√15 + √21 + √35 + 5)/(√3 + 2√5 + √7) = (a√3 + b√5 + c√7). Then (√3.√5 + √3.√7+ √5.√7 + 5) = (√3 + 2√5 + √7).(a√3 + b√5 + c√7). By equating the coefficients of √3.√5, √3.√7, and √5.√7; we get 2a + b = 1, a + c = 1, and b + 2c = 1. So b = 1 - 2a and b = 1 - 2c. Hence 1 - 2a = 1 - 2c and so a=c. Thus a+a =1 & hence a = 1/2. So b = 1 - 2a = 0 and c = a = 1/2. In the end, we will finally get (√15+√21+√35 +5)/(√3 + 2√5 + √7) = (a√3+b√5 +c√7) = (√3+√7)/2. .

[@velimir_ikalovic]

great video, but to slow for my taste, I had to skip over some obvious things that were repeating. That makes it perfect for everyone who is not fluent in math, but want to learn more. good job overal.

[@thinkingaboutreligion2645]

14 minutes of algebra with just one single x. In one tiny corner of the video. That's impressive in itself, but it was a great watch too.

[@mustafizrahman2822]

Your solving position is LEGEND LEVEL!(SERIOUSLY). I have about to solve it like your process but failed. Thanks for your nice explanation.

[@PreMath]

You're welcome Rahman dear! Please keep sharing premath channel with your family and friends. Take care dear and all the best😃 Love and prayers from the USA! 😃

[@cookaburra]

If you assume that the numerator has the denominator as a factor and see how that goes, you can write: Numerator = (sqrt(3) + sqrt(7) + 2*sqrt(5)) * (a*sqrt(3) + b*sqrt(7) + c*sqrt(5)) Then multiple out and see that: a + b = 1 c + 2b = 1 c + 2a = 1 3a + 7b + 10c = 5 Solution then follows as: a = b = 1/2 c = 0 so the simplified expression is: (sqrt(3) + sqrt(7))/2

[@RKJha-wm3us]

Very helpful video 👏👍🙏

[@shrikrishnagokhale3557]

I like the Q's and A's both.

[@mathsplus01]

Congratulation of 300k subscribers. Amazing

[@durgatoshpandey5250]

Thanks. I reached upto the reciprocal (1/x) part myself. But learnt from you the trick of multiplying numerator and denominator with the conjugate to rationalise the denominator.

[@stumbling]

I noticed that if you set x = √5 then factoring you get ( x√3 + x√7 + √21 + x² ) / ( √3 + 2x + √7) Then setting numerator to a function, f(x) = x² + x√3 + x√7 + √21 The denominator is in fact the first derivative of the function, f'(x) = 2x + √3 + √7 I am not sure if this can be used to solve the problem but was an interesting observation.

[@ayushshukla4205]

Very nice.

[@bobsinger8637]

Ja this is a cool problem but I think all you can say about the series of terms that this is equivalent to is that the sum of these terms when you examine the positive valuses you see that they grow a little bit and will aproach the highest numerical value that you can easily see using arithmetic however because of the presence of prime numbers the function approaches the sum of their roots but it doesn’t get there so their limit is equal to their sum. The numerator grows faster than the denominator so the expression itself will aproach infinity very slowly and probably not converge to a specific value. It’s hard to say this fir sure because of how many numbers would have to be looked at along this curve, but I think because of the the numerators accelerated growth it has to be a growth function and also a decay function because of negative integer values that are twins present beneath the radical symbols in both numerator and denominator. I really like that you’re showing the starting point for this with the factor method to take out the unseen primes beneath the radical signed in this expression. Again this problem is really great. Thank you very much.

[@md8744]

Learning more to teach my kids at 7th and 8th grade. It is helpful

[@Potencyfunction]

In the 7th grade? Lol mines are learning in the first grade this LOL

[@matthewleitch1]

This is an interesting video but strangely frustrating. I could not see any reason to think each move was getting us closer to an answer until about 8 minutes had passed. What was it about the expression that suggested this method would work, if anything?

[@andywright8803]

Factorising tends to be a good thing in general

[@matthewleitch1]

@@andywright8803 Yes, that's a start. Though in other situations with surds factorisation is not the first move.

[@monikamalinovka9514]

Back then they don't teach us about make a whole number in square root. So 5 =(√5)(√5). That blow my mind!

[@PreMath]

You are awesome Monika 👍 Take care dear and stay blessed😃

[@13kefka]

That IS the definition of square root : the number that you multiply by itself to get the number

[@girolamocapita3321]

Very nice

[@ghmaxiron2639]

Nice and interesting video Thank you

[@aashsyed1277]

VERY NICE VIDEO DEAR. THANK YOU. LOT'S OF LOVE FROM FAR AWAY

[@PreMath]

So nice of you Aaash dear. You are the best Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@victorchoripapa2232]

Nice solution. Great video. Thanks!

[@alittax]

Beautiful, thank you!

[@242math]

got it, thanks for the challenge bro

[@PreMath]

Glad to hear that! You are awesome 242👍 Take care dear and all the best😃 Love and prayers from the USA! 😃

[@cenderelaabenezer7489]

Thank you

[@PreMath]

You're welcome Abenezer! You are awesome 👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃 Love and prayers from the USA! 😃

[@lancege8596]

Very helpful for grade 7 students

[@jesusantoniocarhuashuerta4662]

Excellent

[@PreMath]

Thanks Jesus dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@SrisailamNavuluri]

I am learning something from you.

[@PreMath]

Excellent dear Glad to hear that! You are awesome 👍 Take care dear and all the best😃 Love and prayers from the USA! 😃

[@abdellahaitouahmane1593]

very nice sir ( from morrocco)

[@PreMath]

Thanks dear Thanks Abdellah for the feedback. You are awesome 👍 Take care dear and stay blessed😃 Love and prayers from the USA! 😃

[@erwinracoma1935]

Thank you.

[@vaghelabhikhusing6553]

Good sir

[@andrewrao634]

Impressive, but too complicated. Based on the radicals in the numerator, it's clear that we need to multiply the denominator by some of its own radicals: d*rt3 = 3 +2*rt15 + rt21 d*rt5 = 10 + rt15 + rt35 d*rt7 = 7 + rt21 + 2*rt35 It's then obvious that d*(rt3 + rt7) = 2n, so n/d = (rt3 + rt7)/2

[@teambellavsteamalice]

Wow, nice trick! I tried something similar, but that didn't work.

[@sianneang6893]

thank you.it's very helpful.

[@PreMath]

Glad to hear that Thanks Ang for the feedback. You are awesome 👍 Take care dear and stay blessed😃 Love and prayers from the USA! 😃

[@nicogehren6566]

nice soution sir thanks

[@PreMath]

Welcome Nico So nice of you dear! You are awesome 👍 I'm glad you liked it! Take care dear and stay blessed😃

[@janvigoel5767]

I also solve easily bcz i hve done in my school time

[@MarieAnne.]

That seems a lot simpler than what I did. I did manage to factor numerator, but couldn't with denominator, of course. But it didn't occur to me to split up denominator. Instead I tried to rationalize part of denominator in first step, then rationalize all of the denominator in second step: 1st step: (√15+√35+√21+5) / (√3+2√5+√7) = [(√15+√35+√21+5)(√3+2√5-√7)] / [(√3+2√5+√7)(√3+2√5-√7)] = (8√3+6√5+8√7+2√105) / (16+4√15) = (4√3+3√5+4√7+√105) / [2(4+√15)] 2nd step: = [(4√3+3√5+4√7+√105)(4-√15)] / [2(4+√15)(4-√15)] = (√3+√7)/2 I left out the intermediate calculations, but I believe your method is less prone to errors because you're dealing with very simple fractions.

[@mathematicsolympiadchannel5345]

Pls do olympiad math

[@PreMath]

Sure dear. I'm working on that. Take care dear and all the best😃 Love and prayers from the USA! 😃

[@pareshmehta4670]

Nice

[@naucifacio]

I think this is a little bit faster: a:=sqrt(3)+sqrt (5)+sqrt(7) Then our expression becomes: (a^2-5)/(2*(a+sqrt(5)) Simplifying (a-sqrt(5))/2 Substitute a (sqrt(3)+sqrt(7))/2

[@theophonchana5025]

x = 2 ÷ (square root of 7 - square root of 3)

[@julieng.4375]

It's incredible

[@skyrimguard3501]

Thank you very much

[@manasbhattacharyya4359]

It can be solve by plane division method

[@michaelaseraphine124]

Thank you sir

[@teodoromaranan1415]

very interesting, but very difficult.

[@tudanluo]

Let numerator = [sqrt(3)+2*sqrt(5)+sqrt(7)] * [p*sqrt(3)+q*sqrt(5)+r*sqrt(7)], then p=r=0.5, q=0.

[@jpolowin0]

That was my approach also.

[@lavoiedereussite922]

good

[@user-mi1lg2ih1b]

Great

[@Zorlig]

Couldn't you have just done long division?

[@mokshithreddy1634]

HATSUP TO YOUR DEDICATION BUT ONE THING FROM ME......... By the methods approximation and calculus can we solve.???‽¿‽ I solve it by approximation..... I got same answer but please solve it calculus

[@marciec6862]

👍🏻

[@minka866]

I couldn't understand the erasing signal.

[@Skandalos]

LOL I solved it exactly the same way including the substitution. What are the odds ...

[@HungNguyen-rj3ek]

Why made it that complicated? We have: question = (sqrt(3)+sqrt(5))*(sqrt(5)+sqrt(7))/((sqrt(3)+sqrt(5))+(sqrt(5)+sqrt(7)) Then we multiply both with (sqrt(5)-sqrt(3))*(sqrt(7)-sqrt(3)), simplify then we have the solution

[@philipwebb8297]

Genius.

[@voltape]

jodidazo el problem!

[@user-ed8zf5zv5j]

I FUCKiNG LOVE MATHS!!!

[@kaderdiaw2604]

🥰

[@Ensign_Cthulhu]

7:23 I would have flipped the order and called it sqrt(7) + sqrt (5). It works out the same, but you don't have to deal with negatives.

[@Teamstudy4595]

Under root7+under root3/2

[@PreMath]

Excellent Jayant You are awesome 👍 Take care dear and stay blessed😃

[@dinaris84]

speed Х2

[@langucarl4212]

Why putting a pretence? Why you can write a answers without any brackets

[@ranveeryadav176]

Wow I can't explain how same we think

[@poseidon.M]

💖

[@PreMath]

So nice of you dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃

[@YusufKhan-od4ku]

I can solve it in more easy way Make a+b+c to the whole square in the numerator

[@theophonchana5025]

x = (Square root of 3 + square root of 7) ÷ 2

[@theophonchana5025]

5 - 7 = [Can't substract] (because 5 is too small to substract by 7)

[@123_gooo6]

Lol... What? 😃

[@osamaabdurrahim9146]

-2

[@junkgum]

You owe 7 dollars and only have 5 so you lose 2 dollars from your credit card.

[@theophonchana5025]

2 × square root of 5

[@PreMath]

Great

[@Michael-wc8xl]

42

[@hernando-d]

👍

[@doghousedon1]

Easy Pease, 42

[@skeetorretard]

at this point in our level of technology whats the use in simplifying.

[@Propane_Acccessories]

There's a few reasons, but I'll say just one. When you are creating code, you want to simplify it as much as possible so that it runs faster and it is concise. If you make an error on your code, it can be hard to find it if you have a bunch of unsimplified algebraic expressions.

[@santoshKumar.7ts]

sir how can you write ✓15=✓3✓5.. ✓3✓5 is written for only 15 as you written for 5 √5√ 5

[@madihagul5702]

Never Do Math without tell how will work in practical Life?

[@mintusaren895]

Primary.

[@umedsingh9304]

Very easy question

[@teambellavsteamalice]

The number of steps is a bit annoying, you should really reduce the number of unnecessary ones! At 3:05 you have six rows. Unless your viewers are a mix of math lovers and toddlers you really only show line 3 (√15 +... = √3√5+...) and line six. You just need to encircle √3 in term 1 and 3 and √5 in term 3 and 4. You immediately see both are multiplied by the same pair. In the second part you can also make life simpler by keeping it shorter: always write the higher root first, so √7+√3. when multiplying with √7-√5 you get 7-5=2 and keep all terms positive later on do explain ab/(a+b) = 1/[1/a+1/b] but immediately write it as 1/[ 1/(√7+√5) + 1/(√5+√3) ] explain (a+b)(a-b) =(a^2-b^2) only once So like you did you can now show 1/(√7+√5) = (√7-√5)/2 and 1/(√5+√3) = (√5-√3)/2 adding you can easily see both are divided by 2 and the √5 drop out, so simply as 1 / [ (√7-√3)/2 ] in one step. you might move the two on top right away. finish again applying the trick of multiplying with (√7+√3)/(√7+√3) getting 7-3=4 in the denominator and divide out the 2 for the answer.

[@lechat165]

too complicated. I solved this in mind by making sqrt(3*5)+sqrt(5*7)+sqrt(3*7)+5 = 1/2 * ( (sqrt(3)+sqrt(5)+sqrt(7))^2 - sqrt(5)^2 ). Then I used formula for difference of squares, and (sqrt(3)+2*sqrt(5)+sqrt(7)) expression just cancelled with the denominator. Done.

[@antikertech157]

Nice solution. I did also something like that very very similar, although just a bit indirect.

[@shiva2usa]

Too lengthy to explain well known facts. (A-B)(A+B) is A^2-B2 and this doesn't need to be invented.

[@shadmanhasan4205]

Real answer = 2.19

[@thomasklugh4345]

Yeah, but can you do all that backwards... 🙄 Seriously, I've often wondered what a word problem would look like for such a set-up as this.

[@davidchiles5331]

Please just leave us alone.

[@theophonchana5025]

3 - 5 = [Can't substract ]

[@jcfos6294]

Pas trivial du tout. Aucun intérêt en mathématiques. Jamais au programme de la licence ni du Master en France. C'est nul.

[@arpansit3155]

first

[@PreMath]

You are awesome Arpan👍 Please keep sharing premath channel with your family and friends. Take care dear and all the best😃 Love and prayers from the USA! 😃

[@jaykumarjoshi5173]

I do not understand any thing what you tought I think I have spoil my time I feel sorry for this video

[@theophonchana5025]

Square root of 15 Square root of 35 Square root of 21 Square root of 3 Square root of 5 Square root of 7

[@mouloudimedlarbi2443]

🇲🇦🍉

[@gordonhall3331]

worked it out in three minutes in head this is pathetic