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Solving a tricky sum of square roots (Olympiad practice)

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Thanks to Devesh from India for the suggestion! A version of this problem was given in an Olympiad qualifying test.
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Опубликовано:

22 авг 2024

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Комментарии : 768

@camembertdalembert6323 3 года назад

the end can be done in a simpler way. Just add the two starting equations, then factorise. It becomes (x²+y²)(x+y)=365, then replace x+y by 9.

@zarinakhatun5574 3 года назад

Yeah I also solved it in this way

@deerh2o 3 года назад

me too -- surprised when Presh went the direction he did.

@zdrastvutye 3 года назад

yes because 365=182+183 from both equations

@ihti20 3 года назад

Haven't noticed this comment and posted the same. This step is really natural, I started to solve with it. We can also consider formula x³+y³=(x+y)(x²-xy+y²). After adding (x+y)xy we get it. I always begin solving this sort of problems with adding and subtracting expressions, that helps often.

@warfyaa6143 3 года назад

yup, I did this.

@harshitarora778 3 года назад

It's always fun to learn from you fresh limewater!

@newchannelverygood162 3 года назад

Fresh Limewater.... lol

@specialise7477 3 года назад

It is not fresh limewater but Presh Talwalker

@Vermeil- 3 года назад

Lmao

@FaranAiki 3 года назад

@SPECIALISE, no, it is a Gougu Theorem Talk Water.

@newchannelverygood162 3 года назад

@@specialise7477 I know

@homeliving2462 3 года назад

This is perfection,Presh!!

@chakshsigarg5130 3 года назад

Right

@firebag143 3 года назад

Sir's name is Paresh Talwalkar

@firebag143 3 года назад

And yes he is very perfect

@SriNu_CSD 3 года назад

Yes 👌👍

@vlatkomacici7528 3 года назад

And also is gum

@Maon22 3 года назад

Looked difficult but done with simplicity

@farrier2708 3 года назад

Your idea of simplicity differs from mine, FiL, by, at the very least, a factor of a^2+b^2,.

@SriNu_CSD 3 года назад

Perfection 👍

@finmat95 3 года назад

Everything is simple IF YOU KNOW HOW to resolve it

@alphabeta2589 3 года назад

@@abirhossainshanto4900 It was a PRMO problem from India if I am not wrong.2017 and grade 8ths were asked to solve

@NestorAbad 3 года назад

I love these algebra tricks! 😍 I followed the same steps as you, but with a little variation in the ending: When you get x+y=9 (that is √a+√b=9) then you can do a little trick with the initial equations a√a + b√b = 183 and a√b + b√a = 182: if you add them you get a√a + b√b + a√b + b√a = 365, but the left hand side of this equation factors as (a+b)(√a+√b), so you have (a+b)·9 = 365 and a+b = 365/9. Thanks for sharing!

@MindYourDecisions 3 года назад

Excellent!

@AJain-18 3 года назад

Simple. First add them and take out commom. You get : (a + b)(a^.5 + b^.5) = 365 Then add first eqn and three times the second eqn and take the cube root on both side. You get : (a^.5 + b^.5) = 9 So a + b = 365/9 and final ans is 73

@Skandalos 3 года назад

Yea, just play around and explore the opportunities. Of course experience helps.

@ravineemkarolijoshinainital 3 года назад

@Saumya Bhatt Freedom of speech.

@dimitriskontoleon6787 2 месяца назад

Ouaou really cool solution... Yes need experience to just see this, but is really cool one solution.

@sadeekmuhammad3646 3 года назад

I really love the way you animate your videos to make any concept clear 😃

@walexandre9452 3 года назад

It's really impressive! I'd like to know the program he is using for animate his videos.

@gamingmusicandjokesandabit1240 3 года назад

Apparently partway through I thought we already had the sum of the squares of x and y

@BleuSquid 3 года назад

The sum of the squares isn't the same as the square of the sum. e.g. 2^2 + 3^2 = 4 + 9 = 13, (2 + 3)^2 = 5^2 = 25.

@gamingmusicandjokesandabit1240 3 года назад

@@BleuSquidI didn't confuse it, promise 🙂

@gamerb-fz1oj 3 года назад

@@gamingmusicandjokesandabit1240 are you confused with the sum of the cubed x and y? Cos at 2:23 you see: x^3 +y^3 =183

@andreassiouras6607 3 года назад

This is the most excited and hyped math has ever gotten me Thanks Presh

@AliKhanMaths 3 года назад

A super satisfying solution! I strive to be able to solve problems like that, and I share my maths tricks to help others do the same!

@mosesmuchina1308 2 года назад

i enjoy your easy to understand solutions. Thanks Presh for making mathematics fun and tricky too.

@EatThatLogic 3 года назад

Once we substitute sqrt(a) and sqrt(b), it becomes easy to see where it is headed. Nice solution! 😁

@akshaychopra6679 3 года назад

Yeah......Always try to substitute in rational powers....

@shelleyweiss9920 3 года назад

I enjoy your videos so much. I try to solve each problem before watching the solution then compare my solution to yours. On this one I kept laughing as I watched every step I took faithfully presented on the screen! Alternative final steps, once you reach x+y=9, substitute in x=9-y to the equation (x^2)y + x(y^2)=182, and get the quadratic x^2 - 9x + 182/9 = 0. By the quadratic formula, x=14/3 or 13/3 (and solve that y=13/3 or 14/3 respectively), thus x and y are 14/3 and 13/3, and x^2 and y^2 are 169/9 and 196/9... the rest is easy arithmetic. Excellent problem! Loved it!

@laurendoe168 3 года назад

Wow... what a long, but rather simple, route to getting the answer. I loved how substitution eliminated the need to even care that it involved square roots.

@sharpmind2869 3 года назад

I solved it , this is from prmo 2017. Also √a = 14/3 and √b = 13/3 . 👍👍😊😊😉

@JohnSmith-kj2od 3 года назад

It was the first question of the paper lol

@christopherrice4360 3 года назад

Thank You for saying what square root of a and square root of b equal. I went through the entire RU-vid comment box to see if anyone said what those values were. I'm the type of person who needs every single value solved in order to feel like the Math problem was 100% completely finished.

@christopherrice4360 3 года назад

@@marcnye9221 may i please see the actual values of and b by taking the squares of the square root values? I would do it myself but i feel like the calculations might be tricky. Cool how the values of square root of a and square root of b can be reversed.

@marcnye9221 3 года назад

@@christopherrice4360 After deducing that x+y=9 as in the video, next observe that 1=183-182 =(x^3+y^3)-(x*y^2+x^2*y) =(x+y)*(x-y)^2 =9(x-y)^2 and hence x-y=1/3 or x-y=-1/3. Combine these again with x+y=9 and you can solve for x and y individually. Taking squares gives the corresponding a and b values.

@ex-i9708 3 года назад

2:53 from here there is a shorter approach If we pay attention we can find that the value of (a+b)*(√a+√b) factors to be exactly equal to the sum of the given equations and since we calculated the value of √a+√b (which is 9) then we substitute in the equation We get: (a+b)*9 = 182+183 (a+b)=365/9 9/5 *365/9 the nines cancel out and we are left with 365/5 which is 73. Thanks for reading my answer

@void.vision-beyond 3 года назад

O my god😯😯 I'm in class 10th and answer of maths what prof. presh tell is just goes over my head. But, this I understood crystal clear. ✌️✌️ It was just awesome. 🔥🔥Unbelievable let's and put up's. That was really amazing. Thanks professor 🙏🏻🙏🏻 The answer and explaintion was just astonishing. ❤️❤️

@jimschneider799 3 года назад

Nice. I'm not sure if my approach would have counted as simpler or not, but here goes: After making the substitutions sqrt(a) = x and sqrt(b) = y to obtain the system: (1) x^3 + y^3 = (x + y) (x^2 - x y + y^2) = 183 (2) x y^2 + x^2 y = (x + y) x y = 182 Adding (1) and (2) together gives: (3) x^3 + x y^2 + x^2 y + y^3 = (x + y) (x^2 + y^2) = 182+183 = 365 Then, adding (3) and twice (2) together gives: (4) x^3 + 3 x y^2 + 3 x^2 y + y^3 = (x + y)^3 = 365 + 2*182 = 729 Take the cube root of (4) to obtain x + y = 9, then divide (3) by this result to get x^2 + y^2 = a + b = 365/9. Multiply this by 9/5 to get 9/5 (a + b) = 9/5 * 365/9 = 73.// I probably wouldn't have thought of this if I hadn't factored (1) and (2) first, and realized that I could add multiples of (2) to (1) to obtain expressions in terms of x+y and x^2+y^2.

@thekidslife6523 3 года назад

Man really you are some sort of magician you made this tough problem look easy for people like who aren't even good at math.

@bibhuprasadmahananda6986 3 года назад

I directly went on to add both equations to get (a+b)(sqrt(a) +sqrt(b)) = 365. Then by the binomial formula... (sqrt(a) + sqrt(b))^3 = a(sqrt(a)) + b(sqrt(b)) +3(a^2)b + 3a(b^2). From the 2 given equations... We have that (sqrt(a) +sqrt(b))^3 = 729. So, sqrt(a) + sqrt(b) = 9....then substituting it back to our previously derived equation (that we obtained upon adding the 2 equations), we get 9(a+b) = 365...dividing 5 on both sides gives the desired result.... That is, 9(a + b)/5 = 73.

@agnibeshbasu3089 3 года назад

This is from the first stage of Indian Math Olympiad called Pre-RMO

@epikherolol8189 3 года назад

But I think prmo's name changed?

@prathampatel1740 3 года назад

@@epikherolol8189 it was taken as IOQM for 2020, and this year too

@epikherolol8189 3 года назад

@@prathampatel1740 oh ya

@pratikchavan3875 3 года назад

Yeah, its of 2017 :)

@ishaanjain9449 3 года назад

That's why I was thinking that I've seen this before. 😂

@anandk9220 3 года назад

Easy question but involves little solving using sum of cubes identity & componendo-dividendo properties. After obtaining a/b, substituting the value of (a/b) and its square root gives sqrt(b) = 13/3 and accordingly sqrt(a) = 14/3. Then using those values provides answer as 73. All dear friends who aren't comfortable with componendo-dividendo properties, here's the easiest way to solve this problem. Two given equations have first equation as sum of two cubic terms √a and √b, while multiplying second equation by 3 and adding that with first one, actually completes expansion of (√a + √b)^3, which is 729. So, (√a + √b) = 9 Now factorise second equation to get √(ab) = 182/9 Use above values in first equation and divide both sides by 5 to get answer as 73. General solution for RHS of equation 1 and 2 to be 'p' and 'q' respectively with factor in division to be 'k' (as 5 in this case) Answer- (p + q) ÷ k CAN'T GET BETTER THAN THIS !!! 😊😊😊😊😊😊😊😊😊😊 EDIT : I watched video solution after writing this comment. Happy to see it's the same easiest way to understand for all !!!

@ariesmars29 3 года назад

I love when you say "That's the answer" at the end.

@Wittokun 3 года назад

3:52 At this point, instead of squaring the equation, I factorized the sum of cubes in the given first equation and substitute all known values. Surprisingly, I got the same answers! 😁

@colombus2314 3 года назад

Im a pineapple, you have to respect me (None sense comment)

@TR_Arial 2 года назад

That's the fun part in mathematics, there's a lot of different ways to get to the solution.

@josh-ed4ri 10 месяцев назад

A nice alternative starting point is to use the exchange symmetry (x y) to see the two solutions to the system of equations must lie on the line y = -x + c where c is a constant. Plugging this equation into the two given equations leads to c = 9. The rest of the solutions follows in the same way.

@mcmac8027 3 года назад

I'm just happy I found this channel. The fun always starts at the start of every video

@DavesTreeFarm 3 года назад

Such a convoluted process that when he said "That's the answer!"- I couldn't remember the question.

@gaurav7582 3 года назад

This question is from PRMO(pre regional mathematics olympiad) in India 🇮🇳

@prathampatel1740 3 года назад

yeah i was about to say, i remembered doing this exact some question some time ago

@neerugupta6762 3 года назад

What! I did this question. I am in 9th. And I felt as If I improved in maths and able to solve world problems. But you told it came in prmo. Though I tried to take part but my school is not registered.😞. I wished I could give this exam. Can I in 10th?

@prathampatel1740 3 года назад

@@neerugupta6762 PRMO hasn't happened yet and you can still give the exam, and yes you can give it in 10th 11th and 12th and you don't need your school to be registered, just follow Prashant Jain (PJ) sir on yt, and i'm sure you'll learn a lot thorugh PRMO

@somadas4704 3 года назад

Yes exactly

@deveshswami2739 3 года назад

Perfectly done, presh. I do appreciate your efforts and love your channel videos. Keep going.

@VsukraM 3 года назад

How did you answer 2 months ago

@notauser9063 3 года назад

2 months ago ?!?!

@homeliving2462 3 года назад

@@VsukraM maybe he is the special member of the community getting early-access to the videos.

@deveshswami2739 2 года назад

@@VsukraM I'm the one who gave this problem as suggestion to presh to make video on, so before posting on yt, he gave me the link to this video on yt. This is how... AND I'M NOT ANY SPECIAL MEMBER OF COMMUNITY LOL

@Advocatekamalkumarkarmkar 3 года назад

Maths is the language , art , pattern of the universe. Physics is the law of the universe. Chemistry is the reaction and colour of the universe. :)

@mrhatman675 3 года назад

Nope

@segmentsAndCurves 3 года назад

@@mrhatman675 Agree.

@segmentsAndCurves 3 года назад

Math is a set of axioms, don't mistified stuff like that.

@yu5016 3 года назад

cringe

@antoniopedrofalcaolopesmor6095 3 года назад

@@segmentsAndCurves Still it is incredible of how maths, being a product of the human mind, it is so efficient at explaining the universe.

@vikramkumar3677 3 года назад

Indians know that this question appeared in 2017 prmo first stage of mathematical Olympiad of India

@darshraatparsadhraj5583 3 года назад

If only I can present every Mathematics workings like this instead of writing lines of calculations

@Pak_Tri_Mahardika 3 года назад

All the videos on your channel are really cool, may I know what app do you use for presentation?

@davidjames1684 3 года назад

A computer program can solve this rather easily. I just looped a and b to be between 10 and 50, checking 0.1 intervals, and it told me a is close to 18.8 and b is close to 21.8 so then I checked from 18.7 to 18.8 and 21.7 to 21.8 with much higher precision, and I could quickly see that a was becoming asymptotic to 18 7/9 and b was becoming asymptotic to 21 7/9, and therefore, the final answer is 73, and we can check that Sqrt(18 7/9) = 4 1/3 + Sqrt(21 7/9) = 4 2/3 is indeed 9. Notice that a + 3 = b.

@kusumpandey1754 3 года назад

Every time a new question with new trick encourages my interest of maths 😀

@antoniopedrofalcaolopesmor6095 3 года назад

That is the purpose of these videos, I guess, if everybody feels the same way as you, Presh will certainly have a sense of mission accomplished, thank you Presh for your nice videos!!

@kusumpandey1754 3 года назад

@@antoniopedrofalcaolopesmor6095 yes right

@namishbaranwal3522 3 года назад

@@antoniopedrofalcaolopesmor6095 agreed

@iniyangiri2145 3 года назад

Presh i really appreciate your work!

@andregoncalves4083 3 года назад

I solved it in a different way. I matched the top two expressions by adding 1 to the second. The roots will cancel out and the result will be equal to b=a. Then you can solve one of the 2 expressions and discover a and b.

@jursamaj 2 года назад

From the 2 equations, we know a & b must be positive, and we can easily determine that they must be less than 32.24. The 1st equation can be rearranged to find that b=(183-a^1.5)^(2/3). Plugging a & b into the second equation, an easy numerical approximation in a spreadsheet finds solution at 18.7777777778. Sure enough, plugging in a=18+7/9 makes an exactly solution. Plugging 18+7/9 & 21+7/9 into the final equation yields 73. Never underestimate the value of numerical approximation, then trying the obvious exact value.

@jimmykitty 3 года назад

As Always, You're really Awesome!! Love your works ❤❤ love form Bangladesh 🇧🇩 ❤

@Ankit-vn7lq 2 года назад

Hi there, u know anmol and bprp ?

@Ankit-vn7lq 2 года назад

I saw u there

@jimmykitty 2 года назад

@@Ankit-vn7lq Yes!!! *BlackpenRedpen* and *Anmol The Maths Sailor* !! I'm a Math Enthusiast... What's about you?? ☺😊😎

@Ankit-vn7lq 2 года назад

@@jimmykitty same here 😊😎👍

@jimmykitty 2 года назад

@@Ankit-vn7lq Wow!! Nice to meet you 😊😇🌈

@rahulraveendran5529 3 года назад

I love Mathematics. It's my first video. Really enjoyed the learning. Since you are explaining the concept slowly it's easy to catch. Thank u very much. God bless you. 👍

@edal7066 3 года назад

square both the equations and subtract them from one another. you get (a-b)(a^2 - b^2) = 183^2 - 182 ^2. Obviously a=183 and b=182 is a possible solution. (9/5)*(a+b) =657

@deekshanaik2438 2 года назад

I got 333 as my ans.... So maybe the mistake is in our solution

@edal7066 2 года назад

@@deekshanaik2438 you mean (9/5)(a+b)=333 in your case. what are a and b in your calculations? the initial set of equations is nonlinear meaning it need not have unique solution pair (a,b). so if your pair (a,b) satisfies equation and (9/5)(a+b)=333 as you say then it might be one of possibly many solutions - if your calculated correctly of course.

@deekshanaik2438 2 года назад

@@edal7066 ohh that's true....I'll try again.... Thank you for the knowledge!!

@ShriPopeAlKhalifa 3 года назад

Once l became unable to solve the problem I start feeling low, as I think that I'm good at maths

@evancaillat1234 3 года назад

Well its a good things cause it push you toi study more toi ne better from Yesterday !

@ShriPopeAlKhalifa 3 года назад

@@evancaillat1234 Yeah I did think that but then I procrastinate

@ShriPopeAlKhalifa 3 года назад

@Nita Singh And the irony is that on this type of math I wanted to be a mathematican or a physicist

@evancaillat1234 3 года назад

You know to some point you must question what you're doing ans why are you studying thé thing you're studying do you want to be rich ? Do you want to be succefull and possibly gain a high social rank and make you're parents prouds ? Being the best ? And those are all good and understandable motives but they will not get you very far in the World the only thing than Can stop you from procrastinating IS passion and living what you do and i'm not talking about a you juste like what you're doing but really being ready to suffer for this to sacrifice you're Time and possibly you're youth and this IS hard but it's thé only way ( how Can i just write a whole paragraph and talk about procrastinating)

@ShriPopeAlKhalifa 3 года назад

@@evancaillat1234 Yeah you are right in all sense ... I need to work on it, I don't have to waste time in order to do that. And Thanks for giving your valuable time Seriously nobody write so long paragraph to advise you or correct you

@rmschad5234 3 года назад

At first I added the equations and factored into (a+b)(sqrt(a)+sqrt(b)). Then I added the first equation and three times the second, noticed the binomial expansion and solved for sqrt(a)+sqrt(b). Finally I substituted the later into the former and had the answer.

@manishankarkhetani6378 3 года назад

If (a+b)(√a + √b) = 365 Clearly 365=5x73 Then a+ b = 73 If we put this value as a+b we get 131.4 as ans? I don't know where I went wrong

@rakuraa4773 3 года назад

@@manishankarkhetani6378 they have to be real numbers not whole numbers so a+b=365/9 and (a^0.5) +(b^0.5)=9

@subhammondal3796 3 года назад

This question is from 2017 prmo exam of india...it was a question of 2 marks...i have solved it during my preperation...glad to see it here😁😁

@hirasm.b.sinurat9077 3 года назад

That was fun to learn with him. Always found a simple solution

@ved9402 3 года назад

Problem keeps becomes good, Better, best

@udayptp 3 года назад

Which software do u use for making such videos bro please let me know

@brightjovanny 3 года назад

Hello Presh, please which software do you use to make your maths videos?

@karangupta1825 3 года назад

Namaste, Presh. My name is Karan Gupta, I am from Ranchi, India. Could you please try solving this problem and if possible, then make a video: If 2^x=3^y=6^z, then what is the value of: 1)1/x + 1/y 2)x, y and z respectively. I used logarithms and my answers were: 1)1/x + 1/y = 1/{Log(2)Log(3)} 2)x = Log(6)Log(3) y = Log(6)Log(2) z = Log(2)Log(3) I also found that: 1/x + 1/y = 1/z.

@tobiaskyrion6019 3 года назад

Multiplying the second equation with sqrt(a/b) gives a*sqrt(a) + a*sqrt(b) = 182*sqrt(a/b), similarly b*sqrt(a) + b*sqrt(b) = 182*sqrt(b/a). Adding and using the first equation gives 365 = 182*(a + b)/sqrt(ab) (*). Adding the first and the second equation gives (a + b)*(sqrt(a) + sqrt(b)) = 365, i.e. a + 2*sqrt(ab) + b = 365^2/(a + b)^2, which yields sqrt(ab) = (1/2)*(365^2/(a + b)^2 - (a +b)). Substituting sqrt(ab) in(*) gives 365*(1/2)*(365^2/(a + b)^2 - (a +b)) = 182*(a + b), hence 365^3 = 729*(a + b)^3. The real root of the latter is a + b = 365/9, i.e. (9/5)*(a + b) = 73.

@ihti20 3 года назад

Second part of solution can be done easier: just sum initial expressions and get (x²+y²)(x+y)=365. I came up with more complicated solution first, I got quadratic equation with variable √a/√b. It gives the same answer but can't be evaluated without calculator. I tried to find easy way and managed only when I got annoyed by square roots and got rid of them by the same substitution. Then I saw that Del Ferro's structure and it was done.

@mcmac8027 3 года назад

The initial equations a√a + b√b = 183 (E1) and b√a + a√b = 182 (E2) and the required unknown: 9/5 (a + b) = ? (U1) show a close relationship, and luckily they resemble two-point linear equations that usually start by adding them on both sides after substituting x = √a (E3) and y = √b (E4) respectively. Then it follows that a = x² and b = y² Then the sum of equations E1 and E2 is: (x²·x + y²·y) + (x²·y + x·y²) = 183 + 182 It becomes easier now since, after careful rearranging, this resembles the algebraic terms in a binomial expansion of (x + y) to degree 3, except the numerical coefficients do not satisfy this condition. To resolve this, add the terms reflexively to give: (x²·x + y²·y) + 3(x²·y + x·y²) = 183 + 3(182) x³ + 3x²y + 3xy² + y³ = 729, or (x + y)³ = 729 Also, the required unknown 9/5(a + b)=? already gave a clue which directs the focus to the simplified (x + y)³ = 9³ x + y = 9 (E5) Putting back E3 and E4, √a + √b = 9 or 9 = √a + √b (E6) *9 will be useful later It slowly reveals how the problem was constructed, i.e. it evokes the use substitution (oddly between variables and constants) and algebraic factoring in surprisingly simple ways. Finally, substituting x and y back to √a and √b will lead to the useful relationship between the first two equations to help find the value of the required unknown. Applying E1 until E6, U1 becomes: 9/5 (a + b) = (√a + √b)/5 · (a + b) = 1/5 (√a + √b) (a + b) = 1/5 [ (a√a + b√b) + (b√a + a√b) ] = 1/5 [ (183) + (182) ] = 1/5 (365) = 73 ∴ --- The problems presented in this channel requires knowledge and technique combined, while pattern recognition and systematic manipulation come in handy. It is clear how some problems do not require to find the values of some variables.

@atakantrg7895 Месяц назад

there is an alternative solution but harder : sum of both equation (a+b)(sqrta+sqrtb)=365 and (a-b).(sqrta-sqrtb)=1 and after substituting x=a+b and y=a-b we get (xy^2=365) by multiplying last 2 equation and (a-b).(sqrta-sqrtb)=1 turns into sqrt(365/x).(sqrt(x-sqrt(x^2-y^2))=1 and after using xy^2=365 in the last equation we get x^3 =(365^3)/729= 365/9 and hence 9/5x=73

@sudoheckbegula 3 года назад

HI PRESH, PLZ CONTINUE THIS OLYMPIAD PRACTICE SERIES AND PROVIDE SOME HARDER PROBLEMS AS YOU DID EARLIER, THAT MIGHT PROVE TO BE VERY USEFUL FOR THE AUDIENCE

@dharmendrakishor5483 3 года назад

Pj sir student Ntse stage 1 qualified Class 11 th Right ?

@sudoheckbegula 3 года назад

@@dharmendrakishor5483 chilla chilla ke sabko scheme bata do xd

@dharmendrakishor5483 3 года назад

@@sudoheckbegula XD Bhiya ma abhi 10th ma hu Delhi me Kya aap mujhe bhi guide karenge Ntse ke liye? Aur aap PRMO bhi kiye h kya?

@sudoheckbegula 3 года назад

@@dharmendrakishor5483 nhi bhai prmo nhi kiya h aur guidance ke liye to pjsorop unhone higuide kiya tha ntseke liye aur prmo gawwd to wo hai hi

@dharmendrakishor5483 3 года назад

@@sudoheckbegula but pj sir mera phone kabhi nhi uthate and Message ka reply bhi nhi karte, aur es saal unhone ntse ke liye padhaya bhi nhi h! kese baar kru class me baar krte nhi.

@rmela4501 3 года назад

a=196/9 and b=169/9 (also interchangeable, due to symmetry). Credit: wolfram alpha

@p.girijapuchhakayala8855 3 года назад

OOOO...Cool problem. Loved it. THANK u PRESH for bringing us these problems.

@oishiknandi693 3 года назад

First time stopped the video and figured out the answer.. Great content !! 👍

@antoniopedrofalcaolopesmor6095 3 года назад

This was not the easiest problem to come up with the right ideas, congratulate you

@roshannishani1607 3 года назад

Presh sir, Can you please answer to this question? (1) . (3√2-√3) (4√3-√2) (2). 4/7+4√3 (this question is in P upon Q form)

@kanguru_ 11 месяцев назад

Let u=sqrta + sqrtb, v=sqrta-sqrtb; then uv=a-b, u^2+v^2=2(a+b). From the sum and difference of the original 2 equations: uv^2=1 and u(u^2+v^2)/2=365. So u^3/2=365-1/2 and u=9, so u=+-1/3. Then a+b= (81+1/9)/2, so (9/5)*(a+b)= (9/5)*730/2/9=73

@andreisergeyev273 3 года назад

This is rather simple equation, which does not require any tricks. The system of two equations (2.23 sec): x^3 + y^3 = 183 and xy^2 + yx^2 = 182 is a symmetric system with respect x and y. It has a regular solution via the substitution x+y = p and xy = q. After this substitution we get: x^3+y^3= (x+y)(x^2-xy +y^2) = p(p^2-q) = 182 and pq = 183. Thus, p^3 = 182 + 3pq = 729. Therefore, p = 9 and q = 182/9. Then, a+b = x^2 +y^2 = p^2 - 2q = 81 - 364/9 = 365/9.

@PHYSICSSIRJEE 3 года назад

A highly efficient solution 🙂👍🏼

@sudoheckbegula 3 года назад

ARE SIR AAP __/\__ ME NS SIR STUDENT

@kcpal5863 2 года назад

Indians are really best in mathematics.. Thanks dear Devesh.. And dear Paresh sir..

@playingmathswithparul8141 2 года назад

Math Trick for all students ru-vid.com7MPaOMUnjpM?feature=share

@yogtanko 3 года назад

Briliant approach

@au2424 3 года назад

the second part can be easily done. the method given here is bit more complicated. if we add the initial two equations we will get an expression like this (a^(1/2)+b^(1/2))(a+b)=365 from the first part we know, a^(1/2)+b^(1/2)=x+y=9 So, 9(a+b)=365 (9/5)(a+b)=73

@Daniel-ef6gg 3 года назад

After you get x+y=9, the easier way to continue is to add both equations. xxx+xxy+xyy+yyy = xx(x+y)+yy(x+y) = (xx+yy)(x+y) = (a+b)×9=365. Then you can divide both sides to get the answer.

@angeluomo 3 года назад

I worked out that, to satisfy both equations, sqrt(a) and sqrt(b) are equal to 39/9 and 42/9 (and vice-versa). The sum of sqrt(a) and sqrt(b) has to equal 9. I don't know if these are the only solutions for sqrt(a) and sqrt(b), but these definitely work. Working through the first equation, you get 74,088/729 + 59,319/729 = 133,407/729 = 183

@angeluomo 3 года назад

This can be simplified further: sqrt(a) and sqrt(b) are equal to 13/3 and 14/3 (interchangeably).

@pavelkotsev1542 2 года назад

You can subtract both equations. Then on the right side you get 1. Square the result and you get (a-b)^2.(a+b-2sqrt(ab))=1 You can first square both equations and then subtract them. You get (a-b)^2.(a+b)=365 Now we know that 365/(a+b) = 1/(a+b+sqrt(ab)) Which is easily simplified to (a+b)/sqrt(ab) = 365/182 And that is something :) SUPPOSE, just suppose, that a+b=365 AND ab=182^2 182^2=(2.7.13)^2 The only way a+b=365 is if a=13^2/const and b=14^2/const. This const is here because of us WRONGDOING when we SUPPOSED. At the end, it will turn out that const=9 but that's not important right now. Nevertheless, it is true that sqrt(a)/sqrt(b) = 13/14 And that is something BIG :) Now solve as you like - it's easy from now on. Find the actual values. There are no super big numbers from now on and there's also a lot of things that cancel out. Hint - it's easy if you use the second equation, because it has 182 and that's why a lot of things cancel out. More difficult that the solution in the video, however I thin it requires less thinking IF you spot that 182 = 2.7.13 and then you cannot "distribute" those numbers arbitrary between a and b. A tiny little bit of "numbers theory" actually helped a lot :)

@iamjustahair1315 3 года назад

Thanks devesh from india

@azfarahsan 3 года назад

the auto subtitles almost got your name right

@masterhalim8263 Год назад

I like the way you explain it. Students will like this. Thanks a lot.

@shravanichawathe1358 3 года назад

This is the second question from prmo exam 2017 . Prmo exam is the first stage in section of the indian team from students of std 8 to 12

@prabhakar4654 3 года назад

Please solve : Integral from 0 to 1 of {ln(X+1)}/x^2 + 1 dx

@arm9180 3 года назад

These are so satisfying

@PolarO_O 3 года назад

First question can also be answered as 65 for a and 110 for b

@MajdAf 3 года назад

i mean if you substract the two equations you can very simply find that a=b, then replace b by a in the first equation and solve it for a. Then just substitute a and b by their value and get the value of 9/5(a+b).

@mathskafunda4383 2 года назад

By the way, we can also solve for the values of A and B, which are a=196/9 and b=169/9

@BroseMusic 3 года назад

Wow, I never get these right and somehow managed to with a different approach. If you square the first and second equations and subtract the second from the first, you can factor the combined equation to: (a+b)(a-b)²=365. We know that (9/5)(a+b) = x Solve for (a+b) (a+b) = (5/9)x Substitute this into the factored equation we have above and you get: (5/9)x(a-b)²=365. Divide both sides by 5/9 and you end up with: x(a-b)²=73*9 Since 9 is a square, it must divide (a-b)² which means that 73 must divide x. Since 73 is a prime, x must be equal to 73.

@FightAgainstHate Год назад

Add two main equations, taking common and then put value of √a+√b=9 further simplifying equation then divide both side by 9/5 . Finally a+b= 73

@ravitejathatikonda1359 3 года назад

*Assuming a,b as x², y² looks little wierd. Instead we can show the solution in an easy manner too. *Let: 1st equ as ¥ 2nd equation as $ Now, substitute the result from (¥+3$) [Factorise] in (¥+$) [Factorise]. => There you go the answer.

@luffytaro689 3 года назад

This was one of the easiest one to solve I knew how to do it immediately after seeing the problem...

@kcool3483 3 года назад

Cap

@iakhon-thegeniusgamer7288 3 года назад

There is another way of solving this. After adding the two equations, we will get this equation, (a+b)(root a+ root b)=365. From the second equation (Which is b root a+ a root b= 182), we get this equation, root ab (root a+ root b)=182. We will multiply this equation by 2. The new equation will be 2 root ab (root a+ root b)=364. We will add the first and the third newly found equations. So this will be alike this» (a+b) (root a+ root b)+ 2 root ab (root a+ root b)=365+364 or (root a+ root b) (a+ b+ 2 root ab)= 729. a+ b+ 2 root ab is equal to (root a+ root b) square. So the equation will be (root a + root b) (root a+ root b) square= 729 or (root a+ root b) cube= 729 or root a+ root b =9. We know that (a+b) (root a+ root b)=365. We will put 9 at the spot of root a+ root b. So (a+b) 9=365 or a+b=365/9. The question wanted to know the value of 9/5 (a+b). We will put 365/9 at the spot of a+b. So we will multiply the equation by 9/5. 9/5 (a+b)= 9/5 (365/9) and we will see that 9/5 (a+b) is equal to 73.