Solving a tricky sum of square roots (Olympiad practice)
Вставка
- Опубліковано 8 вер 2024
- Thanks to Devesh from India for the suggestion! A version of this problem was given in an Olympiad qualifying test.
Socratic
socratic.org/q...
Math StackExchange
math.stackexch...
Brainly
brainly.in/que...
Quora
www.quora.com/...
Subscribe: www.youtube.co...
Send me suggestions by email (address at end of many videos). I may not reply but I do consider all ideas!
If you buy from the links below I may receive a commission for sales. (As an Amazon Associate I earn from qualifying purchases.) This has no effect on the price for you.
My Books (worldwide links)
mindyourdecisi...
My Books (US links)
Mind Your Decisions: Five Book Compilation
amzn.to/2pbJ4wR
A collection of 5 books:
"The Joy of Game Theory" rated 4.2/5 stars on 194 reviews
amzn.to/1uQvA20
"The Irrationality Illusion: How To Make Smart Decisions And Overcome Bias" rated 3.9/5 stars on 16 reviews
amzn.to/1o3FaAg
"40 Paradoxes in Logic, Probability, and Game Theory" rated 4/5 stars on 29 reviews
amzn.to/1LOCI4U
"The Best Mental Math Tricks" rated 4.2/5 stars on 54 reviews
amzn.to/18maAdo
"Multiply Numbers By Drawing Lines" rated 4.1/5 stars on 22 reviews
amzn.to/XRm7M4
Mind Your Puzzles: Collection Of Volumes 1 To 3
amzn.to/2mMdrJr
A collection of 3 books:
"Math Puzzles Volume 1" rated 4.4/5 stars on 72 reviews
amzn.to/1GhUUSH
"Math Puzzles Volume 2" rated 4.2/5 stars on 20 reviews
amzn.to/1NKbyCs
"Math Puzzles Volume 3" rated 4.2/5 stars on 16 reviews
amzn.to/1NKbGlp
2017 Shorty Awards Nominee. Mind Your Decisions was nominated in the STEM category (Science, Technology, Engineering, and Math) along with eventual winner Bill Nye; finalists Adam Savage, Dr. Sandra Lee, Simone Giertz, Tim Peake, Unbox Therapy; and other nominees Elon Musk, Gizmoslip, Hope Jahren, Life Noggin, and Nerdwriter.
My Blog
mindyourdecisi...
Twitter
/ preshtalwalkar
Merch
teespring.com/...
Patreon
/ mindyourdecisions
Press
mindyourdecisi...
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.
Yeah I also solved it in this way
me too -- surprised when Presh went the direction he did.
yes because 365=182+183 from both equations
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.
yup, I did this.
It's always fun to learn from you fresh limewater!
Fresh Limewater.... lol
It is not fresh limewater but Presh Talwalker
Lmao
@SPECIALISE, no, it is a Gougu Theorem Talk Water.
@@specialise7477 I know
Looked difficult but done with simplicity
Your idea of simplicity differs from mine, FiL, by, at the very least, a factor of a^2+b^2,.
Perfection 👍
Everything is simple IF YOU KNOW HOW to resolve it
@@abirhossainshanto4900 It was a PRMO problem from India if I am not wrong.2017 and grade 8ths were asked to solve
This is perfection,Presh!!
Right
Sir's name is Paresh Talwalkar
And yes he is very perfect
Yes 👌👍
And also is gum
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!
Excellent!
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
Yea, just play around and explore the opportunities. Of course experience helps.
@Saumya Bhatt Freedom of speech.
Ouaou really cool solution... Yes need experience to just see this, but is really cool one solution.
I really love the way you animate your videos to make any concept clear 😃
It's really impressive! I'd like to know the program he is using for animate his videos.
i enjoy your easy to understand solutions. Thanks Presh for making mathematics fun and tricky too.
This is the most excited and hyped math has ever gotten me
Thanks Presh
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!
Once we substitute sqrt(a) and sqrt(b), it becomes easy to see where it is headed. Nice solution! 😁
Yeah......Always try to substitute in rational powers....
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. ❤️❤️
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
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!
Apparently partway through I thought we already had the sum of the squares of x and y
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.
@@BleuSquidI didn't confuse it, promise 🙂
@@gamingmusicandjokesandabit1240 are you confused with the sum of the cubed x and y?
Cos at 2:23 you see: x^3 +y^3 =183
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.
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.
I solved it , this is from prmo 2017.
Also √a = 14/3 and √b = 13/3 . 👍👍😊😊😉
It was the first question of the paper lol
Thank You for saying what square root of a and square root of b equal. I went through the entire UA-cam 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.
@@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.
@@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.
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.
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.
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 !!!
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.
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! 😁
Im a pineapple, you have to respect me
(None sense comment)
That's the fun part in mathematics, there's a lot of different ways to get to the solution.
This is from the first stage of Indian Math Olympiad called Pre-RMO
But I think prmo's name changed?
@@epikherolol8189 it was taken as IOQM for 2020, and this year too
@@prathampatel1740 oh ya
Yeah, its of 2017 :)
That's why I was thinking that I've seen this before. 😂
I love when you say "That's the answer" at the end.
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.
I'm just happy I found this channel. The fun always starts at the start of every video
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.
Presh i really appreciate your work!
If only I can present every Mathematics workings like this instead of writing lines of calculations
Such a convoluted process that when he said "That's the answer!"- I couldn't remember the question.
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.
OOOO...Cool problem. Loved it. THANK u PRESH for bringing us these problems.
Every time a new question with new trick encourages my interest of maths 😀
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!!
@@antoniopedrofalcaolopesmor6095 yes right
@@antoniopedrofalcaolopesmor6095 agreed
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.
Perfectly done, presh.
I do appreciate your efforts and love your channel videos. Keep going.
How did you answer 2 months ago
2 months ago ?!?!
@@VsukraM maybe he is the special member of the community getting early-access to the videos.
.
@@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
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. 👍
As Always, You're really Awesome!! Love your works ❤❤ love form Bangladesh 🇧🇩 ❤
Hi there, u know anmol and bprp ?
I saw u there
@@Ankit-vn7lq Yes!!! *BlackpenRedpen* and *Anmol The Maths Sailor* !! I'm a Math Enthusiast... What's about you?? ☺😊😎
@@jimmykitty same here 😊😎👍
@@Ankit-vn7lq Wow!! Nice to meet you 😊😇🌈
Problem keeps becomes good, Better, best
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
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
Please solve :
Integral from 0 to 1 of {ln(X+1)}/x^2 + 1 dx
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.
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
@@manishankarkhetani6378 they have to be real numbers not whole numbers so a+b=365/9 and (a^0.5) +(b^0.5)=9
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.
Briliant approach
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
I got 333 as my ans.... So maybe the mistake is in our solution
@@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.
@@edal7066 ohh that's true....I'll try again.... Thank you for the knowledge!!
All the videos on your channel are really cool, may I know what app do you use for presentation?
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
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.
Indians know that this question appeared in 2017 prmo first stage of mathematical Olympiad of India
A highly efficient solution 🙂👍🏼
ARE SIR AAP __/\__ ME NS SIR STUDENT
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
This question is from PRMO(pre regional mathematics olympiad) in India 🇮🇳
yeah i was about to say, i remembered doing this exact some question some time ago
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?
@@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
Yes exactly
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.
Indians are really best in mathematics..
Thanks dear Devesh..
And dear Paresh sir..
Math Trick for all students
ua-cam.com/users/shorts7MPaOMUnjpM?feature=share
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.
I like the way you explain it. Students will like this. Thanks a lot.
It's from PRMO 2017 in India
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)
Very beautiful equation
First question can also be answered as 65 for a and 110 for b
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
Pj sir student
Ntse stage 1 qualified
Class 11 th
Right ?
@@dharmendrakishor5483 chilla chilla ke sabko scheme bata do xd
@@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?
@@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
@@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.
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).
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😁😁
That was fun to learn with him. Always found a simple solution
I got the same answer pausing the video. Thanks for the question.
Which software do u use for making such videos bro please let me know
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
This can be simplified further: sqrt(a) and sqrt(b) are equal to 13/3 and 14/3 (interchangeably).
First time stopped the video and figured out the answer.. Great content !! 👍
This was not the easiest problem to come up with the right ideas, congratulate you
Thanks devesh from india
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.
wao,,, what a question..!!... Nice strategy used...(and also beautiful video effects to show whats going on..!!)
a=196/9 and b=169/9 (also interchangeable, due to symmetry). Credit: wolfram alpha
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 :)
We can square on both sides and subtracting we will get a and b directly
?
Great videos!
Hello Presh, please which software do you use to make your maths videos?
Love from India sir
U are great
a delicious problem with a neat solution
Wow , That's awesome!!!
I took a much more ...haphazard... path to the same answer:
made a rebol2 function
f: func[a b][reduce[((a ** 1.5) + (b ** 1.5))((a * square-root b) + (b * square-root a))]]
then guessed at a and b until function f spit out [183 182]
(it took me 41 guesses, but when I guessed it, a was 21.77777777777 and b was 18.77777777777)
after that, the (9 / 5) * (a + b) returned 73
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.
:)
Nope
@@mrhatman675 Agree.
Math is a set of axioms, don't mistified stuff like that.
cringe
@@segmentsAndCurves Still it is incredible of how maths, being a product of the human mind, it is so efficient at explaining the universe.
These are so satisfying
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
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.
the auto subtitles almost got your name right
This was a very good question I solved by pausing the video. Thank you for explaining it really well..
It came in PRMO 2017
Excellent!
Heart filled with joy .Explained very well . A lot of thanks to you , dear professor.
You would have added two equations after finding that √a + √b =9
after adding we get a√a+b√b+a√b+b√a=365
Fatoring we get
(√a+√b)(a+b)=365
9(a+b)=365
9(a+b)/5=73
This was less complicated.
Согласна.
Nice question and solution!
Nice grindy problem, very fun, thank you!
*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.
I CAN'T BELIEVE I DID IT!!!!!
This is great You made my day
Thank you❤
That was WAAYYY easier than I thought!
I dug this the most!
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.
Your videos are amazing. 🤩🤩🤩
Why the unnecessarily long solution? Simply add the two expressions, and factor it.
You get: (a+b)(sqrt(a)+sqrt(b)) = 365.
Observe 365 has only two factors 5 and 73, thus a+b has to be 73, and sqrt(a)+sqrt(b) has to be 5, which is not correct.
However given the extra information of (9/5)(a+b), we can scale (sqrt(a)+sqrt(b)) by (5/9) and get an answer that makes more sense in two steps!
73 can't be the other factor as the sqrt() terms are smaller than the (a+b) terms.
"observe 365 has only only two factors 5 and 73". Why assume factoring over the integers?
@@matthewjames7513 because I can, and assumed that they would want a clean answer for the question. You are correct, the answer need not be integers.