Solving a tricky sum of square roots (Olympiad practice)

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

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[harshitarora]

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

[camembertdalembert]

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.

[Maon]

Looked difficult but done with simplicity

[AJain-]

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

[sadeekmuhammad]

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

[NestorAbad]

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!

[andreassiouras]

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

[mosesmuchina]

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

[AliKhanMaths]

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!

[shelleyweiss]

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!

[ex-i]

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

[laurendoe]

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.

[sharpmind]

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

[jimschneider]

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.

[gamingmusicandjokesandabit]

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

[void.vision-beyond]

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. ❤️❤️

[EatThatLogic]

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

[anandk]

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 !!!

[bibhuprasadmahananda]

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.

[thekidslife]

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.