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@@prabaln4752 bhai skm sir ne question thodi na banaya hai question (and i dont know who he is anyways) originally was in an olympiad aur aise question ka koi easy tarika nhi hota bas answer anna chahiye aur diagram bna kr ye tareeka is like of 2-3 steps
Here is the purely algebraic solution if anyone is interested: Label the equations as: X^2 + Y^2 + XY= 25 - (1) Y^2 + Z^2 + ZY = 49 - (2) Z^2 + X^2 + XZ = 64 -(3) Subtract the equation (1) from (2) We get (Z-X)(X+Y+Z) = 24 -(4) Again, Subtract equation (2) from (3) We get (X-Y)(X+Y+Z) = 15 -(5) Dividing equation (4) by equation (5) (Z-X)/(X-Y) = 8/5 Solving, we get 5Z = 13X - 8Y -(6) Substitute the value of Z in equation (3) ((13X-8Y)/5)^2 + ((13X-8Y)/5)*X + X^2 = 64 Simplify to get 259X^2 + 64Y^2 - 248XY= 64*25 -(7) Divide (7) by (1), (259X^2 + 64Y^2 -248XY) / (X^2 + Y^2 + XY) = 64 Now, take Y^2 common from both the numerator and denominator, and label (X/Y) = T wherever it occurs. Hence (259T^2 - 248T + 64)/(T^2 + T + 1) = 64 Simplify to get 195T^2 -312T = 0 As T or X/Y is not equal to zero, hence T = 312/195 =8/5. Therefore, T= X/Y = 8/5 Simply substitute the value of X back in equation (1) to get 129Y^2 = 625 Hence Y = +- 25/sqrt(129) Now as Y is only positive, hence Y = 25/sqrt(129) Hence X = 8Y/5 = 40/sqrt(129) And Z = (13X -8Y)/5 = 64/sqrt(129) Hence X+Y+Z = 129/sqrt(129) = sqrt(129) Therefore (X+Y+Z)^2 - 100 = (sqrt(129))^2 - 100 = 129 - 100 = 29 QED. Note: Refer to Chapter -10 “Miscellaneous Equations” in Hall and Knight’s book: “Higher Algebra” to solve such intriguing problems using only pure algebra. A very similar problem can be found in Art. 137 Example 3.
luckily i have that book, and i was able to deduct the equations 4 and 5 but did not think of solving it because it was not linear and thought their must be another easier way, thank you very much for this solution
My dumb ass thought as eq 1 and eq 2 both have ( x+y+z), it would be a common factor of 24 and 15. Hence I got my answer as 3. The reason it is wrong because x,y,z are real numbers. If the question said x,y,z are integer it would have been correct.
It's about amount of time ...in exam there is hardly 5 mins to commit on a question even if you are only targetting selected questions.... given proper time i think most students can solve it ...
I directly calculated values of x,y,z because three equation and three variables can be solved .Therefore x=40/√129 , y=25/√129 , z=64/√129 so Answer = 29
Taking two equations at a time ek ko dusre mei se subtract kra aur common leke teen eq mili is format ki (y-z)(x+y+z) = constant X+y+z ki term ko Maan liya k aur xyz mei teen equation mil gyi with k. Unhe k ke terms mei nikal liya. c/k type aayegi Saari. X+y+z=k mei put kra k² ki value mil gyi.
im a 2025 batch and literally kal ka class mai we learnt this formula for angle between intersection of circles, and usme humne cos theta ka formula likha tha, our sir said we'd learn the proof towards the end of the year
did it by completing 5,7,8 triangle and labelling it ABC (AB=5,BC=8,CA=7) and then by pure euclid geo creating circumcircle of triangle ABC (lets call the point where lengths x,y,z meet P now extend AP and BP to meet the circumcircle at A' and B' respectively now observe that angle ABC is 60 degrees(by cosine rule) therefore by cyclic quadilaterals (x^2 = yz) now replacing yz by x^2 in equation 2 we get the value of x^2+y^2+z^2 = 49 and by adding all three eq we get the value of xy+yz+zx=40 finally we get (x+y+z)^2=138 (Edit: this was not my first attempt as in my first attempt i just used pure algebra and subtracted eq1 from e2 and so on to individually take out the values of x y and z, as this method will strike by brain first, i did second method just for fun :)
i was thinking ki vaise koi tareeka hoga. I was able to think of the triangle part and then with a little pondering i foudn the area using vector rule and equated to get sum taken two at a time. can you tell me how you used linear algebra
This question tells us that for solving we don't always have to do hardwork smartwork is equally important for time saving. but maza bhut aya aur mai aise hi koi channel maths ke liye hi dhund rha tha, mil gya !
I tried making (x+y+z)² with these 3 equations and easily made it in 2 steps then i got (x+y+z)²= 69 + 3/2{xy+yz+zx} To get the thing in {} I use AM-GM concept as these all are +ve in which i used '=' sign only in place of inequality as at x=y all x² = y² = xy same for the other 2 equation by AM-GM i got the values of xy,yz,zx =25/3,49/3,64/3 on putting them in very first equation i got 38 as my answer Now i don't know what am i doing wrong??
Bhai mai sach bata rha hu bas Nv sir se theory krle aur fiitjee ka module more than sufficient hai bohot ache sawal gai module mein joh iit se match krte hai baaki cengage mat krio thodi irrelevant hai agar 9 10 mein geometry achi ki thi toh kr lena
Elegant way to get x y + y z + z x = 40 Thereafter 2 (x^2 + y^2 + z^2) + x y + y z + z x = 25 + 49 + 64 = 138 2(x + y + z)^2=138 + 3(x y + y z + z x) (x + y + z)^2=69 + 60 (x + y + z)^2 - 100 = 29
this was also solved by PJ sir named amazing manipulation of algebra and Geometry... bhaiya can you solve some selected tough questions from resonance HLP because their solutions of some chapters are not available anywhere..
We can take any 2 equation at a time and subtract,and by taking 2 each time we get 3(x+y+z) =78 So x+y+z =26 (x+y+z)^2 =676 -100= 576. Is anything wrong in my process.?
Bro ese qnas ko 1st approach me kaise solve kare?(Ab ye mat kaho bhai ki practice se aa jaega ya abhi dekha hai toh baad me karloge na, aisa nahi, Ramanujan jaise 1st approach me new sums ko think kaise kare vo pls batao acche se)
Mai literally iske pehle trigonometry padh raha tha.... Isko Maine 1min 30 sec me solve kiya kyunki cosine rule mann me baitha tha 😅😅 thoda sa dekhne ke baad teeno add krke solve kr diya
Well i am a did a bit different i ploted them 😂 just the in the positive quadrant and got something like a lateral surface of cylinder (user desmosfor conforming only x-y plane rest was then confirmed as well) and got ans at intersection
Bhaiya kuch alag tareke se kiye hai Maine 6 line mein answer aa gaya I divided first equation by x2 and last equation by x2 and subtracted both then by logic you get value of x+y+z directly Put this in asked equation and get the answer
@@Knightrider118y-z common lele (x+y+z) aa jayega Saath mei. Usse k lele. Xyz teeno mei equation mil jayegi k ke Saath unhe k ke terms me solve krle. Fir unhe x+y+z=k mei rakh de usse milega k²=constant. Bas k hi to chahiye tha mil gya. Aise hi kra mene bhi
Bhaiya, ek similar question 2 variable me kuchh aisa sa hi numerical value tha rhs me wo maine try kiya tha 9th me but mere se nhi bana tha infact mere school ke keval ek maths teacher se bana tga jinhone IITK se Bsc kiya hua tha
Bhai maine cauchy swartz inequality lagane ki sochi par inequality equate nahi ho pai kya hum us method se solve kar sakte hai? Btw this method is amazing
