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3 solutions
@Samuraiwarm Tsunayoshi Perfect solution, did this problem so many time and could not get it, but i now how to do it now.
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I'm not sure if we can use matrices to solve that.
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You can (use matrices) to solve for the values of x , y , z , w , but the point is that is not necessary.
There are several variants of this question, which involve finding a suitable equation and then manipulating it.
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@Calvin Lin – Yep, I won't use matrices to solve x , y , z , w individually. I was thinking about getting x 2 + y 2 + z 2 + w 2 directly.
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@Samuraiwarm Tsunayoshi – That's how I would do it if I had 1 hour to waste inverting a 4x4 matrix.
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@Finn Hulse – ha ha i tried it and takes a lot of time ->so much time wasted.
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@Mardokay Mosazghi – There's no such thing as wasting time for math lol.
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@Samuraiwarm Tsunayoshi – i mean when you have the easier method why waste time especially on contests. But yeah i agree.
how did you come up with co-efficient of (2) equation? .. shouldnt it be
− ( 2 2 + 4 2 + 6 2 + 8 2 )
the co-efficient of t^3 in (t-1)(t-2)(t-3)(t-4) is negative and it is being multiplied by -1.. so It should be 1+9+25+49 .... the overall co-efficient should be x 2 + y 2 + z 2 + w 2 + 1 + 2 + 4 + 9 AND on the right hand side it should be − ( 2 2 + 4 2 + 6 2 + 8 2 )
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Oh god, I made a mistake a little bit. P ( t ) was supposed to be ( t − 1 ) ( t − 4 ) ( t − 9 ) ( t − 2 5 ) − ( t − 9 ) ( t − 2 5 ) ( t − 4 9 ) x 2 − ( t − 1 ) ( t − 2 5 ) ( t − 4 9 ) y 2 − ( t − 1 ) ( t − 9 ) ( t − 4 9 ) z 2 − ( t − 1 ) ( t − 9 ) ( t − 2 5 ) w 2 . Thank you for pointing me out!
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Taking the previous assumption was not permissible as it resulted in the answer as ( -36 ) which isnt possible.
Amazing solution this one, but I actually solved it in 2 minutes, maybe by accident. Could someone please point me out where my solution could have gone wrong? This is how it goes: As with the first equation we definitely get a positive number(no matter the 1) and in the last definitely a negative, we can assume that in the first y^2=z^2=w^2=0, and then x^2/(2^2-1^2)=1 So x^2= 4-1=3 In the second equation, accordingly, we say that x^2=z^2=w^2=0, and, accordingly, find out that y^2=7 The same thing with the last two, and there we have 3+7+11+15=36 Not a very nice explanation, I would be very grateful if you explained me why my solution actually worked.
this ques took hell of time. after many faults I was finally able to solve this one. Too good question ... !
thnaks Rohan
@Mardokay Mosazghi !!! I posted this problem before. AIME 1984 !!!
i didn't know sorry.
I actually studied this problem just a year ago, twice, and my head exploded, twice.
t − 1 2 x 2 + t − 3 2 y 2 + t − 5 2 z 2 + t − 7 2 w 2 − 1 = 0 for t = 2 2 , 4 2 , 6 2 , 8 2 .
Multiply ( t − 1 ) ( t − 9 ) ( t − 2 5 ) ( t − 4 9 ) both sides.
( t − 9 ) ( t − 2 5 ) ( t − 4 9 ) x 2 + ( t − 1 ) ( t − 2 5 ) ( t − 4 9 ) y 2 + ( t − 1 ) ( t − 9 ) ( t − 4 9 ) z 2 + ( t − 1 ) ( t − 9 ) ( t − 2 5 ) w 2 − ( t − 1 ) ( t − 9 ) ( t − 2 5 ) ( t − 4 9 ) = 0 (1)
Let P ( t ) = ( t − 1 ) ( t − 4 ) ( t − 9 ) ( t − 2 5 ) − ( t − 9 ) ( t − 2 5 ) ( t − 4 9 ) x 2 − ( t − 1 ) ( t − 2 5 ) ( t − 4 9 ) y 2 − ( t − 1 ) ( t − 9 ) ( t − 4 9 ) z 2 − ( t − 1 ) ( t − 9 ) ( t − 2 5 ) w 2
We know that t = 2 2 , 4 2 , 6 2 , 8 2 are the roots of P ( t ) and de g ( P ) = 4 .
Therefore, P ( t ) = ( t − 2 2 ) ( t − 4 2 ) ( t − 6 2 ) ( t − 8 2 ) . (2)
Compare the coefficients of t 3 from (1) and (2) we get
x 2 + y 2 + z 2 + w 2 + 1 + 9 + 2 5 + 4 9 = 2 2 + 4 2 + 6 2 + 8 2
Which gives x 2 + y 2 + z 2 + w 2 = 3 6 ~~~