Guess The Series
If x , y , z are real and
4 x 2 + 9 y 2 + 1 6 z 2 − 6 x y − 1 2 y z − 8 x z = 0
then x , y , z are in:
(Choose the option that applies to all solutions.)
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3 solutions
How did u get that?
Rearranging the equation we get, 2 ( 2 x − 3 y ) 2 + 2 ( 2 x − 4 z ) 2 + 2 ( 3 y − 4 z ) 2 = 0
Hence the solution of above equation will be: 2 x = 3 y = 4 z
Therefore x,y,z ⇒ x , 3 2 x , 2 x
Clearly in harmonic progression..
Assume, 2x=a , 3y=b , 4z=c , then, we have [ a 2 + b 2 + c 2 = a b + b c + c a ] and this yields a = b = c .
So, 2x = 3y = 4z ⇒ x = 2 k , y = 3 k , z = 4 k , for any real k .
Cleary,
k
2
,
k
3
,
k
4
are in
AP
⇒
2
k
,
3
k
,
4
k
are in
HP
i.e.,
x, y, z
are in HP.
EXcept for the part where I multiplied the entire equation by 2 and then factorized it...did it the same way!!!
true
assume 2x=a, 3y=b & 4z =c so the equation is a^2 + b^2 + c^2 -ab -bc -ca =0 or (a + b + c) (a^2 + b^2 + c^2 -ab -bc -ca) = 0 or a^3 + b^3 + c^3 - 3abc =0 or a^2/bc + b^2/ca + c^2/ab =3 is a HP
I solved like this: ( 2 x − 3 y ) 2 = 4 x 2 + 9 y 2 − 1 2 x y ( 2 x − 4 z ) 2 = 4 x 2 + 1 6 z 2 − 1 6 x z ( 3 y − 4 z ) 2 = 9 y 2 + 1 6 z 2 − 2 4 y z
After adding this 3 equations you get:
( 2 x − 3 y ) 2 + ( 2 x − 4 z ) 2 + ( 3 y − 4 z ) 2 = 2 ( 4 x 2 + 9 y 2 + 1 6 z 2 − 6 x y − 1 2 y z − 8 x z )
From this we get:
( 4 x 2 + 9 y 2 + 1 6 z 2 − 6 x y − 1 2 y z − 8 x z ) = 2 1 ( ( 2 x − 3 y ) 2 + ( 2 x − 4 z ) 2 + ( 3 y − 4 z ) 2 )
If ( 4 x 2 + 9 y 2 + 1 6 z 2 − 6 x y − 1 2 y z − 8 x z ) is zero we get:
2 1 ( ( 2 x − 3 y ) 2 + ( 2 x − 4 z ) 2 + ( 3 y − 4 z ) 2 ) =0
This expression is zero only when all summands are zero. And from this we get:
2 x = 3 y = 4 z ⇒ ( x , 3 2 x , 2 1 x ) so this is harmonic progression...