Let x 1 , x 2 , x 3 , x 4 , and x 5 be real numbers satisfying
K = x 1 − 1 + 2 x 2 − 4 + 3 x 3 − 9 + 4 x 4 − 1 6 + 5 x 5 − 2 5 = 2 x 1 + x 2 + x 3 + x 4 + x 5
Find the value of K 2 .
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Hi - nice problem! I think you have a typo in your solution, though; on the second line, your equation says ( x 5 − 2 5 − 1 ) 2 = 0 from which you infer x 5 = 5 0 . Could you explain how you get from your first step to your second?
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The real equation is actually
( x 1 − 1 − 1 ) 2 + ( x 2 − 4 − 2 ) 2 + ( x 3 − 9 − 3 ) 2 + ( x 4 − 1 6 − 4 ) 2 + ( x 5 − 2 5 − 5 ) 2 = 0
Since squares can’t be negative, all of them have to be equal zero
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Yup, I agree that's what it should say. The solution needs a few edits.
Oh extremely sorry, those are gonna be 1, 2,3,4,5....
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From the given equation, ( x 1 − 2 x 1 − 1 ) + ( x 2 − 4 x 2 − 4 ) + ( x 3 − 6 x 3 − 9 ) + ( x 4 − 8 x 4 − 1 6 ) + ( x 5 − 1 0 x 5 − 2 5 ) =0
( x 1 − 1 − 1 ) 2 + ( x 2 − 4 − 2 ) 2 + ( x 3 − 9 − 3 ) 2 + ( x 4 − 1 6 − 4 ) 2 = ( x 5 − 2 5 − 5 ) 2 =0
So, x 1 =2, x 2 =8, x 3 =18, x 4 =32, x 5 =50.
2 x 1 + x 2 + x 3 + x 4 + x 5 =55, 5 5 2 =3025 .
Hope this helps. :) Special Thanks to Chris Lewis for correcting my mistake.