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The real numbers x,y,z satisfy 0<=x<=y<=z<=y . If their squares form an arithmetic progression with common difference 2, determine the minimum possible value of

|x-y|+|y-z|.

 Jan 17, 2021
 #1
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0<=x<=y<=z<=y .      can you clarify this part?

 Jan 17, 2021
 #2
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I meant 0<=x<=y<=z<=4. sorry

 Jan 17, 2021

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