D is between C and E, CD = x 2 , CE = 32-2x, and DE = 12x. Find CD, DE, and CE.

Step-by-step explanation:

The segment addition theorem tells you ...

  CD +DE = CE

  x^2 +12x = 32 -2x

Subtract the right side to put this in standard form.

  x^2 +14x -32 = 0

  (x +16)(x -2) = 0

  x = -16 or 2

In order for DE to have a positive length, we must have x > 0. So ...

  CD = x^2 = 2^2 = 4

  DE = 12x = 12(2) = 24

  CE = 32 -2x = 32 -2(2) = 28

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step-by-step explanation: