Question

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

Answer 1

Answer:

• CD = 4
• DE = 24
• CE = 28

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