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