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Loiza02go
Answered

Find x so that x-3, 2x-4, and 3x+8 are terms of a geometric sequence.


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Sagot :

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Answer:

x/(x-4) = (3x-8)/x

x^2=(x-4)(3x-8)

=3x^2-20x+32

2x^2-20x+32=0

x^2-10x+16=0

(x-8)(x-2)=0

x=2 or. x=8

Step-by-step explanation:

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