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Solve for x in the proportion

Mathematics
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\[\frac {4x^{2}}{8x^{2}-8x} = \frac {2}{x}\]
Last question for my assignment.
Have you learned cross-multiplication? If so: (x)(4x^2) = (2)(8x^2 - 8x) -expand -collect like terms -isolate for x Does that help?

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Other answers:

No.
x = 16 and x = −16
x = 2!
(x)(4x^2) = (2)(8x^2 - 8x) 4x^3 = 16x^2 - 16x 4x^3 - 16x^2 + 16x = 0 4x(x^2 - 4x + 4) = 0 For 4x = 0: x = 0 For x^2 - 4x + 4 = 0: -what multiplies to 4 and adds to -4: -2, -2 x^2 - 4x + 4 = (x - 2)^2 (x - 2)^2 = 0 x - 2 = 0 x = 2
Yes!
and x = 0!
:]
But do you understand!?!?!?!

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