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careless850

  • 3 years ago

Solve for x in the proportion

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  1. careless850
    • 3 years ago
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    \[\frac {4x^{2}}{8x^{2}-8x} = \frac {2}{x}\]

  2. careless850
    • 3 years ago
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    Last question for my assignment.

  3. redshift
    • 3 years ago
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    Have you learned cross-multiplication? If so: (x)(4x^2) = (2)(8x^2 - 8x) -expand -collect like terms -isolate for x Does that help?

  4. careless850
    • 3 years ago
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    No.

  5. careless850
    • 3 years ago
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    x = 16 and x = −16

  6. careless850
    • 3 years ago
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    x = 2!

  7. redshift
    • 3 years ago
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    (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

  8. careless850
    • 3 years ago
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    Yes!

  9. redshift
    • 3 years ago
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    and x = 0!

  10. careless850
    • 3 years ago
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    :]

  11. redshift
    • 3 years ago
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    But do you understand!?!?!?!

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