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a123

  • 3 years ago

(x+3)^2 - (3x-1)^2 = x^3 + 4? i've tried this three times and I still get the wrong answer!

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  1. a123
    • 3 years ago
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    \[(x+3)^2 - (3x - 1)^2 = x^3 + 4\]

  2. a123
    • 3 years ago
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    just to clarify

  3. jim_thompson5910
    • 3 years ago
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    unless you know the cubic formula, your best bet is to use a graphing calculator to find the approximate solutions.

  4. a123
    • 3 years ago
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    we're not allowed to use a calculator. the answer is supposed to come out to be - 2/3

  5. a123
    • 3 years ago
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    but I don't know how to get there

  6. jim_thompson5910
    • 3 years ago
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    ok so there's a typo somewhere in the problem then, my guess is that x^3 is one of the typos

  7. a123
    • 3 years ago
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    wait, I made a mistake typing out the problem. it's actually \[(x+3)^3\] as the first one

  8. jim_thompson5910
    • 3 years ago
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    oh ok thx, one sec

  9. a123
    • 3 years ago
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    so \[(x+3)^3 - (3x - 1)^2 = x^3 + 4\]

  10. jim_thompson5910
    • 3 years ago
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    \[\large (x+3)^3- (3x-1)^2 = x^3 + 4\] \[\large x^3+9x^2+27x+27- (9x^2-6x+1) = x^3 + 4\] \[\large x^3+9x^2+27x+27- 9x^2+6x-1 = x^3 + 4\] \[\large x^3+9x^2+27x+27- 9x^2+6x-1 - x^3 - 4=0\] \[\large 33x+22=0\] \[\large 33x=-22\] \[\large x=-\frac{22}{33}\] \[\large x=-\frac{2}{3}\]

  11. a123
    • 3 years ago
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    thank you! i see exactly where I went wrong.

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