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anonymous

  • 5 years ago

[\sqrt{d ^{2}-19} -2d + 11=0

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  1. bahrom7893
    • 5 years ago
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    wait can u repost the question?

  2. bahrom7893
    • 5 years ago
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    is it the same that i answered?

  3. anonymous
    • 5 years ago
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    no i typed it wrong the first time

  4. bahrom7893
    • 5 years ago
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    So can u repost it, i still dont see the right question

  5. anonymous
    • 5 years ago
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    \[\sqrt{d ^{2}-19}-2d+11=0\]

  6. bahrom7893
    • 5 years ago
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    okay

  7. bahrom7893
    • 5 years ago
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    \[\sqrt{d^2 - 19} = 2d - 11\] Move -2d and 11 to the other side

  8. bahrom7893
    • 5 years ago
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    Square both sides: \[d^2 - 19 = (2d - 11)^2\]

  9. bahrom7893
    • 5 years ago
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    Rewrite the right side as: \[d^2 - 19 = 4d^2 - 44d + 121\]

  10. bahrom7893
    • 5 years ago
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    Rewrite in form of a regular quadratic equation and simplify: \[3d^2 - 44d +130 = 0\]

  11. bahrom7893
    • 5 years ago
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    Can u finish solving for d?

  12. anonymous
    • 5 years ago
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    yes, Do you have time for one more. It is similar with a cube root?

  13. bahrom7893
    • 5 years ago
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    yeah

  14. bahrom7893
    • 5 years ago
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    in that case u cube both sides, same method

  15. anonymous
    • 5 years ago
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    \[\sqrt[3]{8v ^{2}-6v}+1=0\]

  16. bahrom7893
    • 5 years ago
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    so move one over to the right and cube both sides to get: 8v^2 - 6v = (-1)^3 8v^2 - 6v = -1. Solve for V

  17. anonymous
    • 5 years ago
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    I get \[8v^{2}-6v+1\]

  18. bahrom7893
    • 5 years ago
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    yeah 8v^2 - 5v + 1 = 0

  19. bahrom7893
    • 5 years ago
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    solve for v

  20. anonymous
    • 5 years ago
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    thanks

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