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anonymous

  • 5 years ago

I have a question for using quadratic equation again. I have 3^2-5x-7=0 Can any one help me?

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  1. anonymous
    • 5 years ago
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    Do you mean 3x^2?

  2. anonymous
    • 5 years ago
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    is nt in the form of quadartic

  3. anonymous
    • 5 years ago
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    yes I meant 3x^2-5x-7=0

  4. anonymous
    • 5 years ago
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    (3x+7)(x-1)

  5. anonymous
    • 5 years ago
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    no sorry

  6. anonymous
    • 5 years ago
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    x[1] = 1/6 (5-sqrt(109)) x[2] = 1/6 (5+sqrt(109))

  7. anonymous
    • 5 years ago
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    OK doesn't factorise so use formula: x =( 5 +/- sqrt(25 - 4 times 3 times -7))/3 I get x = (5 +/- sqrt109)/3

  8. anonymous
    • 5 years ago
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    ?

  9. anonymous
    • 5 years ago
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    Do you know the quadratic formula? The solution of ax^2 + bx + c = 0 is \[x = -b +/- \sqrt{(b^2 - 4ac}) \over 2a\]

  10. anonymous
    • 5 years ago
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    yes I know that

  11. anonymous
    • 5 years ago
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    @ suzi20 would it not be this x=-5+sqrt109/6, x=-5-sqrt109/6

  12. anonymous
    • 5 years ago
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    OK so look at suzi20's correct solution and see if you can get the same answer by substituting for a, b and c in the formula, where a = 3, b = -5 and c = -7

  13. anonymous
    • 5 years ago
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    leahp x=-5+sqrt109/6, x=-5-sqrt109/6 is a correct answer

  14. anonymous
    • 5 years ago
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    thank you, again

  15. anonymous
    • 5 years ago
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    I have one more question does the letter i is not also use for the symbol +- sign together

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