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anonymous

  • 5 years ago

Solve each problem by using the quadratic formula. Write solutions in simplest radical form. 2x^2 + 2x- 1 = 0

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  1. anonymous
    • 5 years ago
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    so ax^2 + bx + c = 0

  2. anonymous
    • 5 years ago
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    \[(-b \pm \sqrt{b^2 - 4 a c}) /2a\]

  3. anonymous
    • 5 years ago
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    just plug in the values your equation is 2x^2 + 2x - 1 = 0 ax^2 + bx + c = 0

  4. anonymous
    • 5 years ago
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    tell me what a = , b = , and c =

  5. anonymous
    • 5 years ago
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    are they a=2 b=2 c= -1 ?

  6. anonymous
    • 5 years ago
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    perfect now plug them into the quadratic equaiton

  7. anonymous
    • 5 years ago
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    and you must rememeber the pm

  8. anonymous
    • 5 years ago
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    pm -> plus and minus

  9. anonymous
    • 5 years ago
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    because you will have two solutions one for a positive square root, and one for a negative square root

  10. anonymous
    • 5 years ago
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    \[-2\pm \sqrt{12} /4\] this is what im getting so far am i on the right track

  11. anonymous
    • 5 years ago
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    yeah. see that plus and minus you have to look at it like this -2 + sqrt(12)/4 and -2 - sqrt(12)/4

  12. anonymous
    • 5 years ago
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    so you will get two answers

  13. anonymous
    • 5 years ago
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    would the first answer be 1 and the second answer be -2

  14. anonymous
    • 5 years ago
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    you will get -1 +- sqrt(3)/2

  15. anonymous
    • 5 years ago
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    i have to go

  16. anonymous
    • 5 years ago
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    you will get -1 +- sqrt(3)/2

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