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ninashabrang

  • one year ago

Solve 2x2 - 8x = -7. I just want to know if my answer is correct. I got x=2+- 2 sqr 2

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  1. swissgirl
    • one year ago
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    \(2x^2-8x+7=0\) \(A=2, B=-8, C=7\) \(\large \frac{-(-8) \pm \sqrt{(-8)^2-4(2)(7)}}{2(2)}\) \(\large \frac{8 \pm \sqrt{64-56}}{4}\) \(\large \frac{8 \pm \sqrt{8}}{4}\) \(\large \frac{8 \pm 2\sqrt{2}}{4}\) \(\large \frac{4\pm \sqrt{2}}{2}\) OR \(\large 2\frac{\pm \sqrt{2}}{2}\)

  2. swissgirl
    • one year ago
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    hmmm are you sure this is the right equation?

  3. ninashabrang
    • one year ago
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    oh sorry these are the answer choices: negative 2 plus or minus square root of 2 negative 2 plus or minus 2 square root of 2 quantity of 2 plus or minus square root of 2 all over 2 2 plus or minus square root of 2 end root over 2

  4. swissgirl
    • one year ago
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    So what do you think?

  5. ninashabrang
    • one year ago
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    im not sure I cant understand what you typed up there..haha

  6. swissgirl
    • one year ago
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    So how did you get your answer?

  7. ninashabrang
    • one year ago
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    my own answer is x=2+-2 sqr 2

  8. swissgirl
    • one year ago
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    How did you get that?

  9. ninashabrang
    • one year ago
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    its long in my notebook haha I just need to know if my answer is correct or not

  10. swissgirl
    • one year ago
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    Your answer is incorrect

  11. swissgirl
    • one year ago
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    Look at my answer above

  12. ninashabrang
    • one year ago
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    OH it shows now! thanks! ill review it and find my mistake!

  13. swissgirl
    • one year ago
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    \( \large 2 \pm \frac{\sqrt{2}}{2}\)

  14. ninashabrang
    • one year ago
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    Thanks!

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