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ladiesman217

  • 3 years ago

i give medals and fans

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  1. ladiesman217
    • 3 years ago
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    solve \[2x ^{2}-8x=-7\]

  2. .Sam.
    • 3 years ago
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    Apply http://openstudy.com/users/.sam.#/updates/5181023ce4b0aaab28b79617

  3. Shadowys
    • 3 years ago
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    add 7 to both sides. \(2x^2 - 8x +7=0\) then apply the general formula.

  4. ladiesman217
    • 3 years ago
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    \[\frac{ -(-8)\sqrt{8x ^{2}-2x*7} }{ 2*2 }\]

  5. Shadowys
    • 3 years ago
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    it's \(x=\frac{-(-8)\pm \sqrt{(-8)^2-4(2)(7)}}{2(2)}\) though

  6. ladiesman217
    • 3 years ago
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    ok so simplify all that its \[\frac{ 8\sqrt{64-14} }{ 4 }\]

  7. Shadowys
    • 3 years ago
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    u forgot to multiply the 14 by 4? the others are right

  8. ladiesman217
    • 3 years ago
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    so \[\frac{ 8\sqrt{64-56} }{ 4 }\]

  9. Shadowys
    • 3 years ago
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    yup! you've got it. :)

  10. uri
    • 3 years ago
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    Its soo easy :D

  11. .Sam.
    • 3 years ago
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    loool

  12. ladiesman217
    • 3 years ago
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    so simplify that its \[\frac{ 2\sqrt{2} }{ 2 }\]

  13. Shadowys
    • 3 years ago
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    wait. you've missed the plus minus out. \(\frac{8 \pm \sqrt{8}}{4}\)=\(\frac{8 \pm 2\sqrt{2}}{4}=\frac{4 \pm \sqrt{2}}{2}\)

  14. ladiesman217
    • 3 years ago
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    ok thnx

  15. Shadowys
    • 3 years ago
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    you’re welcome :)

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