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anonymous

  • one year ago

Solve x2 − 3x = −8

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  1. anonymous
    • one year ago
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    @Michele_Laino

  2. pooja195
    • one year ago
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    a ,b , c values are \[\huge\rm Ax^2+Bx+C=0\] where a =leading coefficient b= leading coefficient c= constant term So we need to add 8 to both sides \[\huge~\rm~ x^2 − 3x+8 \] Now factor from here What numbers multiply to make 8 but add to make -3? if there are none the answer would be prime

  3. anonymous
    • one year ago
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    @ganeshie8 @Hero

  4. pooja195
    • one year ago
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    ?

  5. anonymous
    • one year ago
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    idk what numbers multiply to 8 and add to -3

  6. anonymous
    • one year ago
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    @pooja195

  7. pooja195
    • one year ago
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    Then clearly there is not solution :) reread what i wrote are you given options?

  8. anonymous
    • one year ago
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    no solution isnt an option

  9. pooja195
    • one year ago
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    What are the options?

  10. anonymous
    • one year ago
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    Um I don't know how to right them out

  11. pooja195
    • one year ago
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    Just type them out

  12. pooja195
    • one year ago
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    Or scrnshot

  13. anonymous
    • one year ago
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    okay hold on

  14. anonymous
    • one year ago
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  15. pooja195
    • one year ago
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    I havent learned "i" yet.... @UsukiDoll

  16. anonymous
    • one year ago
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    Use the quadratic formula.

  17. UsukiDoll
    • one year ago
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    HUH?! O_O! that was random

  18. UsukiDoll
    • one year ago
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    yeah use quadratics on this equation... use discriminant formula b^2-4ac to see if we have a perfect square or not perfect square - we can factor not a perfect square - use quadratic formula

  19. anonymous
    • one year ago
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    wait what do i do?

  20. UsukiDoll
    • one year ago
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    \[\huge~\rm~ x^2 − 3x+8 \] where a = 1, b = -3, and c = 8 plug it into the discriminant formula which is b^2-4ac \[\LARGE (-3)^2-4(1)(8) \]

  21. anonymous
    • one year ago
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    okay what next?

  22. UsukiDoll
    • one year ago
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    \[\LARGE (-3)^2-4(1)(8) \] \[\LARGE 9-32 = -23 \] not a perfect square and a negative... ok looks like we're going to have complex roots.

  23. UsukiDoll
    • one year ago
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    now we solved the b^2-4ac already so that makes the quadratic formula a bit easier \[\LARGE \frac{ -b \pm \sqrt{b^2-4ac}}{2a}\]

  24. UsukiDoll
    • one year ago
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    since a = 1, b = -3, and c = 8 \[\LARGE \frac{ -(-3) \pm \sqrt{-23}}{2}\] negatives aren't allowed in the radical...so we write an i which stands for imaginary. \[\LARGE \frac{ 3 \pm \sqrt{23}i}{2}\]

  25. anonymous
    • one year ago
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    so the answer would be b?

  26. UsukiDoll
    • one year ago
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    wait... I haven't looked at that portion yet.

  27. UsukiDoll
    • one year ago
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    .docx hates me :(

  28. anonymous
    • one year ago
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    okay ill try to get it on something else

  29. UsukiDoll
    • one year ago
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    finally got through to it.. it is the second choice.

  30. anonymous
    • one year ago
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    Thank you!!!!

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