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anonymous

  • one year ago

if -2 is a root of x^3-2x^2-3x+q, then q=?

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  1. Nnesha
    • one year ago
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    root(solution , zeros) is a point where line cross x-axis when y =0 so substitute y for 0 and x for -2 solve for q :)

  2. anonymous
    • one year ago
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    ?

  3. Nnesha
    • one year ago
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    what did you get ?

  4. anonymous
    • one year ago
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    im going to explain it to someone but idk how lol

  5. anonymous
    • one year ago
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    im not sure how you did it

  6. Nnesha
    • one year ago
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    alright roots is a point where line intersect x-axis |dw:1440260821340:dw|

  7. anonymous
    • one year ago
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    i think -10?

  8. anonymous
    • one year ago
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    i replaced -2 with x and solved for q

  9. Nnesha
    • one year ago
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    i didn't get negative 10

  10. dinamix
    • one year ago
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    q its not solution to this formal

  11. Nnesha
    • one year ago
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    q is constant not solution i know that but ti's not *negative 10

  12. Nnesha
    • one year ago
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    you can graph this equation x^3-2x^2-3x-10 graph it if graph of this equation intersect x-axis at -2 then it would be correct ....

  13. dinamix
    • one year ago
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    u find q=10 right

  14. Nnesha
    • one year ago
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    i graphed both equations x^3-2x-3x+10 and -10

  15. Nnesha
    • one year ago
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    http://prntscr.com/87n277

  16. Nnesha
    • one year ago
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    \[\huge\rm (-2)^3-2(-2)^2-3(-2)+q=0\] solve for q

  17. dinamix
    • one year ago
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    q=10

  18. Nnesha
    • one year ago
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    don't give ot direct answer it's against the coc .....

  19. anonymous
    • one year ago
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    tnx no i need solution anyways

  20. Nnesha
    • one year ago
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    solve that equation for q what did you get ?

  21. anonymous
    • one year ago
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    cause i need to explain it to someone

  22. dinamix
    • one year ago
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    lol -8-8+6+q=0 , q-10=0 , q=10

  23. Nnesha
    • one year ago
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    i hope you can explain it now :=)

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