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KG1234 Group Title

solve for x: (4x-3)^2=-15

  • one year ago
  • one year ago

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  1. NickR Group Title
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    so (4x-3)^2=(4x-3)(4x-3) right

    • one year ago
  2. KG1234 Group Title
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    yes

    • one year ago
  3. NickR Group Title
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    so take that and foil it

    • one year ago
  4. Hero Group Title
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    (4x-3)(4x-3) = 4x(4x - 3) - 3(4x - 3) = 16x^2 - 12x - 12x +9 = 16x^2 - 24x + 9

    • one year ago
  5. KG1234 Group Title
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    16x^2-24x+24=0

    • one year ago
  6. NickR Group Title
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    but its not = to 0 its =-15

    • one year ago
  7. Hero Group Title
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    (4x-3)(4x-3) = 4x(4x - 3) - 3(4x - 3) = 16x^2 - 12x - 12x +9 = 16x^2 - 24x + 9

    • one year ago
  8. NickR Group Title
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    my bad see you already did that

    • one year ago
  9. NickR Group Title
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    so now is there a common number that they are all divisible by?

    • one year ago
  10. KG1234 Group Title
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    wait im at 16x^2-24x+9=-15 right?

    • one year ago
  11. NickR Group Title
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    ya then add the 15 to get 16x^2-24x+24=0

    • one year ago
  12. KG1234 Group Title
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    okay yeah i did that

    • one year ago
  13. NickR Group Title
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    now find a common denominator

    • one year ago
  14. KG1234 Group Title
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    8?

    • one year ago
  15. NickR Group Title
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    yup

    • one year ago
  16. KG1234 Group Title
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    so does that then turn into 2x^2-3x=3=0?

    • one year ago
  17. KG1234 Group Title
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    2x^2-3x+3 sorry

    • one year ago
  18. Hero Group Title
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    16x^2-24x+9=-15 16x^2 - 24x + 24 = 0 8(2x^2 - 3x + 3) = 0 2x^2 - 3x + 3 = 0 x^2 - 3/2 + 3/2 = 0 x^2 - 3/2x = -3/2 x^2 -3/2x + 9/16 = -3/2 + 9/16 (x - 3/4)^2 = -15/16 Solution is imaginary

    • one year ago
  19. NickR Group Title
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    yes

    • one year ago
  20. KG1234 Group Title
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    then what do i do?

    • one year ago
  21. KG1234 Group Title
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    the answer is 3 + or - the square root of 15 divided by4 but idk how to get it

    • one year ago
  22. NickR Group Title
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    i would do the quadratic formula

    • one year ago
  23. KG1234 Group Title
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    -b + or - the square root of b^2 -4ac divided by 2a?

    • one year ago
  24. Hero Group Title
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    \[\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

    • one year ago
  25. NickR Group Title
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    (-b+-sqrt(b^2-4ac))/2a

    • one year ago
  26. NickR Group Title
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    yup

    • one year ago
  27. KG1234 Group Title
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    i got it thanks to both of you!

    • one year ago
  28. NickR Group Title
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    not a problem

    • one year ago
  29. Hero Group Title
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    You got what? Mind sharing what you got?

    • one year ago
  30. KG1234 Group Title
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    3 + or - the square root of 15 divided by 4

    • one year ago
  31. KG1234 Group Title
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    3+ or - i square root 15 divided by 4 sorry i forgot the "i"

    • one year ago
  32. NickR Group Title
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    yup

    • one year ago
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