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Adi3

  • one year ago

Will Medal Please help!! Use Quadratic formula to solve for x: (x+2)(x-1) = 5

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  1. anonymous
    • one year ago
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    Start off by putting into \[ax^2+bx+c=0\] form

  2. Adi3
    • one year ago
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    ok, wait let me solve it first

  3. anonymous
    • one year ago
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    Take your time

  4. Adi3
    • one year ago
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    ok, x^2 +1x -7

  5. anonymous
    • one year ago
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    Good now we have the quadratic formula which is the following \[x = \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] where a = 1, b= 1, and c = -7

  6. Adi3
    • one year ago
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    ok, wait again sorry

  7. Adi3
    • one year ago
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    Answer is 2,-1

  8. anonymous
    • one year ago
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    Are you sure?

  9. Adi3
    • one year ago
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    yes

  10. Adi3
    • one year ago
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    no no i am wrong wait

  11. Adi3
    • one year ago
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    the answer is 2.15 and -3.15

  12. Adi3
    • one year ago
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    i took the wrong equation

  13. anonymous
    • one year ago
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    \[x= \frac{ -1 \pm \sqrt{1^2-4(1)(-7)} }{ 2(1) } \implies x = \frac{ -1 \pm \sqrt{29} }{ 2 }\] correct?

  14. Adi3
    • one year ago
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    yes

  15. anonymous
    • one year ago
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    That should \[x \approx -3.19,~~ \text{and}~~ x \approx 2.19\]

  16. Adi3
    • one year ago
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    can i ask you an another quetion pls

  17. anonymous
    • one year ago
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    Sure thing

  18. Adi3
    • one year ago
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    Use Quadratic formula to solve for x: \[x + \frac{ 1 }{ x+2 } = 4\]

  19. anonymous
    • one year ago
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    Same thing as earlier, try putting it into \[ax^2+bx+c=0\]

  20. Adi3
    • one year ago
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    how will i do that

  21. anonymous
    • one year ago
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    You can multiply through by (x+2)

  22. Adi3
    • one year ago
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    ??

  23. anonymous
    • one year ago
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    I mean you can find a common denominator which in this case is (x+2) if you like

  24. Adi3
    • one year ago
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    can we cross multiply

  25. anonymous
    • one year ago
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    You can do as following \[\frac{ 1 }{ x+2 }=4-x \] now you may do as you like

  26. anonymous
    • one year ago
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    \[\frac{ 1 }{ (x+2) } = (4-x)\]

  27. Adi3
    • one year ago
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    ok

  28. Adi3
    • one year ago
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    wait

  29. Adi3
    • one year ago
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    is it = to 4x-x-5??

  30. anonymous
    • one year ago
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    Hmm, not quite, \[\frac{ 1 }{ (x+2) } = (4-x) \implies 1 = (4-x)(x+2)\]

  31. Adi3
    • one year ago
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    yeah then we multiply (4-x)(x+2) = 4x - x +6

  32. Adi3
    • one year ago
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    an we bring it to the other side

  33. Adi3
    • one year ago
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    -4x + x -6 + 1

  34. Adi3
    • one year ago
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    = -4x + x -5 = 0

  35. Adi3
    • one year ago
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    right??

  36. anonymous
    • one year ago
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    Remember, we want our equation in \[ax^2+bx+c=0\] form, so what you wrote, does that make sense? Think of what you did in the previous problem

  37. Adi3
    • one year ago
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    x^2-4x-5

  38. Adi3
    • one year ago
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    = 0

  39. anonymous
    • one year ago
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    \[(4-x)(x+2)=1 \implies 4x+8-x^2-2x =1 \implies 2x+8-x^2=1 \] so we can set this up as \[x^2-2x-7=0\]

  40. Adi3
    • one year ago
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    ok. sorry i was offline, i had a lag

  41. Adi3
    • one year ago
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    answer is = 1.9, -0.9 right??

  42. Adi3
    • one year ago
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    @iambatman

  43. anonymous
    • one year ago
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    Can you show how you got that?

  44. Adi3
    • one year ago
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    ok |dw:1444558479751:dw|

  45. Adi3
    • one year ago
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    |dw:1444558581896:dw|

  46. anonymous
    • one year ago
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    What is a, b, and c?

  47. Adi3
    • one year ago
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    b =-2, c -7

  48. anonymous
    • one year ago
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    \[x = \frac{ -(-2) \pm \sqrt{(-2)^2-4(1)(-7)} }{ 2(1) } = \frac{ 2 \pm \sqrt{32} }{ 2 }\]

  49. anonymous
    • one year ago
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    a = 1, b = -2, c = -7

  50. Adi3
    • one year ago
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    thnxs

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