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adnanchowdhury

How do I show the real roots in for this graph? http://d.pr/i/COS9

  • one year ago
  • one year ago

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  1. dpaInc
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    be truthful to yourself.

    • one year ago
  2. adnanchowdhury
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    |dw:1334914514106:dw| This is my drawing.

    • one year ago
  3. Lumenaire
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    Real roots means x-axis intercepts

    • one year ago
  4. adnanchowdhury
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    So is it 2 and where line y= x+16 intercepts the x axis?

    • one year ago
  5. Lumenaire
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    adnanchowdhury, bad drawing, it should only touch the x-axis

    • one year ago
  6. Lumenaire
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    and yes, it is 2 and -16

    • one year ago
  7. adnanchowdhury
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    Are you sure? because when x=0, y=16?

    • one year ago
  8. Lumenaire
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    does not make a difference, you want y = 0

    • one year ago
  9. Lumenaire
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    and y = 0. when x = -16

    • one year ago
  10. Lumenaire
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    sorry, just read the question again, the roots are the solutions to the question, which means the intercepts between the 2 graphs: |dw:1334914789804:dw|

    • one year ago
  11. Lumenaire
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    clearly one of the roots is at x = 0 and the only other root is where the line extends to meet the curve

    • one year ago
  12. adnanchowdhury
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    okay, so there are 3 roots?

    • one year ago
  13. Lumenaire
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    (x-2)^4 = x+16 so x^4 -8x^3+24x^2-32x+16 = x-16

    • one year ago
  14. Lumenaire
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    16 cancels, and you can factor out x, so one solution is at x = 0

    • one year ago
  15. Lumenaire
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    now, x^3 -8x^2+24x-32 = 0

    • one year ago
  16. Lumenaire
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    this is a third order polynomial with one x-intercept, so the solution to this equation, and x = 0, are the only 2 roots

    • one year ago
  17. adnanchowdhury
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    and so how do i show this on the graph?

    • one year ago
  18. Lumenaire
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    |dw:1334915153788:dw|

    • one year ago
  19. Lumenaire
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    like that :-)

    • one year ago
  20. adnanchowdhury
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    oh i see. Thanks.

    • one year ago
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