anonymous
  • anonymous
How do I show the real roots in for this graph? http://d.pr/i/COS9
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
be truthful to yourself.
anonymous
  • anonymous
|dw:1334914514106:dw| This is my drawing.
anonymous
  • anonymous
Real roots means x-axis intercepts

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anonymous
  • anonymous
So is it 2 and where line y= x+16 intercepts the x axis?
anonymous
  • anonymous
adnanchowdhury, bad drawing, it should only touch the x-axis
anonymous
  • anonymous
and yes, it is 2 and -16
anonymous
  • anonymous
Are you sure? because when x=0, y=16?
anonymous
  • anonymous
does not make a difference, you want y = 0
anonymous
  • anonymous
and y = 0. when x = -16
anonymous
  • anonymous
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|
anonymous
  • anonymous
clearly one of the roots is at x = 0 and the only other root is where the line extends to meet the curve
anonymous
  • anonymous
okay, so there are 3 roots?
anonymous
  • anonymous
(x-2)^4 = x+16 so x^4 -8x^3+24x^2-32x+16 = x-16
anonymous
  • anonymous
16 cancels, and you can factor out x, so one solution is at x = 0
anonymous
  • anonymous
now, x^3 -8x^2+24x-32 = 0
anonymous
  • anonymous
this is a third order polynomial with one x-intercept, so the solution to this equation, and x = 0, are the only 2 roots
anonymous
  • anonymous
and so how do i show this on the graph?
anonymous
  • anonymous
|dw:1334915153788:dw|
anonymous
  • anonymous
like that :-)
anonymous
  • anonymous
oh i see. Thanks.

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