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dpaInc
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be truthful to yourself.
adnanchowdhury
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|dw:1334914514106:dw|
This is my drawing.
Lumenaire
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Real roots means x-axis intercepts
adnanchowdhury
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So is it 2 and where line y= x+16 intercepts the x axis?
Lumenaire
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adnanchowdhury, bad drawing, it should only touch the x-axis
Lumenaire
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and yes, it is 2 and -16
adnanchowdhury
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Are you sure? because when x=0, y=16?
Lumenaire
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does not make a difference, you want y = 0
Lumenaire
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and y = 0. when x = -16
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|
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
adnanchowdhury
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okay, so there are 3 roots?
Lumenaire
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(x-2)^4 = x+16
so x^4 -8x^3+24x^2-32x+16 = x-16
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16 cancels, and you can factor out x, so one solution is at x = 0
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now, x^3 -8x^2+24x-32 = 0
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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
adnanchowdhury
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and so how do i show this on the graph?
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|dw:1334915153788:dw|
Lumenaire
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like that :-)
adnanchowdhury
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oh i see. Thanks.