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DOJO
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#1
Report Thread starter 14 years ago
#1
Find , as surds, the roots of the equation

2(x+1)(x-4)-(x-2)^2=0

I was not that sure on how to approach this question, so I just expanded out of the brackets and then use the quadratic formulae. But still got it wrong.

Any ideas? Thanks
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C4>O7
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#2
Report 14 years ago
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2x^2 - 6x - 8 - x^2 + 4x - 4 = 0
x^2 - 2x -12 = 0
x = (2 +/- rt52)/2
x = 1 +/- rt13
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DOJO
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#3
Report Thread starter 14 years ago
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(Original post by C4>O7)
2x^2 - 6x - 8 - x^2 + 4x - 4 = 0
x^2 - 2x -12 = 0
x = (2 +/- rt52)/2
x = 1 +/- rt13
Thanks, however how did you get sqrt 12, as sqrt52/2 is not 13 but 26
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Christophicus
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#4
Report 14 years ago
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x = (2 +/- rt52)/2
x = [2 +/- root(13*4)]/2
x = [2 +/- 2root13]/2
x = 1 +/- root13
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DOJO
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#5
Report Thread starter 14 years ago
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(Original post by Widowmaker)
x = (2 +/- rt52)/2
x = [2 +/- root(13*4)]/2
x = [2 +/- 2root13]/2
x = 1 +/- root13
Oh so thats what he did, turned it into a surd.
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