Mad_e13
find a quadratic equation having 3 + or  root 3 as roots



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dan815
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u want an equation with roots
3 or sqrt3 ?

Mad_e13
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no 3+/ sqrt3

dan815
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ok

dan815
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so work backwards from factor form

amistre64
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when given a set of roots: {r1,r2,r3,...,rn}
you can set up a poly by subtracting x from each root: (r1x)(r2x)(r3x)...(rnx)

dan815
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oh wait u mean 3+/root3

whpalmer4
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@dan815 2 roots, this is a quadratic, so \(x= 3\pm\sqrt{3}\)

Mad_e13
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yes

dan815
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dw:1371070589089:dw

dan815
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dw:1371070608153:dw

dan815
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multiply em out and get ur equation

Mad_e13
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ok thanks so much

amistre64
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or \[x=\frac{b\pm\sqrt{b^24ac}}{2a}\]
\[x=\frac{3\pm\sqrt{3}}{1}\]
a = 1/2
b = 3
94c = 3
c = 6/4
\[\frac12x^23x+\frac64\]
\[2x^212x+6\]

amistre64
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ugh ... i did a=1 in my determinant :/

amistre64
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94c/2 = 3
92c = 3
c = 3

amistre64
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1/2 x^2 3x +3
x^2 6x + 6 sounds better

Mad_e13
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ok thank you!