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dan815
 one year ago
Best ResponseYou've already chosen the best response.0u want an equation with roots 3 or sqrt3 ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0so work backwards from factor form

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1when 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
 one year ago
Best ResponseYou've already chosen the best response.0oh wait u mean 3+/root3

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 2 roots, this is a quadratic, so \(x= 3\pm\sqrt{3}\)

dan815
 one year ago
Best ResponseYou've already chosen the best response.0multiply em out and get ur equation

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1or \[x=\frac{b\pm\sqrt{b^24ac}}{2a}\] \[x=\frac{3\pm\sqrt{3}}{1}\] a = 1/2 b = 3 94c = 3 c = 6/4 \[\frac12x^23x+\frac64\] \[2x^212x+6\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1ugh ... i did a=1 in my determinant :/

amistre64
 one year ago
Best ResponseYou've already chosen the best response.194c/2 = 3 92c = 3 c = 3

amistre64
 one year ago
Best ResponseYou've already chosen the best response.11/2 x^2 3x +3 x^2 6x + 6 sounds better
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