anonymous
  • anonymous
find a quadratic equation having 3 + or - root 3 as roots
Mathematics
katieb
  • katieb
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dan815
  • dan815
u want an equation with roots 3 or -sqrt3 ?
anonymous
  • anonymous
no 3+/- sqrt3
dan815
  • dan815
ok

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dan815
  • dan815
so work backwards from factor form
amistre64
  • amistre64
when given a set of roots: {r1,r2,r3,...,rn} you can set up a poly by subtracting x from each root: (r1-x)(r2-x)(r3-x)...(rn-x)
dan815
  • dan815
oh wait u mean 3+/-root3
whpalmer4
  • whpalmer4
@dan815 2 roots, this is a quadratic, so \(x= 3\pm\sqrt{3}\)
anonymous
  • anonymous
yes
dan815
  • dan815
|dw:1371070589089:dw|
dan815
  • dan815
|dw:1371070608153:dw|
dan815
  • dan815
multiply em out and get ur equation
anonymous
  • anonymous
ok thanks so much
amistre64
  • amistre64
or \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] \[x=\frac{3\pm\sqrt{3}}{1}\] a = 1/2 b = -3 9-4c = 3 c = 6/4 \[\frac12x^2-3x+\frac64\] \[2x^2-12x+6\]
amistre64
  • amistre64
ugh ... i did a=1 in my determinant :/
amistre64
  • amistre64
9-4c/2 = 3 9-2c = 3 c = 3
amistre64
  • amistre64
1/2 x^2 -3x +3 x^2 -6x + 6 sounds better
anonymous
  • anonymous
ok thank you!

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