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Mad_e13

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

  • 10 months ago
  • 10 months ago

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  1. dan815
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    u want an equation with roots 3 or -sqrt3 ?

    • 10 months ago
  2. Mad_e13
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    no 3+/- sqrt3

    • 10 months ago
  3. dan815
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    ok

    • 10 months ago
  4. dan815
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    so work backwards from factor form

    • 10 months ago
  5. 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: (r1-x)(r2-x)(r3-x)...(rn-x)

    • 10 months ago
  6. dan815
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    oh wait u mean 3+/-root3

    • 10 months ago
  7. whpalmer4
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    @dan815 2 roots, this is a quadratic, so \(x= 3\pm\sqrt{3}\)

    • 10 months ago
  8. Mad_e13
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    yes

    • 10 months ago
  9. dan815
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    |dw:1371070589089:dw|

    • 10 months ago
  10. dan815
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    |dw:1371070608153:dw|

    • 10 months ago
  11. dan815
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    multiply em out and get ur equation

    • 10 months ago
  12. Mad_e13
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    ok thanks so much

    • 10 months ago
  13. amistre64
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    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 \[\frac12x^2-3x+\frac64\] \[2x^2-12x+6\]

    • 10 months ago
  14. amistre64
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    ugh ... i did a=1 in my determinant :/

    • 10 months ago
  15. amistre64
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    9-4c/2 = 3 9-2c = 3 c = 3

    • 10 months ago
  16. amistre64
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    1/2 x^2 -3x +3 x^2 -6x + 6 sounds better

    • 10 months ago
  17. Mad_e13
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    ok thank you!

    • 10 months ago
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