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theslytherinhelper Group Title

Algebra 2, Solving Polynomial Equations Determine the zeros of f(x) = x4 – x3 + 7x2 – 9x – 18.

  • one year ago
  • one year ago

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  1. theslytherinhelper Group Title
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    I've gotten most of the work down, if you'd like to see it. I'm stuck on the last step :(

    • one year ago
  2. amistre64 Group Title
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    youre not doing ferrari are you?

    • one year ago
  3. theslytherinhelper Group Title
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    ferrari?

    • one year ago
  4. amistre64 Group Title
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    x4 – x3 + 7x2 – 9x – 18=0 to get rid of the x3 term, let x=(y+1/4) (y+1/4)^4 – (y+1/4)^3 + (y+1/4)^2 – 9(y+1/4) – 18=0 \[y^4+\frac58y^2-\frac{69}8y-\frac{5171}{256}=0\] put the linear parts on the other side \[y^4+\frac58y^2=\frac{69}8y+\frac{5171}{256}\] complete the square on the left by adding 5y^2/8+25/64 to each side \[(y^2+\frac58)^2=\frac58y^2+\frac{69}8y+\frac{5271}{256}\] this is as far as we can go without introducing another variable, like z adding a z into the left side gives us more room to play with, but we have to adjust the right side by the same amount: (a+b+c)^2 = a^2+2ab+b^2+(c^2+2c(a+b)) so lets input the z on the left, and add z^2 + 2z(y^2+5/8) to the right side \[(y^2+\frac58+z)^2=\frac58y^2+\frac{69}8y+\frac{5271}{256}+z^2 + 2z(y^2+\frac58)\] lets clean up the right side to see it in its quadraitc form for y \[(y^2+\frac58+z)^2=(2z+\frac58)y^2+\frac{69}8y+(z^2+\frac{10}{8}z+\frac{5271}{256}) \] solving the right side for yusing the quadratic formula we get: \[\Large y=\frac{-\frac{69}{8}\pm\sqrt{(\frac{69}{8})^2-4(2z+\frac58)(z^2+\frac{10}{8}z+\frac{5271}{256})}}{2(2z+\frac58)}\] if we can get the discriminant to equal zero, for some reason, the method will give us roots ... so \[(\frac{69}{8})^2-4(2z+\frac58)(z^2+\frac{10}{8}z+\frac{5271}{256})=0\] which simplifies (lol) to:\[8z^3+\frac{25}{2}z^2+\frac{5371}{32}z-\frac{11733}{512}=0\]which is just Cardanos method now.

    • one year ago
  5. amistre64 Group Title
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    Cardanos method is to reduse it by removing the z^2; let z=(u-25/48) 8(u-25/48)^3+25(u-25/48)^2/2+5371(u-25/48)/32-11733/512 which reduces to:\[u^3+\frac{484}{24}u=\frac{2918}{216}\] hence: \[u=\sqrt[3]{\sqrt{\frac{(2918/216)^2}{4}+\frac{(484/24)^3}{27}}+\frac{(2918/216)}{2}}\\~~~~~-\sqrt[3]{\sqrt{\frac{(2918/216)^2}{4}+\frac{(484/24)^3}{27}}-\frac{(2918/216)}{2}}\] \[u=...\]

    • one year ago
  6. amistre64 Group Title
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    lol, im pretty sure i might have mistyped a number or two, but if youve never heard of ferrari then im sure theres a simpler way they want you to look at this

    • one year ago
  7. amistre64 Group Title
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    what is your last step?

    • one year ago
  8. theslytherinhelper Group Title
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    Thanks sooo much! The last step is simplifying the equation. ^_^

    • one year ago
  9. amistre64 Group Title
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    im not sure of the method you used, so ill need to know a little more about how you approached this

    • one year ago
  10. amistre64 Group Title
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    my u wasnt all the way simplified \[u^3+\frac{121}{6}u=\frac{1459}{108}\] might make life on my method here a smidge simpler \[u = \sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}=abt~~0.6559\] \[z=u-\frac{25}{48}=\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48}\] plugging this into our (y^2...)^2 = y^2+... setup with the zs gives us \[(y^2+\frac58+z)^2=(2z+\frac58)y^2+\frac{69}8y+(z^2+\frac{10}{8}z+\frac{5271}{256})\] \[(y^2+\frac58+(\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48}))^2\\=(2(\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48})+\frac58)y^2+\frac{69}8y\\+((\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48})^2\\+\frac{10}{8}(\sqrt[3]{\frac{\sqrt{22361}}{8}+\frac{1459}{216}}-\sqrt[3]{\frac{\sqrt{22361}}{8}-\frac{1459}{216}}-\frac{25}{48})+\frac{5271}{256}\]

    • one year ago
  11. amistre64 Group Title
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    im sooo glad for the wolfram :)

    • one year ago
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