anonymous
  • anonymous
find all real and imaginary solutions to w^4+216w=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[w^4+216w=0\] \[w(w^3+216)=0\] one solution is 0 and 216 is the cube of 6, so another solution is -6
anonymous
  • anonymous
now factor \[w^3+216\] as the sum of two cubes. it is \[(w+6)(w^2-6w+36)\]
anonymous
  • anonymous
the first factor we know give a zero of -6 and the other two we find by using the quadratic formula unless you are supposed to know how to fine the cube roots of -1 using polar form of complex numbers. if not just use the formula.

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anonymous
  • anonymous
\[w^2-6w+36=0\] \[w=\frac{-6+\sqrt{6^2-4\times 36}}{2}\] \[w=\frac{-6+\sqrt{-108}}{2}\] \[w=\frac{-6+6\sqrt{3}i}{2}\] \[w=-3+3\sqrt{3}i\] and of course \[w=-3-3\sqrt{3}i\]

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