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December 21, 2014

December 21, 2014

Posted by **Joey** on Saturday, April 5, 2008 at 4:53pm.

Find a polynomial equation with real coefficients that has the given roots.

4i, sqrt5

my answer: x^4-22x^2+80=0

correct answer: x^4+11x^2-80=0

- Algebra -
**Reiny**, Saturday, April 5, 2008 at 4:59pmIf 4i is a root, there must have been a -4i

and if √5 was a root there must have been a -√5

so the factors were (x-4i)(x+4i)(x+√5)x-√5)

= (x^2 + 16)(x^2-5)

= x^4 - 5x^2 + 16x^2 - 80

= x^4 + 11x^2 - 80

- Algebra -
**bobpursley**, Saturday, April 5, 2008 at 4:59pm(x+4i)(x-4i)(x-sqrt5)(x+sqrt5)

(x^2 + 16)(x^2-5)

x^4 +11x^2 - 80

I dont know what you did.

- Algebra -
**Reiny**, Saturday, April 5, 2008 at 5:05pmWhat is that saying?

something about "great minds ...." .lol

notice even the posing time was the same

- Algebra -

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