I thought the answer would be: $$(x^2 - 4)(x^2 - 8) = 0$$ but it has $4$ roots the positive and negative values. Which is the correct answer?
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1Your answer is correct. Both $2$ and $2\sqrt{2}$ are roots of your equation, which is a biquadratic. As you observed, they are not the only roots. – N. F. Taussig Feb 23 '15 at 19:15
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If you take biquadratic as a synonym for quartic then any of these work:
$$(x-2)^3(x-2\sqrt{2}) \\ (x-2)^2(x-2\sqrt{2})^2 \\ (x-2)(x-2\sqrt{2})^3$$
But otherwise I don't see how to get only those two roots in a quartic equation without odd powers (the more common definition of biquadratic).
However, as N.F. Taussig points out, what you have is correct, as long as you don't care if there are other roots besides $2$ and $2\sqrt{2}$. Also, it satisfies the more common definition of what a biquadratic equation is.
John
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The equations you created are biquadratic or not? As I see you managed to create (biquadratic) equations with only 2 roots. – prishila Feb 23 '15 at 19:04
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It might be biquadratic, but it might not. I linked to the Wolfram Alpha page in my answer that talks about it. My definition doesn't matter as much as yours (or your professor's, I assume). Regardless, it needs to be degree $4$, which means four roots (including multiplicities). So to get two distinct roots, I have to make one or both multiple roots. – John Feb 23 '15 at 19:07
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In my book a before the x^4 should be different from 0 so IN THIS CASE I assume that in this exercise in any solution will be 2 more roots from the ones needed, am I right? – prishila Feb 23 '15 at 19:14
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