Is there a simple way to prove that solution for $ax^4 + bx^3 + x^2 + 1 = 0$ always has at least one imaginary root?
Prove that $ax^4 + bx^3 + x^2 + 1 = 0$ always has imaginary solution
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algebra-precalculus
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0What do you know about $a$ and $b$? If they are both $0$, the solution is real. – 2012-12-13
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0Sorry, the equation was wrong, updated it now – 2012-12-13
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0$-10x^4+x^2+1=0$ has 2 real and 2 complex solutions? Did you mean "there is always at least 1 imaginary/complex solution"? – 2012-12-13
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0@CBenni Yup, you are correct - I will update the question – 2012-12-13
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0Are you asking for a proof that there are at least two imaginary roots? It is easy to get two real ones-$a=-1, b=0$ for instance. – 2012-12-13
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0@RossMillikan exactly, I need a proof to say that at least 1 of the 4 roots of the above equation is imaginary. – 2012-12-13
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1Are, $a,b\in\mathbb{C}$? Are you looking for Complex solutions or Imaginary solutions? Remember imaginary numbers have no real part. – 2012-12-13
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0@CameronDerwin Complex solutions. I believe Imaginary numbers would form a subset of it. – 2012-12-13
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0If $x$ is imaginary $x^4$ and $x^2$ will be real, so if coefficients are real then $b=0$ by equating the real and imaginary parts of each side. – 2012-12-13
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0@MarkBennet does it mean that for x to be imaginary b must be equal to $0$? – 2012-12-13
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0Even if $b=0$ the solutions can be complex, not pure imaginary. Please distinguish between these. $x^4+x^2+1=0$ has roots $\pm \frac {\sqrt 3}2 \pm \frac i2$ – 2012-12-13