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
1 Answers
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Hint: Assuming $a$ and $b$ are real, it is quite easy to identify the $x$-values at the turning points of this function (left hand side of equation), and also to identify which order they come in and the general shape of the function. It is also very easy to compute the value of the function at one of the turning points. See how far this takes you.