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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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.