# Prove that $ax^4 + bx^3 + x^2 + 1 = 0$ always has imaginary solution

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?

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What do you know about $a$ and $b$? If they are both $0$, the solution is real. –  TMM Dec 13 '12 at 16:02
Sorry, the equation was wrong, updated it now –  mathkid Dec 13 '12 at 16:04
$-10x^4+x^2+1=0$ has 2 real and 2 complex solutions? Did you mean "there is always at least 1 imaginary/complex solution"? –  CBenni Dec 13 '12 at 16:08
@CBenni Yup, you are correct - I will update the question –  mathkid Dec 13 '12 at 16:10
Are, $a,b\in\mathbb{C}$? Are you looking for Complex solutions or Imaginary solutions? Remember imaginary numbers have no real part. –  dirty derwin Dec 13 '12 at 16:18

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.