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\begin{align*} P(z) &= 2z^{4}+5z^{3}+7z^{2}+5z+2 \\ &= 2z^{4}+5z^{3}+\frac{7}{2}z^{2}+\frac{7}{2}z^{2}+5z+2 \\ &= z^{2}(2z^{2}+5z+\frac{7}{2})+(\frac{7}{2}z^{2}+5z+2) \end{align*} \linebreak \text{Both quadratics enclosed in brackets have positive leading coefficient and discriminant equal to}...

Yeah when I originally wrote this question, I knew the roots were easily attainable through division of the equation P(z) = 0 by z^{2} , and since I had my own alternative method for proving the roots aren't real, I wanted to deny the reader the option of solving P(z) = 0 so as to make the...

Yeah I thought so, but otherwise your proof is acceptable.

Are you assuming the points (-2, q(-2)) and (2, q(2)) lie on opposite sides of the axis of symmetry of the parabola y = q(t)? Because if they happen to lie on the same side then the minimum value of q(t) will occur away from these endpoints (but still within the domain of q), and so you needed...

I originally thought of expressing it as a sum of two positive definite quadratics (as opposed to your suggestion of writing it as a product) but was unable to find two that were suitable. By quadratics I mean a quadratic and a biquadratic.

\text{It is given that } P(z) = 2z^{4} + 5z^{3} + 7z^{2} + 5z + 2. \\ \text{To carry out the proof, we will need some preliminary results.} \text{Lemma 1: If } \alpha \text{ is a root of } P, \text{ then } \frac{1}{\alpha} \text{ is also a root}. \\ \text{Proof: See fan96's reply.}...

I believe you are correct which thus completes the proof, though my solution is still different from what you, fan96, and integral95 have proposed, but that which still utilizes some of the ideas you guys have put forth. Well done for still managing to solve it otherwise.

Lol but you were definitely right in observing the implication that the roots occur in reciprocal pairs due to the symmetry of the coefficients, as this is integral to the proof (unless there's some shortcut way I haven't picked up on).

Are you sure? You have not accounted for the possibility that \alpha lies on the unit circle which would give \overline{\alpha} = \alpha^{-1} and \alpha = \overline{\alpha^{-1}} , meaning only two roots are recovered and the possibility that the remaining two being purely real is still...

Lol for a second there I thought your original solution was correct (though I think I spotted the error), but either way it still differs from my proof.

\large \text{Let } P(z) = 2z^{4} + 5z^{3} + 7z^{2} + 5z + 2. \\ \text{ Without solving } P(z) = 0, \text{ prove that } P \text{ has no real roots.}

I found these 2003 ones: http://www.angelfire.com/ab7/fourunit/sol.pdf I think the rest can be found here: http://www.angelfire.com/ab7/fourunit/

Unfortunately these are the only ones I've completed. I wish I had more

Thanks so much :) They took me a bit over a month during semester break. Well, I have been offered a job to produce something similar by a certain company who will publish my work with my name on it so I suppose I have been given recognition already. Thanks for your kind review by the way :)

Hi all, Attached are my set of fully worked solutions for the 2004 HSC MX2 exam. They are complete with detailed explanations, multiple methods to selected questions, and remarks proceeding an answer or result to highlight certain subtleties that students would perhaps miss, or simply to...