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Re: HSC 2013 4U Marathon $I get this:$ \left ( x^{2}+\frac{1}{x^{2}} \right )+a\left ( x+\frac{1}{x} \right )+b=0 x+\frac{1}{x}=\frac{-a\pm \sqrt{a^{2}-4b}}{2}

Re: HSC 2013 4U Marathon If it has at least 1 real root doesn't that mean it should have 2 real roots due to the conjugate complex root theorem and the fact that it is a quartic? Or are we not counting multiplicities here? I am sure my solution is wrong because when I sub a=1 and b=0 I get...

Re: HSC 2013 4U Marathon $I used a very dodgy method lol$ $I wanted to find the discriminant of$ x^{4}+ax^{3}+bx^{2}+ax+1 $to be$ \geq 0 $so I made a=0$ \Rightarrow x^{4}+bx^{2}+1\therefore b^{2}\geq 4\therefore a^{2}+b^{2}\geq 4