Suppose $a$ and $b$ are real numbers such that the quadratic polynomial $f(x) = x^2 + ax + b$has no non-negative real roots. Prove that ther exist two polynomials g,h, whose coefficients are non-negative real numbers, such that $$f(x)= \frac{g(x)}{h(x)}$$

for all real numbers $x$.

Irish MO 2007


closed as off-topic by user147263, Daniel W. Farlow, Adam Hughes, Claude Leibovici, Gabriel Romon Apr 3 '15 at 16:49

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If both roots of $f(x)$ are real then take $g(x)=f(x)$ and $h(x)=1$.

If both are complex with non-negative real part, then $f(x)=(x-c+di)(x-c-di)=x^2-2cx+(c^2+d^2)$. Therefore we can take $h(x)=x^2+2cx+(c^2+d^2)$ and $g(x)=f(x)h(x)=\left[x^2+(c^2+d^2)\right]^2+4c^2x^2$.

If both roots have negative real parts then take again $g(x)=f(x)$ and $h(x)=1$.

  • $\begingroup$ This is correct, but you are reusing $a$ and $b$ incorrectly, they are already the coefficients of $f(x)$ - please switch to $c \pm di$ $\endgroup$ – ivancho Apr 3 '15 at 2:46

Let $$ g(x)=f(x) $$ $$ h(x)=1 $$


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