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Problem Of The Week #437 October 5th, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

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The cubic equation $x^3+ax^2+bx+c=0$ has real coefficients and three real roots. Show that $a^2-3b\ge 0$ and that $\sqrt{a^2-3b}$ is less than or equal to the difference between the largest and the smallest roots.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
No one answered last week's POTW, (Sadface) but you can read the suggested solution as follows:

Let the three real roots be $r,\,s$ and $t$ such that $r\ge s \ge t$.

$a^2-3b=(r+s+t)^2-3(rs+st+tr)=r^2+s^2+t^2-rs-st-tr=\dfrac{(r-s)^2}{2}+\dfrac{(s-t)^2}{2}+\dfrac{(t-r)^2}{2}\ge 0$, with equality when $r=s=t$.

Also, $r\ge s \ge t$ implies

$(s-r)(s-t)\le 0\\ s^2-rs-st+rt \le 0 \\ (r^2+s^2+t^2)-(rs+st+tr) \le r^2+t^2-2rt \\ a^2-3b \le (r-t)^2 \\ \sqrt{a^2-3b} \le r-t $
 
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