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Problem Of The Week #420 June 8th, 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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For real numbers $a$ and $b$ that satisfy $a^3+12a^2+49a+69=0$ and $b^3-9b^2+28b-31=0$, find $a+b$.

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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
Congratulations to the following members for their correct solution! (Cool)
1. topsquark
2. castor28
3. kaliprasad
4. Opalg

If $a+b=k$ then $a = k-b$. The equation with solution $-b$ is $x^3 + 9x^2 + 28x + 31 = 0$. Comparing this with the equation for $a$, it looks as though it would be best to write this in terms of $x-1$. Then it becomes $(x-1)^3 + 12(x-1)^2 + 49(x-1) + 69 = 0$. That is exactly the equation satisfied by $a$. So with $x = -b$ and $x-1 = a$ it follows that $-b-1=a$, hence $a+b = -1$.
 
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