(In reply to
Solution by Brian Smith)
I came to the same conclusion that a double root must occur at x=1 or x=-1 and also decided to work on the case x=-1
If there is indeed a double root at x=-1, then synthetic division by (x+1) twice will yield a remainder that must equal zero. This remainder works out to 6-6a-3b which means b=2a-2
Using this substitution a^2+b^2 = 5a^2-8a+4 which is quadratic so has minimum when a=4/5 and then b=-2/5 and a^2+b^2=4/5.
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Posted by Jer
on 2021-10-02 10:58:34 |
re: Solution
