Roots of a Quartic
Let r 1 , r 2 , r 3 , r 4 be roots of the polynomial x 4 − 4 x 3 + 6 x 2 + 8 x + 1 0 .
r 2 + r 3 + r 4 1 + r 1 + r 3 + r 4 1 + r 1 + r 2 + r 4 1 + r 1 + r 2 + r 3 1 = B A
where A and B are co-prime positive integers. Find A + B .
The answer is 43.
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2 solutions
Nice solution,solved it similarly.
I can't understand your method. How you differentiated both sides. I actually transformed the equation.
This is quite similar to the problem that you shared. i mean the one with the 5th roots of unity
Awesome solution brother......learned a lot from it
I use exactly the same way.. :)
f(x)= (x^2-2x-2)^2+6x^2+6>0 for all x, so I think there should be no roots?
Really nice solution, using log and differentiating then. I started the same way, but used only Vietta theorem. It's longer, but does not require calculus staff. Nice problem. Thanks
by the property of polynomial f'(x)/f(x)=∑1/(x-alphaɻȫ) and put x=4
By Vieta r 1 + r 2 + r 3 + r 4 = 4 The required expression becomes 4 − r 1 1 + 4 − r 2 1 + 4 − r 3 1 + 4 − r 4 1 Now x 4 − 4 x 3 + 6 x 2 + 8 x + 1 0 = ( x − r 1 ) ( x − r 2 ) ( x − r 3 ) ( x − r 4 ) Taking l o g l o g ( x 4 − 4 x 3 + 6 x 2 + 8 x + 1 0 ) = l o g ( x − r 1 ) + l o g ( x − r 2 ) + l o g ( x − r 3 ) + l o g ( x − r 4 ) Differentiating both sides, we finally get, x 4 − 4 x 3 + 6 x 2 + 8 x + 1 0 4 x 3 − 1 2 x 2 + 1 2 x + 8 = x − r 1 1 + x − r 2 1 + x − r 3 1 + x − r 4 1 Putting x = 4 4 − r 1 1 + 4 − r 2 1 + 4 − r 3 1 + 4 − r 4 1 = 2 3 2 0 Hence answer is 2 0 + 2 3 = 4 3