Vieta's might be helpful!
Let be a real number satisfying , then find the sum of all possible integer values of for which has 6 real and distinct roots.
The answer is 98.
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1 solution
Well for x+1/x to have real and distinct roots, we must have t > 2 or t < − 2
Now, let the given cubic polynomial be p ( t ) .
For p ( t ) to have 3 real roots(that are distinct), p ( 3 ) p ( − 1 ) < 0 , where 3 and -1 are the roots of p ′ ( t ) .
From this we will get − 5 < c < 2 7 .
p(t) must not cut the x-axis between -2 and 2.
Also,turning points are t=3 and t=-1.Also note that leading coefficient is positive.
Now,using these bits of information, our extremely rough graph should look like the graph above.
From graph,p(-2)>0,p(2)>0 and p(0) also >0.
Thus combining all inequalities, we can easily get c ∈ 2 3 , 2 4 , 2 5 , 2 6 making final answer to be 9 8 .