+. @bullet-for-1, Therefore, p(·, k) has either two reals roots of the same sign or two conjugated complex roots And p(·, k) has one real root going to ?sign (a 1 a 2 ) ? = ?? when k goes to 1 + M . Therefore, there exists ? > 0 such that for 1 + M < k < 1 + M + ?, p(·, k) has two negative real roots, and x 0 (k) is greater than both of them. (This is also the case for large enough k due to the parabolic branch mentioned above.) As before, none of them yields a root of f, general, p has conjugated complex roots on some interval [K, L] with K ? 1 + M + ?. But these cannot give rise to any complex zero of f , by Rouché's theorem and a connectedness argument

+. @bullet-for-1, k) has either two reals roots of the same sign or two conjugated complex roots This time p(·, k) has one real root going to ?sign (a 1 a 2 ) ? = +? when k goes to 1 + M . Therefore, there exists ? > 0 such that for 1 + M < k < 1 + M + ?, p(·, k) has two positive real roots, and x 0 (k) is less than both of them. As before, none of them yields a root of f . For larger k, the conclusion that f keeps having no roots follows in the same way as in Case 1a

<. @bullet-for-1-+-m-<-k, (k) p(x 0 (k), k) < 0. Therefore, p(·, k) has positive real roots and x 0 (k) is in between. Only the smallest one yields a (real) root of f, because ( k ? ( 1 ? M 2 ) ) ( x ? x 0 ) < 0

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