Find-all-values-of-k-such-that-the-expression-x-3-kx-2-7x-6-can-be-resolved-into-three-linear-real-factors-

Question Number 205051 by mr W last updated on 06/Mar/24

Find all values of k such that the

expr e ssion x^{3} + {kx}^{2} - 7 x + 6 can be

re s olved into three linear real factors .

Answered by Frix last updated on 06/Mar/24

k < c with c being the real solution of

c^{3} - \frac{49 c^{2}}{24} + \frac{63 c}{2} - \frac{50}{3} = 0

c \approx . 543134153288

Commented by mr W last updated on 06/Mar/24

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Answered by mr W last updated on 07/Mar/24

f (x) = x^{3} + {k x}^{2} - 7 x + 6

f^{'} (x) = 3 x^{2} + 2 k x - 7

a t x = p : f (p) = 0, f^{'} (p) = 0

p^{3} + {k p}^{2} - 7 p + 6 = 0 \dots (i)

3 p^{2} + 2 k p - 7 = 0 \dots (i i)

p^{3} + 7 p - 12 = 0

p = \sqrt[3]{\frac{\sqrt{3945}}{9} + 6} - \sqrt[3]{\frac{\sqrt{3945}}{9} - 6}

k = \frac{1}{2} (\frac{7}{p} - 3 p) \approx 0.54313415

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