If-x-4-px-3-qx-2-rx-5-0-has-four-real-roots-then-find-the-minimum-value-of-pr-

Question Number 44898 by ajfour last updated on 06/Oct/18

I f x^{4} + {p x}^{3} + {q x}^{2} + r x + 5 = 0

h a s f o u r r e a l r o o t s, t h e n f i n d

t h e m i n i m u m v a l u e o f p r .

Commented by MJS last updated on 06/Oct/18

let me try \dots

Answered by ajfour last updated on 06/Oct/18

x_{1} x_{2} x_{3} x_{4} = 5

Σ x_{1} x_{2} x_{3} = - r

x_{1} + x_{2} + x_{3} + x_{4} = - p

p r = 5 (\frac{1}{x_{4}} + \frac{1}{x_{2}} + \frac{1}{x_{3}} + \frac{1}{x_{4}}) Σ x_{1}

> 5 \times \frac{4}{x_{c}} \times 4 x_{c} (= 80) .

H e n c e m i n (p r) = 80 .

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