if-x-4-x-3-0-S-amp-P-are-amp-real-roots-what-is-value-of-4P-2-S-2-

Question Number 202873 by lorance last updated on 05/Jan/24

i f x^{4} - x - 3 = 0

S & P a r e + & \times r e a l r o o t s

w h a t i s v a l u e o f 4 P^{2} + S^{2} ?

Commented by Frix last updated on 05/Jan/24

x^{4} + p x + q = 0

x_{1} = a

x_{2} = b

x_{3, 4} = - \frac{a + b}{2} \pm \frac{\sqrt{3 a^{2} + 2 a b + 3 b^{2}}}{2} i

\Rightarrow

x^{4} - (a + b) (a^{2} + b^{2}) x + a (a^{2} + a b + b^{2}) b = 0

p = - (a + b) (a^{2} + b^{2}) \land q = a (a^{2} + a b + b^{2}) b

You ask for 4 a^{2} b^{2} + {(a + b)}^{2} = r

Let u = a + b \land v = a b

p = 2 u v - u^{3}

q = u^{2} v - v^{2}

This {doesn}^{'} t lead to any nice solution for

r = u^{2} + 4 v^{2} generally and also not with the

given p = - 1 \land q = - 3 :

2 u v - u^{3} = - 1 \Rightarrow v = \frac{u^{3} - 1}{2 u}

u^{2} v - v^{2} = - 3 \Rightarrow u^{6} + 12 u^{3} - 1 = 0 \Rightarrow

u = \pm \sqrt{\sqrt[3]{\frac{\sqrt{257}}{2} + \frac{1}{2}} - \sqrt[3]{\frac{\sqrt{257}}{2} - \frac{1}{2}}}

r = u^{4} + u^{2} - 2 u + \frac{1}{u^{2}}

No chance for something useable \dots

Answered by Rasheed.Sindhi last updated on 05/Jan/24

l e t a, b a r e r e a l a n d c, d a r e i m a g i n a r y r o o t s .

(x - a) (x - b) (x - c) (x - d) = x^{4} - x - 3 = 0

x^{4} - (a + b + c + d) x^{3} + (a b + a c + a d + b c + b d + c d) x^{2}

- (a b c + a b d + a c d + b c d) x + a b c d = x^{4} - x - 3 = 0

a + b + c + d = 0 \land a b + a c + a d + b c + b d + c d = 0

\land a b c + a b d + a c d + b c d = 1 \land a b c d = - 3

{(a + b + c + d)}^{2} = 0^{2}

a^{2} + b^{2} + c^{2} + d^{2} + 2 (a b + a c + a d + b c + b d + c d) = 0

a^{2} + b^{2} + c^{2} + d^{2} = 0

\dots . .

\dots