let p(x)= x^3 px +q 1) prove that p have double roots⇔ 4p^3 +27q^2 =0 3) let suppose p have 3 real roots differnts prove that 4p^3 +27q^2 <0.


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Question Number 30220 by abdo imad last updated on 18/Feb/18

$l e t p (x) = x^{3} p x + q$ $1) p r o v e t h a t p h a v e d o u b l e r o o t s \Leftrightarrow 4 p^{3} + 27 q^{2} = 0$ $3) l e t s u p p o s e p h a v e 3 r e a l r o o t s d i f f e r n t s p r o v e t h a t$ $4 p^{3} + 27 q^{2} < 0 .$
Commented byabdo imad last updated on 18/Feb/18

$p (x) = x^{3} + p x + q$
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