If x^4 +ax^2 +bx+c=0 ⇒ t^4 +At^2 +B=0 Find A and B.


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Question Number 65062 by ajfour last updated on 24/Jul/19

$I f x^{4} + {a x}^{2} + b x + c = 0$ $\Rightarrow t^{4} + {A t}^{2} + B = 0$ $F i n d A a n d B .$
Commented by MJS last updated on 24/Jul/19

$I once posted that matrix method by$ $Tschirnhaus . . . cannot find it right now,$ $must be somewhere on this forum$ $it should help in this case$
Commented by MJS last updated on 24/Jul/19

$found it : question 54775$
Commented by MJS last updated on 24/Jul/19

${you}^{'} ll have to adapt it to remove the linear$ $term but it looks possible$
Commented by MJS last updated on 25/Jul/19

$. . . you also need to learn how to use the$ $method with a polynome of 4^{th} degree . . .$ $maybe you can find something on the web$
Commented by ajfour last updated on 25/Jul/19

$M j S S i r, t h a n k s f o r y o u r$ $s i n c e r e s u g g e s t i o n s a n d y e s$ $m y m e t h o d i s h o p e l e s s l y$ $c o m p l i c a t e d . .$
	Answered by ajfour last updated on 26/Jul/19

	$l e t x = \frac{p t + q}{t + 1}$ $\Rightarrow p^{4} t^{4} + 4 p^{3} {q t}^{3} + 6 p^{2} q^{2} t^{2} + 4 {p q}^{3} t + q^{4}$ $+ a (p^{2} t^{2} + 2 p q t + q^{2}) (t^{2} + 2 t + 1)$ $+ b (p t + q) (t^{3} + 3 t^{2} + 3 t + 1)$ $+ c (t^{4} + 4 t^{3} + 6 t^{2} + 4 t + 1) = 0$ $\Rightarrow$ $(p^{4} + {a p}^{2} + b p + c) t^{4} +$ $(4 p^{3} q + 2 {a p}^{2} + 2 a p q + 3 b p + b q + 4 c) t^{3}$ $+ (6 p^{2} q^{2} + {a p}^{2} + 4 a p q + {a q}^{2} +$ $3 b p + 3 b q + 6 c) t^{2} +$ $(4 {p q}^{3} + 2 a p q + 2 {a q}^{2} + b p + 3 b q + 4 c) t$ $+ (q^{4} + {a q}^{2} + b q + c) = 0$ $s i n c e c o e f f i c i e n t s o f t^{3}, t h a v e$ $t o b e z e r o, \Rightarrow$ $4 p^{3} q + 2 {a p}^{2} + 2 a p q + 3 b p + b q + 4 c = 0$ $. . . . (I)$ $4 {p q}^{3} + 2 a p q + 2 {a q}^{2} + b p + 3 b q + 4 c = 0$ $. . . . (I I)$ $s u b t r a c t i n g \Rightarrow$ $2 p q (p^{2} - q^{2}) + a (p^{2} - q^{2}) + b (p - q) = 0$ $A n d a s p \neq q, \Rightarrow$ $2 p q (p + q) + a (p + q) + b = 0 . . . (i)$ $A d d i n g (I), (I I)$ $2 p q (p^{2} + q^{2}) + a (p^{2} + q^{2}) + 2 a p q +$ $2 b (p + q) + 4 c = 0 . . . . . (i i)$ $l e t p + q = s, p q = m$ $t r a n s f o r m i n g (i) & (i i)$ $________________________$ $s (2 m + a) + b = 0 . . . . (A)$ $(2 m + a) (s^{2} - 2 m) +$ $2 a m + 2 b s + 4 c = 0 . . . . (B)$ $________________________$ $(A) \Rightarrow m = - (\frac{a s + b}{2 s})$ $S u b s t i t u t i n g i n (B)$ $- \frac{b}{s} (s^{2} + \frac{a s + b}{s}) - \frac{a (a s + b)}{s}$ $+ 2 b s + 4 c = 0$ $\Rightarrow \frac{b (a s + b)}{s^{2}} + \frac{a (a s + b)}{s} = b s + 4 c$ $________________________$ $\Rightarrow {b s}^{3} + (4 c - a^{2}) s^{2} - 2 a b s - b^{2} = 0$ $s i s o b t a i n e d f r o m t h i s e q .$ $m = - (\frac{a s + b}{2 s})$ $p, q a r e t h e n r o o t s o f q u a d r a t i c$ $z^{2} - s z + m = 0$ $________________________$ $N o w$ $(p^{4} + {a p}^{2} + b p + c) t^{4} +$ $+ (6 p^{2} q^{2} + {a p}^{2} + 4 a p q + {a q}^{2} +$ $3 b p + 3 b q + 6 c) t^{2} +$ $+ (q^{4} + {a q}^{2} + b q + c) = 0$ $\Rightarrow$ $t^{4} + (\frac{6 p^{2} q^{2} + {a p}^{2} + 4 a p q + {a q}^{2} + 3 b p + 3 b q + 6 c}{p^{4} + {a p}^{2} + b p + c}) t^{2}$ $+ \frac{q^{4} + {a q}^{2} + b q + c}{p^{4} + {a p}^{2} + b p + c} = 0$ $t^{4} + [\frac{6 p^{2} q^{2} + a {(p + q)}^{2} + 2 a p q + 3 b (p + q)}{p^{4} + {a p}^{2} + b p + c}] t^{2}$ $+ \frac{q^{4} + {a q}^{2} + b q + c}{p^{4} + {a p}^{2} + b p + + c} = 0$ $\Rightarrow t^{4} + [\frac{6 m^{2} + {a s}^{2} + 2 a m + 3 b s}{f (p)}] t^{2}$ $+ \frac{f (q)}{f (p)} = 0$ $x = \frac{p t + q}{t + 1} .$
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