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Question Number 105124 by ajfour last updated on 26/Jul/20

x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0

F i n d x .

Answered by ajfour last updated on 26/Jul/20

l e t (x^{2} + p x + q) (x^{2} + r x + m) = 0

\Rightarrow p + r = a

q + m + p r = b

m p + q r = c

q m = d

L e t m p + (b - m - p r) r = c

\Rightarrow m = \frac{c - r (b - p r)}{p - r}

A n d (b - p r - q) p + q r = c

\Rightarrow q = \frac{c - p (b - p r)}{r - p}

\Rightarrow [\frac{c - r (b - p r)}{p - r}] [\frac{c - p (b - p r)}{r - p}] = d

\Rightarrow c^{2} - a c (b - p r) + p r {(b - p r)}^{2}

+ d (a^{2} - 4 p r) = 0

l e t p r = t \Rightarrow

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

= 0

l e t t = z + s \Rightarrow

(z^{3} + 3 {s z}^{2} + 3 s^{2} z + s^{3}) - 2 b (z^{2} + 2 s z + s^{2})

+ (b^{2} + a c - 4 d) (z + s) + (c^{2} - a b c + a^{2} d) = 0

\Rightarrow

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

+ [s^{3} - 2 {b s}^{2} + s (b^{2} + a c - 4 d) + c^{2} - a b c + a^{2} d]

= 0

I f 3 s = 2 b \Rightarrow

z^{3} + (\frac{4 b^{2}}{3} - \frac{8 b^{2}}{3} + b^{2} + a c - 4 d) z

+ (\frac{8 b^{3}}{27} - \frac{8 b^{3}}{9} + \frac{2 b^{3}}{3} + \frac{2 a b c}{3} - \frac{8 b d}{3}

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

\Rightarrow

z^{3} + (a c - 4 d - \frac{b^{2}}{3}) z + (\frac{2 b^{3}}{27} - \frac{a b c}{3} - \frac{8 b d}{3}

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

f i n d i n g z b y {C a r d a n o}^{'} s f o r m u l a

t = p r = z + \frac{2 b}{3}

a n d a s p + r = a

\Rightarrow p, r = \frac{a}{2} \pm \sqrt{\frac{a^{2}}{4} - z - \frac{2 b}{3}}

q = \frac{d}{m} = \frac{d}{[\frac{c - r (b - p r)}{p - r}]}

N o w,

x^{2} + p x + q = 0

\dots .

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