x-5-ax-4-cx-2-dx-e-0-let-x-rt-s-t-p-Find-r-s-p-such-that-equation-gets-transformed-to-t-5-Dt-E-0-

Question Number 56641 by ajfour last updated on 20/Mar/19

x^{5} + {ax}^{4} + {cx}^{2} + dx + e = 0

let x = \frac{rt + s}{t + p} . Find r, s, p such

that equation gets transformed

to λ t^{5} + Dt + E = 0 .

Commented by ajfour last updated on 20/Mar/19

Granted this is done, we need to

see if the quintic can perhaps now

be solved . .

let

x^{5} + kx + q = 0

equivalently

(x^{2} + fx + g) (x^{3} + {hx}^{2} + mx + n) = 0

\Rightarrow

x^{5} + (h + f) x^{4} + (m + fh + g) x^{3} +

(n + fm + gh) x^{2} + (fn + gm) x + gn = 0

\Rightarrow

h + f = 0

m + fh + g = 0

n + fm + gh = 0 \dots (i)

fn + gm = k \dots . (ii)

gn = q

from (i)

[\frac{q}{g} - h (h^{2} - g) + gh = 0] \times h

& from (ii)

- \frac{hq}{g} + g (h^{2} - g) = k

Adding

{3 gh}^{2} + g^{2} - h^{4} = k] \times g \dots (I) &

{gh}^{3} - {2 g}^{2} h = q] \times h

Adding

g^{2} h^{2} + g^{3} = gk + hq \dots (II)

from (I)

h^{2} = \frac{3 g \pm \sqrt{{9 g}^{2} - 4 (k - g^{2})}}{2}

\Rightarrow h^{2} = \frac{3 g}{2} \pm \frac{\sqrt{{13 g}^{2} - 4 k}}{2}

while from (II)

h = \frac{q \pm \sqrt{q^{2} - {4 g}^{3} (g^{2} - k)}}{{2 g}^{2}}

\Rightarrow \frac{3 g}{2} \pm \frac{\sqrt{{13 g}^{2} - 4 k}}{2} = {[\frac{q \pm \sqrt{q^{2} - {4 g}^{3} (g^{2} - k)}}{{2 g}^{2}}]}^{2}

+ \times ? % % / Degree 5 is degree 5 / /

Commented by Tawa1 last updated on 20/Mar/19

Wow, God bless you sir .

Answered by ajfour last updated on 20/Mar/19

{(rt + s)}^{5} + a {(rt + s)}^{4} (t + p) + c {(rt + s)}^{2} {(t + p)}^{3}

+ d (rt + s) {(t + p)}^{4} + {(t + p)}^{5} = 0

r^{5} t^{5} + {5 r}^{4} {st}^{4} + {10 r}^{3} s^{2} t^{3} + {10 r}^{2} s^{3} t^{2} +

+ {5 rs}^{4} t + s^{5}

+ a (r^{4} t^{4} + {4 r}^{3} {st}^{3} + {6 r}^{2} s^{2} t^{2} + {4 rs}^{3} t + s^{4}) (t + p)

+ c (r^{2} t^{2} + 2 rst + s^{2}) (t^{3} + {3 pt}^{2} + {3 p}^{2} t + p^{3})

+ d (rt + s) (t^{4} + {4 pt}^{3} + {6 p}^{2} t^{2} + {4 p}^{3} t + p^{4})

+ e (t^{5} + {5 pt}^{4} + {10 p}^{2} t^{3} + {10 p}^{3} t^{2} + {5 p}^{4} t + p^{5})

= 0

\Rightarrow

(r^{5} + {ar}^{4} + {cr}^{2} + dr + e) t^{5} +

({5 r}^{4} s + {apr}^{4} + {4 ar}^{3} s + {3 cpr}^{2} + 2 crs +

4 dpr + ds + 5 ep) t^{4} +

({10 r}^{3} s^{2} + {4 apr}^{3} s + {6 ar}^{2} s^{2} + {3 cp}^{2} r^{2}

+ 6 cprs + {cs}^{2} + {6 dp}^{2} r + 4 dps + {10 ep}^{2}) t^{3}

({10 r}^{2} s^{3} + {6 apr}^{2} s^{2} + {4 ars}^{3} + {cp}^{3} r^{2} + {6 cp}^{2} rs

+ {3 cps}^{2} + {4 dp}^{3} r + {6 dp}^{2} s + {10 ep}^{3}) t^{2}

+ Dt + E = 0

If coefficients of t^{4}, t^{3}, t^{2} are to be zero;

then

{5 r}^{4} s + {apr}^{4} + {4 ar}^{3} s + {3 cpr}^{2} + 2 crs +

4 dpr + ds + 5 pe = 0

&

{10 r}^{3} s^{2} + {4 apr}^{3} s + {6 ar}^{2} s^{2} + {3 cp}^{2} r^{2} +

6 cprs + {cs}^{2} + {6 dp}^{2} r + 4 dps + {10 ep}^{2} = 0

&

{10 r}^{2} s^{3} + {6 apr}^{2} s^{2} + {4 ars}^{3} + {cp}^{3} r^{2} + {6 cp}^{2} rs

+ {3 cps}^{2} + {4 dp}^{3} r + {6 dp}^{2} s + {10 ep}^{3} = 0

\dots . .

Commented by Tawa1 last updated on 20/Mar/19

Wow, God bless you sir . Weldone

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