x-4-16x-3-9x-2-256x-256-0-Find-the-values-of-x-

Question Number 184787 by Mastermind last updated on 11/Jan/23

x^{4} + {16 x}^{3} + {9 x}^{2} + 256 x + 256 = 0

Find the values of x ?

Commented by MJS_new last updated on 11/Jan/23

I explained this many times before

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

(1)

try \pm factors of the constant d

if you {don}^{'} t succeed :

(2)

let x = t - \frac{a}{4} to get

t^{4} + {p t}^{2} + q t + r = 0

{we}^{'} re trying to find 2 square factors

(t^{2} - α t - β) (t^{2} + α t - γ) = 0

\Leftrightarrow

t^{4} - (α^{2} + β + γ) t^{2} + α (γ - β) t + β γ = 0

by comparing the constants we get

{\begin{cases} - (α^{2} + β + γ) = p \\ α (γ - β) = q \\ β γ = r \end{cases}

solve (1) and (2) for β and γ

then insert in (3) and transform to get

{(α^{2})}^{3} + j {(α^{2})}^{2} + k (α^{2}) + l = 0

if this has got at least one nice solution for

α^{2} we get nice solutions for t^{2} - α t - β = 0 and

t^{2} + α t - γ = 0 and also for x . if {there}^{'} s no nice

solution for α^{2} {it}^{'} s better to approximate

for x from the given equation .

Commented by MJS_new last updated on 11/Jan/23

for the given equation we get

x^{4} + 16 x^{3} + 9 x^{2} + 256 x + 256 = 0

x = t - 4

t^{4} + 87 t^{2} + 696 t - 1392 = 0

α = \sqrt{87}

β = - 4 \sqrt{87}

γ = 4 \sqrt{87}

\dots

x_{1, 2} = - 4 - \frac{\sqrt{87}}{2} \pm \frac{\sqrt{87 + 16 \sqrt{87}}}{2}

x_{3, 4} = - \frac{8 - \sqrt{87}}{2} \pm \frac{\sqrt{- 87 + 16 \sqrt{87}}}{2} i

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