Solve-for-x-in-i-2-x-3-3-x-2-2x-1-0-ii-x-1-2x-3-x-1-x-3-1-

Question Number 77035 by necxxx last updated on 02/Jan/20

S o l v e f o r x i n :

(i) (2 (x + 3) - 3 (x - 2)) (2 x - 1) ⩾ 0

(i i) (x - 1) (2 x + 3) (x + 1) (x + 3) ⩽ 1

Answered by MJS last updated on 03/Jan/20

(i)

- 2 x^{2} + 25 x - 12 ⩾ 0

- 2 (x - 12) (x - \frac{1}{2}) ⩾ 0

(x - 12) (x - \frac{1}{2}) ⩽ 0 \Rightarrow \frac{1}{2} ⩽ x ⩽ 12

(i i)

2 x^{4} + 9 x^{3} + 7 x^{2} + 9 x - 10 ⩽ 0

zeros at x \approx - 3.03943 \land x \approx 1.02431

\Rightarrow \approx - 3.03943 ⩽ x ⩽\approx 1.02431

Commented by necxxx last updated on 03/Jan/20

T h a n k y o u s o m u c h M J S . H o w e v e r o n e

o f t h e m a j o r p l a c e s I h a d i s s u e i n w a s

i n s o l v i n g q u e s t i o n (i i) . I s t h e r e a w a y t o

s o l v e s u c h q u a r t i c e q u a t i o n s ? P l e a s e h e l p

m e o u t w i t h t h a t . T h a n k y o u s o m u c h s i r .

Commented by MJS last updated on 03/Jan/20

in fact we can only try to find exact solutions

(1) (x - a) (x - b) (x - c) (x - d) = x^{4} + \dots + a b c d

\Rightarrow try all factors of \pm the constant

(2) try to find 2 square factors

x^{4} + {a x}^{3} + {b x}^{2} + c x + d = 0; first let x = z - \frac{a}{4}

leading to z^{4} + {p z}^{2} + q z + r = 0

now try to find (z^{2} - α z - β) (z^{2} + α z - γ) = 0

by solving the system for α, β, γ

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

starting with (i) γ = \dots and (ii) β = \dots this

leads to (iii) a polynome of degree 3 for (α^{2})

which sometimes leads to a “ {nice}^{″} solution

if not, this {doesn}^{'} t help

(3) some pokynomes of degree 4 can be solved

using one of the following “ {models}^{″} . you

have to expand and match the constants

which leads to systems of 4 equations for

α, β, γ, δ . again leading to polynomes of

degree 3 for (α^{2}) or (γ^{2}) which might have

“ {nice}^{″} solutions

(x - α - \sqrt{β}) (x - α + \sqrt{β}) (x - γ - \sqrt{δ}) (x - γ + \sqrt{δ}) = 0

(x - α - \sqrt{β} - \sqrt{γ} - \sqrt{δ}) (x - α - \sqrt{β} + \sqrt{γ} + \sqrt{δ}) (x - α + \sqrt{β} - \sqrt{γ} + \sqrt{δ}) (x - α + \sqrt{β} + \sqrt{γ} - \sqrt{δ}) = 0

(x + α + \sqrt{β} + \sqrt{γ} + \sqrt{δ}) (x + α + \sqrt{β} - \sqrt{γ} - \sqrt{δ}) (x + α - \sqrt{β} + \sqrt{γ} - \sqrt{δ}) (x + α - \sqrt{β} + > - \sqrt{γ} + \sqrt{δ}) = 0

if all these {don}^{'} t work you can only approximate

because {Ferrari}^{'} s solution is always possible

to find but never possible to handle

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