a-b-c-R-ax-4-bx-3-cx-2-bx-a-0-sulation-set-of-the-equation-

Question Number 123451 by Fikret last updated on 25/Nov/20

a, b, c \in R

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

s u l a t i o n s e t o f t h e e q u a t i o n ?

Answered by TANMAY PANACEA last updated on 25/Nov/20

{a x}^{2} + \frac{a}{x^{2}} + b x + \frac{b}{x} + c = 0

a (x^{2} + \frac{1}{x^{2}}) + b (x + \frac{1}{x}) + c = 0

x + \frac{1}{x} = y

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

{a y}^{2} + b y + c - 2 a = 0

y = \frac{- b \pm \sqrt{b^{2} - 4 a (c - 2 a)}}{2 a} = m a n d n

m = c o n s i d e r i n g + v e s i g n b e f o r e \sqrt{} s i g n

n = c o n s i d e r i n g - v e s i g n b e f o r e \sqrt{} s i g n

x + \frac{1}{x} = m

x^{2} - m x + 1 = 0

x = \frac{m \pm \sqrt{m^{2} - 4}}{2}

x = \frac{n \pm \sqrt{n^{2} - 4}}{2}

Commented by MJS_new last updated on 25/Nov/20

good!

Commented by TANMAY PANACEA last updated on 25/Nov/20

t h a n k y o u s i r

Answered by Bird last updated on 25/Nov/20

e \Rightarrow x^{4} + \frac{b}{a} x^{3} + \frac{c}{a} x^{2} + \frac{b}{a} x + 1 = 0

(a \neq 0) e q u a t i o n a t f o r m

x^{4} + α x^{3} + β x^{2} + α x + 1 = 0 \Rightarrow

x^{2} + α x + β + \frac{α}{x} + \frac{1}{x^{2}} = 0 \Rightarrow

{(x + \frac{1}{x})}^{2} - 2 + α (x + \frac{1}{x}) + β = 0

w e p u t x + \frac{1}{x} = z \Rightarrow

z^{2} + α z + β - 2 = 0

Δ = α^{2} - 4 (β - 2) = α^{2} - 4 β + 8

Δ > 0 \Rightarrow z_{1} = \frac{- α + \sqrt{α^{2} - 4 β + 8}}{2}

z_{2} = \frac{- α - \sqrt{α^{2} - 4 β + 8}}{2}

a f t e r w e s o l v e x + \frac{1}{x} = z_{i}

i f Δ < 0 \Rightarrow z_{1} = \frac{- α + i \sqrt{- Δ}}{2}

z_{2} = \frac{- α - i \sqrt{- Δ}}{2} \dots .

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