Find-the-value-of-x-by-using-Cardon-s-Method-1-x-2-x-1-3-x-2-1-

Question Number 156579 by Mr.D.N. last updated on 12/Oct/21

Find the value of x by using {Cardon}^{'} s Method .

\frac{1}{x} + \frac{2}{x + 1} + \frac{3}{x + 2} = 1

Answered by mr W last updated on 13/Oct/21

\frac{(x + 1) (x + 2) + 2 x (x + 2) + 3 x (x + 1)}{x (x + 1) (x + 2)} = 1

x^{2} + 3 x + 2 + 2 x^{2} + 4 x + 3 x^{2} + 3 x = x^{3} + 3 x^{2} + 2 x

x^{3} - 3 x^{2} - 8 x - 2 = 0

x = t + 1

t^{3} + 3 t^{2} + 3 t + 1 - 3 t^{2} - 6 t - 3 - 8 t - 8 - 2 = 0

t^{3} - 11 t - 12 = 0

t = \frac{2 \sqrt{33}}{3} \sin (- \frac{1}{3} \sin^{- 1} \frac{18 \sqrt{33}}{121} + \frac{2 k π}{3})

\Rightarrow x = 1 + \frac{2 \sqrt{33}}{3} \sin (- \frac{1}{3} \sin^{- 1} \frac{18 \sqrt{33}}{121} + \frac{2 k π}{3}), k = 0, 1, 2

c a r d a n o m e t h o d {c a n}^{'} t b e a p p l i e d!

Commented by mr W last updated on 14/Oct/21

h o w t o s o l v e t^{3} - 11 t - 12 = 0 ?

s e e Q 155945

Commented by Mr.D.N. last updated on 21/Oct/21

thank you mr . W

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