x-3-1-2-2x-1-1-3-x-

Question Number 91491 by jagoll last updated on 01/May/20

x^{3} + 1 = 2 \sqrt[3]{2 x - 1}

x = ?

Answered by MJS last updated on 01/May/20

obviously x = 1 is a solution

{(x^{3} + 1)}^{3} = 8 (2 x - 1)

x^{9} + 3 x^{6} + 3 x^{3} - 16 x + 9 = 0

(x - 1) (x^{8} + x^{7} + x^{6} + 4 x^{5} + 4 x^{4} + 4 x^{3} + 7 x^{2} + 7 x - 9) = 0

approximately solving leads to

x \approx - 1.61803 \lor x \approx . 618034

these seem to be

x = - \frac{1}{2} - \frac{\sqrt{5}}{2} \lor x = - \frac{1}{2} + \frac{\sqrt{5}}{2}

yes!

(x - 1) (x^{2} + x - 1) (x^{6} + 2 x^{4} + 2 x^{3} + 4 x^{2} + 2 x + 9) = 0

now we can only use approximation by an

electronic calculator

x \approx - 1.08891 \pm . 824866 i

x \approx . 123133 \pm 1 . 43020 i

x \approx . 965776 \pm 1 . 18647 i

only these solve the given equation :

x = 1

x = - \frac{1}{2} \pm \frac{\sqrt{5}}{2}

x \approx - 1.08891 \pm . 824866 i

Commented by jagoll last updated on 01/May/20

i s t h e r e a n e l e c t r o n i c c a l c u l a t o r

i n t h e p l a y s t o r e ?

Commented by MJS last updated on 01/May/20

I {don}^{'} t know, I^{'} ve got an old Texas Instruments

TI - 89 device

Commented by jagoll last updated on 01/May/20

t h a n k a l o t s i r

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