sin54-10-x-1-x-1-is-root-ax-2-bx-c-0-a-b-c-is-integer-a-

Question Number 182854 by manxsol last updated on 16/Dec/22

{(s i n 54)}^{10} = x_{1} x_{1} i s r o o t

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

a, b, c i s i n t e g e r

a = ?

Commented by manxsol last updated on 16/Dec/22

o k, I w i l l c o r r e c t i t ., t h a n k s

f o r t h e s u g g e s t i o n .

H o w d i d ? y o u d o i t ?

Commented by Frix last updated on 15/Dec/22

? ? ?

a = 1

(x - x_{1}) (x - x_{2}) = x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2}

b = - (x_{1} + x_{2})

c = x_{1} x_{2}

Commented by manxsol last updated on 15/Dec/22

s o r r y, i a m e d i t p r o b l e m

Commented by manxsol last updated on 15/Dec/22

t h a n k s

Commented by Frix last updated on 15/Dec/22

No need to be sorry .

\sin 54 ° = \frac{1 + \sqrt{5}}{4}

\sin^{10} 54 ° = \frac{123 + 55 \sqrt{5}}{2048}

Commented by manxsol last updated on 15/Dec/22

I s o l v e d i t l i k e t h i s

s i n 54 = \frac{\sqrt{5} + 1}{4} = \frac{1}{2} \frac{\sqrt{5} + 1}{2} = \frac{1}{2} Φ

{(\frac{1}{2})}^{10} Φ^{10} = \frac{1}{1024} (\frac{123 + 55 \sqrt{5}}{2})

= \frac{123 + 55 \sqrt{5}}{2048} = x_{1}

S e e {(★)}^{}

r o o t s i n t e g e r \Rightarrow x_{2 c o n j u g a t e}

x_{2} = \frac{123 - 55 \sqrt{5}}{2048}

x_{1} + x_{2} = \frac{246}{2048}

x_{1} x_{2} = \frac{4}{4194304}

c a l c u l a t e u s i n g e d i t o r t i n k u t a r a

2048^{2}

4194304.0

123^{2} - 5 \times 55^{2}

4.0

246 \times 2048

503808.0

4194304 / 4

1048576.0

503808 / 4

125952.0

i n p u t

4 \times 5

[p u s h] e n t e r

[p u s h] ⋮

[p u s h] c a l c u l a r

20.0

x^{2} - \frac{246}{2048} x + \frac{4}{4194304} = 0

4194304 x^{2} - 503808 x + 4 = 0

1048576 x^{2} - 125952 x + 1 = 0

s o l u t i o n : a = 1048576

(★)

Φ^{10} = {(\frac{1 + \sqrt{5}}{2})}^{10} = \frac{123 + 55 \sqrt{5}}{2}

s e r i e d e f i b o n a c c i

\begin{array}{cc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \\ 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 \end{array}

a = 55

b = 123

d e d u c c i o n h e c h a d e t a b l a d e

p o t e n c i a s d e Φ = (1 + \sqrt{5}) / 2

Commented by Frix last updated on 15/Dec/22

You wrote a i s i n t e g e r .

You should have written a, b, c are integer .

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