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Question Number 20670 by Tinkutara last updated on 31/Aug/17

F o r t h e e q u a t i o n 3 x^{2} + p x + 3 = 0,

f i n d t h e v a l u e (s) o f p i f o n e r o o t i s

(i) s q u a r e o f t h e o t h e r

(i i) f o u r t h p o w e r o f t h e o t h e r .

Answered by dioph last updated on 31/Aug/17

(i) x_{1} = x_{2}^{2}

1 = x_{1} x_{2} = x_{2}^{3}

∴ x_{2} = 1^{1 / 3}, x_{1} = 1^{2 / 3}

p = - 3 (x_{1} + x_{2}) = - 3 \times 1^{1 / 3} - 3 \times 1^{2 / 3}

values of p :

let ω_{3} = e^{i 2 π / 3} = (- 0.5 + i \frac{\sqrt{3}}{2})

{\begin{cases} - 3 \times 1 - 3 \times 1 = - 6 \\ - 3 \times ω_{3} - 3 \times ω_{3}^{2} = 3 \end{cases}

(i i) x_{1} = x_{2}^{4}

1 = x_{1} x_{2} = x_{2}^{5}

∴ x_{2} = 1^{1 / 5}, x_{1} = 1^{4 / 5}

p = - 3 (x_{1} + x_{2}) = - 3 \times 1^{1 / 5} - 3 \times 1^{4 / 5}

values of p :

let ω_{5} = e^{i 2 π / 5} = (\cos \frac{2 π}{5} + i \sin \frac{2 π}{5})

if \cos \frac{2 π}{5} = \frac{\sqrt{5} - 1}{4} (from regular penthagon) :

{\begin{cases} - 3 \times 1 - 3 \times 1 = - 6 \\ - 3 \times ω_{5} - 3 \times ω_{5}^{4} = - 6 \cos \frac{2 π}{5} = - \frac{3 (\sqrt{5} - 1)}{2} \\ - 3 \times ω_{5}^{2} - 3 ω_{5}^{3} = 6 \cos \frac{π}{5} = 3 + \frac{3 (\sqrt{5} - 1)}{2} \end{cases}

Commented by Tinkutara last updated on 31/Aug/17

Thanks for (i) part .

But in (i i) part, answer is 0 and

\frac{3 (\sqrt{5} - 1)}{2} . Can you get it ?

Commented by dioph last updated on 31/Aug/17

I added the substitution

\cos \frac{2 π}{5} = \frac{\sqrt{5} - 1}{4}

but I cannot see how to get these answers

If p = 0 then roots are i and - i,

which are not related by a

fourth power

Commented by Tinkutara last updated on 31/Aug/17

Thank you very much Sir!