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x-8-ax-4-1-0-a-is-x-1-x-2-x-3-x-4-




Question Number 161521 by vvvv last updated on 19/Dec/21
x^8 +ax^4 +1=0  a=?   is  x_1 +x_2 +x_3 +x_4 =?
$$\boldsymbol{\mathrm{x}}^{\mathrm{8}} +\boldsymbol{\mathrm{ax}}^{\mathrm{4}} +\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{{a}}=?\: \\ $$$$\boldsymbol{{is}}\:\:\boldsymbol{{x}}_{\mathrm{1}} +\boldsymbol{{x}}_{\mathrm{2}} +\boldsymbol{{x}}_{\mathrm{3}} +\boldsymbol{{x}}_{\mathrm{4}} =? \\ $$
Commented by mr W last updated on 19/Dec/21
though you used two times the question  mark, but you didn′t put a clear   question. what is your exact question?
$${though}\:{you}\:{used}\:{two}\:{times}\:{the}\:{question} \\ $$$${mark},\:{but}\:{you}\:{didn}'{t}\:{put}\:{a}\:{clear}\: \\ $$$${question}.\:{what}\:{is}\:{your}\:{exact}\:{question}? \\ $$
Commented by Tyller last updated on 19/Dec/21
(x^4 )^2 +ax^4 +1=0  (x^4 +(a/2))^2 =((a^2 −4)/4)  x=±((−(a/2)±(√((a^2 −4)/4)).))^(1/4)   a∈[0,(√2)]  x_1 +x_2 +x_3 +x_4 =0
$$\left({x}^{\mathrm{4}} \right)^{\mathrm{2}} +{ax}^{\mathrm{4}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{4}} +\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}} \\ $$$${x}=\pm\sqrt[{\mathrm{4}}]{−\frac{{a}}{\mathrm{2}}\pm\sqrt{\frac{{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4}}}.} \\ $$$${a}\in\left[\mathrm{0},\sqrt{\mathrm{2}}\right] \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} +{x}_{\mathrm{3}} +{x}_{\mathrm{4}} =\mathrm{0} \\ $$
Commented by mr W last updated on 19/Dec/21
it is not stated that x∈R. so a can be  any value and there are generally 8  roots.  that is why i said the question is not  clear.
$${it}\:{is}\:{not}\:{stated}\:{that}\:{x}\in\mathbb{R}.\:{so}\:{a}\:{can}\:{be} \\ $$$${any}\:{value}\:{and}\:{there}\:{are}\:{generally}\:\mathrm{8} \\ $$$${roots}. \\ $$$${that}\:{is}\:{why}\:{i}\:{said}\:{the}\:{question}\:{is}\:{not} \\ $$$${clear}. \\ $$

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