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Solve-for-real-numbees-a-16-1-a-b-2-b-16-1-b-c-2-c-16-1-c-x-2-




Question Number 138783 by mathdanisur last updated on 18/Apr/21
Solve for real numbees:   { ((a^(16) +1=((a+b)/2))),((b^(16) +1=((b+c)/2))),((c^(16) +1=((c+x)/2))) :}
$${Solve}\:{for}\:{real}\:{numbees}: \\ $$$$\begin{cases}{{a}^{\mathrm{16}} +\mathrm{1}=\frac{{a}+{b}}{\mathrm{2}}}\\{{b}^{\mathrm{16}} +\mathrm{1}=\frac{{b}+{c}}{\mathrm{2}}}\\{{c}^{\mathrm{16}} +\mathrm{1}=\frac{{c}+{x}}{\mathrm{2}}}\end{cases} \\ $$
Commented by mr W last updated on 18/Apr/21
no real roots!
$${no}\:{real}\:{roots}! \\ $$
Commented by mathdanisur last updated on 18/Apr/21
Dear Sir, can you tell me why there is no real root?
$${Dear}\:{Sir},\:{can}\:{you}\:{tell}\:{me}\:{why}\:{there}\:{is}\:{no}\:{real}\:{root}? \\ $$
Commented by mr W last updated on 18/Apr/21
if a=t is a root, then b=t and c=t are  also roots.  t^(16) +1=((t+t)/2)  t^(16) −t+1=0 ?    f(t)=t^(16) −t+1  f′(t)=16t^(15) −1=0 ⇒t=(1/( ((16))^(1/(15)) ))  f′′(t)=16×15t^(14) >0  i.e. f_(min)  is f((1/( ((16))^(1/(15)) )))=(1/(16^((16)/(15)) ))−(1/(16^(1/(15)) ))+1≈0.2207  since t^(16) −t+1≥0.2207  therefore t^(16) −t+1=0 has no real roots.
$${if}\:{a}={t}\:{is}\:{a}\:{root},\:{then}\:{b}={t}\:{and}\:{c}={t}\:{are} \\ $$$${also}\:{roots}. \\ $$$${t}^{\mathrm{16}} +\mathrm{1}=\frac{{t}+{t}}{\mathrm{2}} \\ $$$${t}^{\mathrm{16}} −{t}+\mathrm{1}=\mathrm{0}\:? \\ $$$$ \\ $$$${f}\left({t}\right)={t}^{\mathrm{16}} −{t}+\mathrm{1} \\ $$$${f}'\left({t}\right)=\mathrm{16}{t}^{\mathrm{15}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{t}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{15}}]{\mathrm{16}}} \\ $$$${f}''\left({t}\right)=\mathrm{16}×\mathrm{15}{t}^{\mathrm{14}} >\mathrm{0} \\ $$$${i}.{e}.\:{f}_{{min}} \:{is}\:{f}\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{15}}]{\mathrm{16}}}\right)=\frac{\mathrm{1}}{\mathrm{16}^{\frac{\mathrm{16}}{\mathrm{15}}} }−\frac{\mathrm{1}}{\mathrm{16}^{\frac{\mathrm{1}}{\mathrm{15}}} }+\mathrm{1}\approx\mathrm{0}.\mathrm{2207} \\ $$$${since}\:{t}^{\mathrm{16}} −{t}+\mathrm{1}\geqslant\mathrm{0}.\mathrm{2207} \\ $$$${therefore}\:{t}^{\mathrm{16}} −{t}+\mathrm{1}=\mathrm{0}\:{has}\:{no}\:{real}\:{roots}. \\ $$
Commented by mathdanisur last updated on 18/Apr/21
Thank you very much dear Sir
$${Thank}\:{you}\:{very}\:{much}\:{dear}\:{Sir} \\ $$

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