Question Number 21314 by Tinkutara last updated on 20/Sep/17
$$\mathrm{Let}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{be}\:\mathrm{the}\:\mathrm{root}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{px}\:−\:\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\:=\:\mathrm{0}, \\ $$$${p}\:\in\:{R}.\:\mathrm{The}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:\mathrm{is} \\ $$
Answered by $@ty@m last updated on 21/Sep/17
![α+β=−p & αβ=− (1/(2p^2 )) −−(1) (a+b)^4 =a^4 +4a^3 b+6a^2 b^2 +4ab^3 +b^4 ⇒a^4 +b^4 =(a+b)^4 −4ab(a^2 +b^2 )−6(ab)^2 ⇒a^4 +b^4 =(a+b)^4 −4ab{(a+b)^2 −2ab}−6(ab)^2 ∴ α^4 + β^4 =(−p)^4 −4(− (1/(2p^2 ))){p^2 −2×((−1)/(2p^2 ))}−6(− (1/(2p^2 )))^2 =p^4 +(2/p^2 )(p^2 +(1/p^2 ))−6×(1/(4p^4 )) =p^4 +2+(2/p^4 )−(3/(2p^4 )) ∴ α^4 + β^4 =p^4 +2+(1/(2p^4 )) −−(2) Differentiating both sides wrt p (d/dp)(α^4 + β^4 )=4p^3 −(2/p^5 ) −−(3) For maxima or minima, (d/dp)(α^4 + β^4 )=0 4p^3 −(2/p^5 )=0 4p^8 −2=0 p^8 =(1/2) p^4 =(1/(√2)) −−(4) Differentiating both sides of (3) wrt p (d^2 /dp^2 )(α^4 + β^4 )=12p^2 +((10)/p^6 ) >0 ∴ α^4 + β^4 is minimum when p^4 =(1/(√2)) ⇒ (α^4 + β^4 )_(min.) =[p^4 +2+(1/(2p^4 ))]_(p^4 =(1/(√2))) =(1/(√2))+2+(1/(2×(1/(√2)))) =(1/(√2))+2+(1/(√2)) =(√2)+2](Q21337.png)
$$\alpha+\beta=−{p}\:\:\:\&\:\:\alpha\beta=−\:\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\:\:\:−−\left(\mathrm{1}\right) \\ $$$$\left({a}+{b}\right)^{\mathrm{4}} ={a}^{\mathrm{4}} +\mathrm{4}{a}^{\mathrm{3}} {b}+\mathrm{6}{a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{4}{ab}^{\mathrm{3}} +{b}^{\mathrm{4}} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\left({a}+{b}\right)^{\mathrm{4}} −\mathrm{4}{ab}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−\mathrm{6}\left({ab}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\left({a}+{b}\right)^{\mathrm{4}} −\mathrm{4}{ab}\left\{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}\right\}−\mathrm{6}\left({ab}\right)^{\mathrm{2}} \\ $$$$\therefore\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:=\left(−{p}\right)^{\mathrm{4}} −\mathrm{4}\left(−\:\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\right)\left\{{p}^{\mathrm{2}} −\mathrm{2}×\frac{−\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\right\}−\mathrm{6}\left(−\:\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$={p}^{\mathrm{4}} +\frac{\mathrm{2}}{{p}^{\mathrm{2}} }\left({p}^{\mathrm{2}} +\frac{\mathrm{1}}{{p}^{\mathrm{2}} }\right)−\mathrm{6}×\frac{\mathrm{1}}{\mathrm{4}{p}^{\mathrm{4}} } \\ $$$$={p}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{2}}{{p}^{\mathrm{4}} }−\frac{\mathrm{3}}{\mathrm{2}{p}^{\mathrm{4}} } \\ $$$$\therefore\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:={p}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{4}} }\:\:−−\left(\mathrm{2}\right)\: \\ $$$${Differentiating}\:{both}\:{sides}\:{wrt}\:{p} \\ $$$$\frac{{d}}{{dp}}\left(\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:\right)=\mathrm{4}{p}^{\mathrm{3}} −\frac{\mathrm{2}}{{p}^{\mathrm{5}} }\:\:−−\left(\mathrm{3}\right) \\ $$$${For}\:{maxima}\:{or}\:{minima}, \\ $$$$\frac{{d}}{{dp}}\left(\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:\right)=\mathrm{0} \\ $$$$\mathrm{4}{p}^{\mathrm{3}} −\frac{\mathrm{2}}{{p}^{\mathrm{5}} }=\mathrm{0} \\ $$$$\mathrm{4}{p}^{\mathrm{8}} −\mathrm{2}=\mathrm{0} \\ $$$${p}^{\mathrm{8}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${p}^{\mathrm{4}} =\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\:−−\left(\mathrm{4}\right) \\ $$$${Differentiating}\:{both}\:{sides}\:{of}\:\left(\mathrm{3}\right)\:{wrt}\:{p} \\ $$$$\frac{{d}^{\mathrm{2}} }{{dp}^{\mathrm{2}} }\left(\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:\right)=\mathrm{12}{p}^{\mathrm{2}} +\frac{\mathrm{10}}{{p}^{\mathrm{6}} }\:\:>\mathrm{0} \\ $$$$\therefore\:\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:{is}\:{minimum}\:{when}\:{p}^{\mathrm{4}} =\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\:\left(\alpha^{\mathrm{4}} \:+\:\beta^{\mathrm{4}} \:\right)_{{min}.} =\left[{p}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}{p}^{\mathrm{4}} }\right]_{{p}^{\mathrm{4}} =\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}×\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}+\mathrm{2}+\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$=\sqrt{\mathrm{2}}+\mathrm{2} \\ $$$$ \\ $$$$ \\ $$
Commented by Tinkutara last updated on 21/Sep/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$