Let α and β be the root of x^2 + px − (1/(2p^2 )) = 0, p ∈ R. The minimum value of α^4 + β^4 is


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Question Number 21314 by Tinkutara last updated on 20/Sep/17

$Let α and β be the root of x^{2} + p x - \frac{1}{2 p^{2}} = 0,$ $p \in R . The minimum value of α^{4} + β^{4} is$
	Answered by $@ty@m last updated on 21/Sep/17
	$α+β=−p & αβ=− (1/(2p^2 )) −−(1) (a+b)^4 =a^4 +4a^3 b+6a^2 b^2 +4ab^3 +b^4 ⇒a^4 +b^4 =(a+b)^4 −4ab(a^2 +b^2 )−6(ab)^2 ⇒a^4 +b^4 =(a+b)^4 −4ab{(a+b)^2 −2ab}−6(ab)^2 ∴ α^4 + β^4 =(−p)^4 −4(− (1/(2p^2 ))){p^2 −2×((−1)/(2p^2 ))}−6(− (1/(2p^2 )))^2 =p^4 +(2/p^2 )(p^2 +(1/p^2 ))−6×(1/(4p^4 )) =p^4 +2+(2/p^4 )−(3/(2p^4 )) ∴ α^4 + β^4 =p^4 +2+(1/(2p^4 )) −−(2) Differentiating both sides wrt p (d/dp)(α^4 + β^4 )=4p^3 −(2/p^5 ) −−(3) For maxima or minima, (d/dp)(α^4 + β^4 )=0 4p^3 −(2/p^5 )=0 4p^8 −2=0 p^8 =(1/2) p^4 =(1/(√2)) −−(4) Differentiating both sides of (3) wrt p (d^2 /dp^2 )(α^4 + β^4 )=12p^2 +((10)/p^6 ) >0 ∴ α^4 + β^4 is minimum when p^4 =(1/(√2)) ⇒ (α^4 + β^4 )_(min.) =[p^4 +2+(1/(2p^4 ))]_(p^4 =(1/(√2))) =(1/(√2))+2+(1/(2×(1/(√2)))) =(1/(√2))+2+(1/(√2)) =(√2)+2$
	$α + β = - p & α β = - \frac{1}{2 p^{2}} - - (1)$ ${(a + b)}^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 {a b}^{3} + b^{4}$ $\Rightarrow a^{4} + b^{4} = {(a + b)}^{4} - 4 a b (a^{2} + b^{2}) - 6 {(a b)}^{2}$ $\Rightarrow a^{4} + b^{4} = {(a + b)}^{4} - 4 a b {{(a + b)}^{2} - 2 a b} - 6 {(a b)}^{2}$ $∴ α^{4} + β^{4} = {(- p)}^{4} - 4 (- \frac{1}{2 p^{2}}) {p^{2} - 2 \times \frac{- 1}{2 p^{2}}} - 6 {(- \frac{1}{2 p^{2}})}^{2}$ $= p^{4} + \frac{2}{p^{2}} (p^{2} + \frac{1}{p^{2}}) - 6 \times \frac{1}{4 p^{4}}$ $= p^{4} + 2 + \frac{2}{p^{4}} - \frac{3}{2 p^{4}}$ $∴ α^{4} + β^{4} = p^{4} + 2 + \frac{1}{2 p^{4}} - - (2)$ $D i f f e r e n t i a t i n g b o t h s i d e s w r t p$ $\frac{d}{d p} (α^{4} + β^{4}) = 4 p^{3} - \frac{2}{p^{5}} - - (3)$ $F o r m a x i m a o r m i n i m a,$ $\frac{d}{d p} (α^{4} + β^{4}) = 0$ $4 p^{3} - \frac{2}{p^{5}} = 0$ $4 p^{8} - 2 = 0$ $p^{8} = \frac{1}{2}$ $p^{4} = \frac{1}{\sqrt{2}} - - (4)$ $D i f f e r e n t i a t i n g b o t h s i d e s o f (3) w r t p$ $\frac{d^{2}}{{d p}^{2}} (α^{4} + β^{4}) = 12 p^{2} + \frac{10}{p^{6}} > 0$ $∴ α^{4} + β^{4} i s m i n i m u m w h e n p^{4} = \frac{1}{\sqrt{2}}$ $\Rightarrow {(α^{4} + β^{4})}_{m i n .} = {[p^{4} + 2 + \frac{1}{2 p^{4}}]}_{p^{4} = \frac{1}{\sqrt{2}}}$ $= \frac{1}{\sqrt{2}} + 2 + \frac{1}{2 \times \frac{1}{\sqrt{2}}}$ $= \frac{1}{\sqrt{2}} + 2 + \frac{1}{\sqrt{2}}$ $= \sqrt{2} + 2$
	Commented by Tinkutara last updated on 21/Sep/17

	$Thank you very much Sir!$
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