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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 & 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!