If α,β are the roots of the equation x2+ax−12a2=0, a≠0 then the minimum value of α4+β4 is

x2+ax−12a2=0,α,β α+β=−a,αβ=−12a2⇒ α2+β2=(α+β)2−2αβ=a2+1a2α4+β4=α2+β22−2α2β2=a2+1a22−2⋅14a4=a4+1a4+2−12a4=a4+12a4+2=2+a2−12a22+22∴ minα4+β4=2+2

If α,β are the roots of the equation x2+ax−12a2=0, a≠0 then the minimum value of α4+β4 is

x2+ax−12a2=0,α,β α+β=−a,αβ=−12a2⇒ α2+β2=(α+β)2−2αβ=a2+1a2α4+β4=α2+β22−2α2β2=a2+1a22−2⋅14a4=a4+1a4+2−12a4=a4+12a4+2=2+a2−12a22+22∴ minα4+β4=2+2

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If $α, β$ are the roots of the equation $x^{2} + a x - \frac{1}{2 a^{2}} = 0,$ $a \neq 0$ then the minimum value of $α^{4} + β^{4}$ is

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$2 - \sqrt{2}$

$2 + \sqrt{2}$

$4 - \sqrt{2}$

$4 + \sqrt{2}$

answer is B.

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Detailed Solution

$x^{2} + a x - \frac{1}{2 a^{2}} = 0, α, β$
$\begin{array}{l} α + β = - a, α β = - \frac{1}{2 a^{2}} \\ \Rightarrow α^{2} + β^{2} = (α + β)^{2} - 2 α β = a^{2} + \frac{1}{a^{2}} \\ α^{4} + β^{4} = {(α^{2} + β^{2})}^{2} - 2 α^{2} β^{2} \\ = {(a^{2} + \frac{1}{a^{2}})}^{2} - 2 \cdot \frac{1}{4 a^{4}} \\ = a^{4} + \frac{1}{a^{4}} + 2 - \frac{1}{2 a^{4}} \\ = a^{4} + \frac{1}{2 a^{4}} + 2 \\ = 2 + {(a^{2} - \frac{1}{\sqrt{2} a^{2}})}^{2} + \frac{2}{\sqrt{2}} \end{array}$
$∴ min (α^{4} + β^{4}) = 2 + \sqrt{2}$