If - - are the roots of the equation x 2 -px- 1 2 p 2 -0- where p- R- Then- the minimum value of - 4 - 4 is

$α, β$ are the roots of the equation $x^{2}$ + bx + c = 0 and $α + 2, β + 2$ are the roots of the equation $x^{2}$ + px + q=0. If the minimum value of the expression $x^{2}$ + bx + c is +2, find the minimum value of $x^{2}$ + px + q

Q. If

α, β

are the roots of the equation

x^{2} - p x + q = 0

, find the value of

α^{4} + α^{2} β^{2} + β^{4}

Q. If

α

and

β

are the roots of the equation

x^{2} - p (x + 1) - q = 0

then value of

\frac{α^{2} + 2 α + 1}{α^{2} + 2 α + q} + \frac{β^{2} + 2 β + 1}{β^{2} + 2 β + q}

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If α,β are the roots of the equation x2+px−12p2=0, where p∈R. Then, the minimum value of α4+β4 is

The correct option is B 2+√2We have, α+β=−pαβ=−12p2α4+β4=(α2+β2)2−2α2β2=((α+β)2−2αβ)2−2α2β2=((−p)2+1p2)2−12p4=p4+12p4+2=(p2)2+(1√2p2)2+2⋅p2⋅1√2p2−2⋅p2⋅1√2p2+2=(p2−1√2p2)2+√2+2=(p2−1√2p2)2+2+√2≥2+√2∴minimum value is 2+√2

If $α, β$ are the roots of the equation $x^{2} + p x - \frac{1}{2 p^{2}} = 0$ , where $p \in R$ . Then, the minimum value of $α^{4} + β^{4}$ is