If $x^{8} - (k - 1) x^{4} + 5 = 0,$ then least possible integral value of $k$ so that equation has maximum number of real roots is

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Solution

Let
$x^{4} = t t^{2} - (k - 1) t + 5 = 0$
For maximum number of real roots both the roots of the equation must be positive
so,
$D > 0 \Rightarrow (k - 1)^{2} - 20 > 0 \Rightarrow k > 1 + \sqrt{2} 0$
So least possible integral value of $k = 6$

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