Question

The smallest value of $k$, for which both the roots of the equation $x^{2}-8 k x+16\left(k^{2}-k+1\right)=0$ are real, distinct and have values at least $4$ , is

Answer 1

$x ^{2}-8 kx +16\left( k ^{2}- k +1\right)=0$
$D >0 \Rightarrow k >1 \ldots$(1)
$\frac{- b }{2 a }>4 \Rightarrow \frac{8 k }{2}>4$
$\Rightarrow k >1 \ldots$(2)
$f (4) \geq 0 \Rightarrow 16-32 k +16\left( k ^{2}- k +1\right) \geq 0$
$k ^{2}-3 k +2 \geq 0$
$k \leq 1 \cup k \geq 2 \ldots$(3)
Using $(1),(2)$ and $(3)$
$k _{\min }=2$