Question

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

Answer 1

Answer: 2

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