Question

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

Answer 1

The given quation is
$x^2-8kx+16(k^2-k=1)=0$
$\therefore$ Both the roots are real and distinct
$\therefore D>0\Rightarrow (8k{)}^2-4x16(k^2-k+1)>0\Rightarrow k>1....(i)$
$\therefore$ Both the roots are greater than or equal to 4
$\therefore \alpha +\beta >8andf\left(4\right)\le 0\Rightarrow k>1....\left(ii\right)and16-32k+16(k^2-k+1)\le 0\Rightarrow (k-1)(k-2)\le 0\Rightarrow k^2-3k+16(k^2-k+1)\le 0\Rightarrow kϵ(-\infty ,1]\cup [2,\infty ).....\left(iii\right)$
Combining (i) ,(ii) and (iii) , we get $k\ge 2$ or the smallest value of $k=2$.