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 atleast $4$, is .........

Answer 1

(i) Given, $ x^2 - 8 kx + 16 \, ( k^2 - k + 1) = 0 $
Now, $D = 64 \{ k^2 - ( k^2 - k + 1 ) \} = 64 \, ( k - 1) > 0 $
$k > 1$
(ii) $ - \frac{ b}{ 2a} > 4 \Rightarrow \frac{ 8 k }{ 2} > 4 \Rightarrow k > 1 $
(iii) $ f ( 4) \ge 0 $
$\Rightarrow 16 - 32 k + 16 (k^2 - k + 1) \ge 0 $
$\Rightarrow k^2 - 3k + 2 \ge 0 $
$\Rightarrow ( k - 2) \, ( k - 1) \ge 0 $
$\Rightarrow k \le 1 \, $ or $\, k \ge 2 $
Hence, $k = 2$