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

Answer: 2

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