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

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Solution

$x^{2} - 8 k x + 16 (k^{2} - k + 1) = 0$
As the roots are real and distinct, we have
$D > 0$
$\Rightarrow 64 k^{2} - 4 (16 (k^{2} - k + 1)) > 0$
$\Rightarrow k > 1 \dots (1)$
As the roots have values atleast $4$ , we have
$- \frac{b}{2 a} \geq 4 \Rightarrow \frac{8 k}{2} \geq 4$
$\Rightarrow k \geq 1 \dots (2)$
$f (4) \geq 0 \Rightarrow 16 - 32 k + 16 (k^{2} - k + 1) \geq 0$
$k^{2} - 3 k + 2 \geq 0$
$k \leq 1$ or $\geq 2 \dots (3)$
Using $(1), (2)$ and $(3)$
$k_{min} = 2$

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The smallest value of k, for which both the roots of the equation x2−8kx+16(k2−k+1)=0 are real, distinct and have values at least 4 is

x2−8kx+16(k2−k+1)=0As the roots are real and distinct, we haveD>0 ⇒64k2−4(16(k2−k+1))>0⇒k>1⋯(1)As the roots have values atleast 4, we have−b2a≥4⇒8k2≥4⇒k≥1⋯(2)f(4)≥0⇒16−32k+16(k2−k+1)≥0k2−3k+2≥0k≤1 or ≥2⋯(3)Using (1),(2) and (3)kmin=2

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