For what minimum value of k do both roots of | vedclass

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(B) Let the quadratic equation be $f(x) = x^2 - 8kx + 16(k^2 - k + 1) = 0$. For the roots to be real and distinct,the discriminant $D > 0$: $D = (-8k)

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the quadratic equation be $f(x) = x^2 - 8kx + 16(k^2 - k + 1) = 0$. For the roots to be real and distinct,the discriminant $D > 0$: $D = (-8k)^2 - 4(1)(16(k^2 - k + 1)) > 0$ $64k^2 - 64(k^2 - k + 1) > 0$ $64k^2 - 64k^2 + 64k - 64 > 0$ $64k - 64 > 0 \implies k > 1$. Let the roots be $\alpha$ and $\beta$. We are given $\alpha, \beta \ge 4$. This implies: $1)$ The vertex of the parabola $x = -b/(2a) = 8k/2 = 4k$ must be $\ge 4$,so $4k \ge 4 \implies k \ge 1$. $2)$ $f(4) \ge 0$: $4^2 - 8k(4) + 16(k^2 - k + 1) \ge 0$ $16 - 32k + 16k^2 - 16k + 16 \ge 0$ $16k^2 - 48k + 32 \ge 0$ $k^2 - 3k + 2 \ge 0$ $(k - 1)(k - 2) \ge 0$. This holds for $k \le 1$ or $k \ge 2$. Combining $k > 1$ (from $D > 0$) and $k \ge 2$ (from $f(4) \ge 0$),we get $k \ge 2$. The minimum value of $k$ is $2$.

For what minimum value of $k$ do both roots of the equation $x^2 - 8kx + 16(k^2 - k + 1) = 0$ exist,are real,distinct,and at least $4$?

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