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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 $4$?

Similar Questions

The values of $a$ for which the equation $2x^2 - 2(2a + 1)x + a(a + 1) = 0$ has one root less than $a$ and the other root greater than $a$ are given by:

If $\alpha, \beta$ are the roots of $x^2+bx+c=0$,$\gamma, \delta$ are the roots of $x^2+b_1x+c_1=0$ and $\gamma < \alpha < \delta < \beta$,then $(c-c_1)^2 < $

If both the roots of the equation $x^2-4ax+1-3a+4a^2=0$ exceed $1$,then $a$ lies in the interval

The values of $a$ for which one root of the equation $x^2 - (a + 1)x + a^2 + a - 8 = 0$ exceeds $2$ and the other is lesser than $2$,are given by

If $x$ is real,then the maximum value of $5 + 4x - 4x^2$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module