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

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Aicharya · Accepted Answer

(i) Given, $x^2-8 k x+16\left(k^2-k+1ight)=0$ Now, $\quad D=64\left\{k^2-\left(k^2-k+1ight)ight\}$ $ \begin{aligned} & =64(k-1)>0 \ & 	herefore \quad k>1 \ & 16-32 k+16\left(k^2-k+1ight) \geq 0 \ & \Rightarrow \quad k^2-3 k+2 \geq 0 \ & \Rightarrow(k-2)(k-1) \geq 0 \Rightarrow k \leq 1 	ext { or } k \geq 2 \ & \end{aligned} $ Hence, $k=2$

The smallest value of $k$, for which both the roots of the equation $x^2-8 k x+16\left(k^2-k+1\right)=0$ are…

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