The smallest value of $k$, for which both the roots of the equation $x^{2}-8 k x+16\left(k^{2}-k+1\r

Question

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 at least 4 , is

Professor PC Thomas Classes · Answer

(2) : The equation is $x^{2}-8 k x+16$ $\left(k^{2}-k+1 ight)=0$\ we have, $D=64\left(k^{2}-\left(k^{2}-k+1 ight) ight)$\ $=64(k-1)>0 \Rightarrow \quad k>1$\ Also $4<-\frac{b}{2 a} \Rightarrow 4<\frac{8 k}{2} \Rightarrow k>1$\ $f(4) \geq 0 \Rightarrow 16-32 k+16\left(k^{2}-k+1 ight) \geq 0$\ which yields $k^{2}-3 k+2 \geq 0$\ $\Rightarrow k \leq 1$ or $k \geq 2$.\ Hence $k=2$.\