Quadratic equation, smallest value of k What is the smallest value of $k$, for which both roots of the equation $x^2-8kx+16(k^2-k+1)=0$ are real, distinct and have values at least $4$? 
My Try: I absolutely have no idea how to approach this logically, but I get a feeling $k$ is zero ( I may be wrong ).
Can someone please tell me more of a logical approach to solve similar problems?
 A: Consider the function $y=f(x)=x^2-8kx+16(k^2-k+1)$. This represents an upwards facing parabola. Now we will apply the various conditions given.


*

*For roots to be real, you need the discriminant to be $\geq 0$.

*For roots to be distinct, you need the discriminant to be $\neq 0$.

*For both roots to be $\geq 4$, since the parabola is facing upwards, $f(4) \geq 0$ and the $x-$coordinate of the vertex of the parabola (global min) should lie to the right of $4$. 


In your case, comparing with $ax^2+bx+c=0$, the discriminant is
$$b^2-4ac=64k^2-64(k^2-k+1).$$
So the first two conditions imply that
$$64k^2-64(k^2-k+1) >0 \implies k >1.$$
From the third condition, we get
\begin{align*}
f(4) & \geq 0\\
16-32k+16(k^2-k+1 & \geq 0\\
k^2-3k+2 &\geq 0\\
k & \in (-\infty, 1] \cup [2, \infty)
\end{align*}
Now we impose the vertex condition. The vertex is at $x=\frac{-b}{2a}$, so
\begin{align*}
\frac{-b}{2a} &> 4\\
\frac{8k}{2} &> 4\\
k & > 1.
\end{align*}
Combining all these we get
$$k \in [2, \infty).$$ 
A: By completing the square $\;x^2 −8kx+16(k^2 −k+1)=0⟺(x−4k)^2 =16(k−1)\,$. For the latter to have real and distinct roots, the RHS must be strictly positive, thus $\,k \gt 1\,$, and in that case the two real roots are $\,4k \pm 4 \sqrt{k-1}\,$. The condition that the smaller root of the two is at least $\,4\,$ then translates to $\,4k - 4 \sqrt{k-1} \ge 4\,$ $\iff$ $k-1 \ge \sqrt{k-1}$ $\iff$ $k \ge 2\,$.
