the smallest k...

The answer is correct and it can be shown graphically. In fact, I solved it using graphs before I thought out the solution. This was easily done using a Microsoft Excel spreadsheet. I have plotted the curves for $k=0$ , $k=1$ , $k=1.5$ , $k=2$ and $k=3$ as in the above to show it. It can be seen that the curve shifts to the right as $k$ increases. When $k=0$ , there is no real root. When $k=1$ there is only one root as the curve just touches the $x$ -axis. I also mentioned this in my solution. Of course written equation and graph show the same thing. When $1 < k < 2$ , there are two roots but one of the root is less than $4$ , when $x = 2$ , the smaller root is exactly $4$ , as mentioned in my solution. Therefore, when $k>2$ both roots are larger than $4$ , as shown in the graph with $k=3$ . QED.

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The roots of $f(x) = x^2-8kx+16(k^2-k+1)$ are:

$x = \dfrac {8k \pm \sqrt{64-64(k^2-k+1)}} {2} = 4k \pm 4\sqrt{k-1}$

For the roots to be at least $4$ , then:

$4k - 4\sqrt{k-1} \ge 4\quad \Rightarrow k - \sqrt{k-1} \ge 1 \quad \Rightarrow k - 1 \ge \sqrt{k-1}$

$\Rightarrow k^2 -2k + 1 \ge k-1 \quad \Rightarrow k^2 -3k + 2 \ge 0 \quad \Rightarrow (k-1)(k-2) \ge 0$

$\Rightarrow k \ge 1\space$ or $\space k \ge 2$ .

We find that when $k=1$ there is only one root $4 \pm 0$ , therefore it is not acceptable.

When $k=2$ , the roots are $\{4,12\}$ , therefore the answer is $k \ge \boxed {2}$ .

The roots of this equation can be written as

$\dfrac { 8k\pm \sqrt { 64{ k }^{ 2 }-64{ k }^{ 2 }+64k-64 } }{ 2 } \\ \Rightarrow \dfrac { 8k\pm 8\sqrt { k-1 } }{ 2 } \\ \Rightarrow 4k\pm 4\sqrt { k-1 }$

Now let us put values

• $k$ can't be $1$ as it would result equal roots

• so let $k$ = $2$

Yes it gives us $\Rightarrow 4\times 2\pm 4\sqrt { 2-1 } \\ \Rightarrow ~ 8-4\\ \Rightarrow \color{#69047E}{4}$

So the answer is $\color{#20A900}{\huge\boxed{1}}$

consider the equation graphically: (This picture does not depict the exact same equation given in the question)

Because the quadratic is monic, it will represent a parabola facing upwards with vertex downwards. It has been mentioned that the roots are both greater than 4.

therefore, $\text{the value of the function that represents the quadratic will be greater than 0 at x≤4}$ this is actually very simple: for the typical quadratic $y=x^{2}+bx+c$ ,for any value of $x$ that lies in the interval containing the roots, the curve will be below the X-axis- thus ≤0 and the curve will be above the X-axis anywhere else.

So, plugging in $x=4$ we should get the value of the function≥0

$\therefore f(4)=16-32k+16(k^{2}-k+1)≥0 \implies k^{2}-3k+2≥0 \implies k\in(-\infty,1] \cup [2,+\infty)$

it is also mentioned that the roots are real and distinct, thus, the discriminant $\Delta>0$

checking: $\Delta=64k^{2}-64(k^{2}-k+1)>0 \implies k>1$

therefore from the two results, we get $\large{k≥\boxed{2}}$