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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?

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  • $\begingroup$ I don't know what you mean about "values." For the earlier part, do you know what the discriminant is? $\endgroup$ – Will Jagy Aug 27 '17 at 3:17
  • $\begingroup$ 'Values' means values of the two roots of the equation. $\endgroup$ – Tanuj Aug 27 '17 at 3:21
  • $\begingroup$ You haven't answered to the second, essential, question of @Will Jagy about the discriminant. $\endgroup$ – Jean Marie Aug 27 '17 at 3:23
  • $\begingroup$ I'm sorry yes I know the discriminant, its coming out to be $64(k-1)$. It has to be position for distinct real roots I know that but what other conditions do I need? $\endgroup$ – Tanuj Aug 27 '17 at 3:25
  • $\begingroup$ Remember that a quadratic with $\Delta < 0$ has no real roots. $\endgroup$ – Toby Mak Aug 27 '17 at 3:26
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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.

  1. For roots to be real, you need the discriminant to be $\geq 0$.
  2. For roots to be distinct, you need the discriminant to be $\neq 0$.
  3. 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 $$k \in [2, \infty).$$

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  • $\begingroup$ How did you come up with the final solution? I mean how did you combine all the inequalities? $\endgroup$ – Tanuj Aug 27 '17 at 3:30
  • $\begingroup$ @Tanuj the first and the last inequality both want $k>1$, the second inequality says either $k \leq 1$ OR $k \geq 2$. So just ask yourself what is the interval on which we can have all of them satisfied? $\endgroup$ – Anurag A Aug 27 '17 at 3:32
  • $\begingroup$ Gotchya! Just take the intersection, right? +1 by the way. $\endgroup$ – Tanuj Aug 27 '17 at 3:33
  • $\begingroup$ [+1] I would like to attract your attention on a detail that looks me important. You say $y=x^2-8kx+\cdots$ is a parabola ; maybe it should be better to say that it is the equation of a parabola. $\endgroup$ – Jean Marie Aug 27 '17 at 3:48
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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 $\,4k - 4 \sqrt{k-1} \ge 4\,$ $\iff$ $k-1 \ge \sqrt{k-1}$ $\iff$ $k \ge 2\,$.

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