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$$-1\leq \frac{-k \pm \sqrt{{k^2}-8k}}{4}\leq 1$$

The problem is the square root, as I don't think I can just square both sides. Any ideas?

So this actually comes from a quadratic in $$\sin{\theta}$$

Thanks.

Owl
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  • Then move terms other than the root to the other side. – xbh Oct 17 '19 at 17:15
  • Of course you can square both sides, as long as you understand what are you doing and what will happen to the inequality. – Ivan Neretin Oct 17 '19 at 17:16
  • Or, regard the middle term as the solution to a certain quadratic equation depending on $k$, and find the range of this $k$ via analyzing the corresponding quadratic function. – xbh Oct 17 '19 at 17:18
  • This is the solution set of $2x^2+kx +k=0$ in $x$. – IMOPUTFIE Oct 17 '19 at 17:20
  • Do you mean that the two numbers (the one with "+" and the one with "-") have to be simultaneously in $[-1,1]$ ? – Jean Marie Oct 17 '19 at 17:38
  • Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. See also the section on titles in How to ask a good question. (The part entitled "Make your title your question" is especially relevant to this.) – Shaun Oct 17 '19 at 17:43
  • "I don't think I can just square both sides." Sure you can! If $-1 \le K \le 1$ then $0\le |K| \le 1$ and $0\le K^2 \le 1$. And if you have $-a \le K \le b$ then you can do $K \le \max(a,b)$ and $K^2 \le \max(a^2, b^2)$. – fleablood Oct 17 '19 at 17:47
  • Be so kind as to answer my question. – Jean Marie Oct 17 '19 at 18:07
  • How did you get the $\pm$. Do you mean that both are between $-1$ and $1$ or that one or the other might be. $k \ge 8$ or $k\le 0$ and if both are between $1$ and $-1$ you will find that $k=0$. But if you mean a solution to a quadratic must have an absolute value less than $1$ that is something else. – fleablood Oct 17 '19 at 18:10
  • It means that either minus or plus has to satisfy the inequality. It comes from a quadratic in sin, hence why it is between minus one and one – Owl Oct 17 '19 at 18:33

2 Answers2

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The middle term is actually the two roots $x_{\pm}$ [not necessarily distinct] for the equation $$ 2x^2 + kx + k = 0. $$ Let $f(x) = 2x^2 + kx + k$. Then this is an upward parabola symmetric w.r.t. the line $x = -k/4$. Note that it always passes through the point $(-1, 2)$. Then by the graph of this function, $x_{\pm } \in [-1, 1]$ iff $$ \begin{cases} \Delta = k^2 - 8k \geqslant 0, \\ -\dfrac k 4 \in [-1, 1], \\ f(-1) \geqslant 0, \\ f(1) \geqslant 0, \end{cases} \iff \begin{cases} k \in (-\infty, 0] \cup [8, +\infty), \\ k \in [- 4,4] \\ 2 \geqslant 0, \\ 2 + 2k \geqslant 0, \end{cases} $$ iff $$ \boxed {\boldsymbol {k \in [-1, 0]}}\ . $$

Update

If it means either one lies in the interval $[-1,1]$, then $$ \begin{cases} \Delta = k^2 - 8k \geqslant 0, \\ -\dfrac k 4 \geqslant -1, \\ f(1) \leqslant 0, \end{cases} $$ also works, and the final result is $\boldsymbol {(-\infty, 0]}$ instead.

Community
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xbh
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  • That's the condition for both the expression with $+$ and the expression with $-$ to be in $[-1,1]$. If you want at least one, then $k < -1$ is also allowed. – Robert Israel Oct 17 '19 at 19:53
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Write it as $$ k - 4 \le \pm \sqrt{k^2 - 8k} \le k + 4$$ Now there will be cases depending on whether $k$ is in the interval $(-\infty, -4]$, $[-4,4]$ or $[4,\infty)$.

EDIT: For example, if $k \in (-\infty, -4]$, the $\pm$ must be $-$, and $k-4 \le -\sqrt{k^2 - 8 k} \le k+4 \le 0$ becomes $$0 \le (k+4)^2 \le k^2-8k \le (k-4)^2$$ Expand the squares and subtract $k^2-8k$...

Robert Israel
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