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Find the value range of $p$ if interval $(1,2)$ lies between the roots of $$2^x + 2^{2-x} + p = 0.$$

The answer is $(-\infty,-5)$.

My ''solution'':

Let $t= 2^x$, so $t \in(2,4)=:I$. Now you can write $$f(t) = -t-{4\over t}$$ Clearly since $t+{4\over t}\geq 4$ so $p\leq -4$. Since $f'(t)=-1+4t^{-2}$ we see that $f'(t)<0$ for $t\in I$ so $f$ is decreasing and thus $f_{\min} = f(4) =-5$. So given equation has a solution if $-5\leq p\leq-4$.

Clearly this is not correct answer since I try $p=-6$ which works.

I can't figure out what I'm doing wrong.

nonuser
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    What do you mean when you say that you try $p = -6$ and it works? I would say that if you take care of the less or equal and greater or equal signs, your answer is right. – M4g1ch May 20 '18 at 18:08
  • Okay, so to set the scene here: after multiplying by $2^x$, this is a quadratic equation in $t=2^x$, so it has at most two real roots. Specifically, it has two distinct real roots exactly when $|p|>4$. So we're looking for values of $p$ with $|p|>4$, such that the roots of this quadratic are on either side of the interval $(1, 2)$? – Jack M May 20 '18 at 18:11

1 Answers1

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What you've determined are conditions for which your equation has a solution in the interval $(1, 2)$. But that isn't the problem you're trying to solve. You want to know when $(1, 2)$ is between the roots of your equation, i.e. when the equation has one root in $(-\infty, 1]$ and one root in $[2, +\infty)$.

Jack M
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