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I have the double inequality:

$-x^2 + x(2n+1) - 2n \leq u < -x^2 + x(2n-1)$

and I am trying to get it into the form

$x \leq \text{ anything } < x+1$

Or at least solve for x as the smallest term. I know I need to use the quadratic formula but I don't understand how I can solve for two quadratics at once? How would this work?

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If you write $$f_1(x)=x^2-(2n+1)x+2n+u\ ,\quad f_2(x)=x^2-(2n-1)x+u$$ then you want to solve $$f_1(x)\ge0\ ,\quad f_2(x)<0\ .$$ The important thing to notice is that $$f_2(x)=f_1(x+1)\ .$$ Each quadratic has discriminant $$\Delta=(2n-1)^2-4u\ .$$

Firstly, if $\Delta\le0$ then $f_2(x)$ cannot be negative, and so there is no solution.

Now consider graphing $y=f_1(x)$ and $y=f_2(x)$, both on the same axes. (Please draw it yourself, I am not good at posting diagrams online.) The graph of $f_1$ is just that of $f_2$, shifted $1$ unit to the right. Suppose that $f_1$ has roots $\alpha_1<\alpha_2$ and $f_2$ has roots $\beta_1<\beta_2$. Please mark these on your graph. There are two cases.

  • Case I: if $\Delta\le1$ then $\beta_2\le\alpha_1$ and from the graph you can see that the solution is $\beta_1<x<\beta_2$.
  • Case II: if $\Delta>1$ then $\beta_2>\alpha_1$ and the solution is $\beta_1<x\le\alpha_1$.

So, the solutions are:

  • if $\Delta\le0$, no solution;
  • if $0<\Delta\le1$ then $n-\frac12-\frac{\sqrt\Delta}2<x<n-\frac12+\frac{\sqrt\Delta}2$;
  • if $\Delta>1$ then $n-\frac12-\frac{\sqrt\Delta}2<x<n+\frac12-\frac{\sqrt\Delta}2$.

Putting the answer in the form you requested, this implies $$x<n-\frac12|1-\sqrt\Delta|<x+1\ ,$$ provided that $\Delta>0$.

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  • $\begingroup$ thank you for the great response, how are you getting from the solution to the form I requested? It seems like you'd get the sum of the two external terms in the middle instead of one neat one. $\endgroup$ – alex Dec 16 '14 at 2:30
  • $\begingroup$ For example if $\Delta>1$ then we have$$x<n+\frac12-\frac{\sqrt\Delta}2\quad\hbox{and}\quad n-\frac12-\frac{\sqrt\Delta}2<x$$and adding $1$ to the second inequality gives$$n+\frac12-\frac{\sqrt\Delta}2<x+1\ .$$ $\endgroup$ – David Dec 16 '14 at 4:08
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Manipulate original inequality to get:

$(x)(2n+1) - 2n \leq u + x^2 < 2nx -x $

Manipulate further to get:

$2n \geq (2n+1)x - x^2 - u > 0$

From the first inequality, we get:

$x^2 - (2n+1)x + u - 2n \geq 0$

Use quadratic formula on that and choose the solution or solutions that satisfy the second inequality $(2n+1)x - x^2 - u > 0$

I hope it's okay that we're not going for $x \leq anything < x + 1$ anymore?

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You don't solve two quadratics at once. You divide the double inequality into two separate ones:

$$-x^2+x(2n+1)-2n\le u$$ $$u<-x^2+x(2n-1)$$

From then on, you can use whatever method you want to isolate $x$. It looks like you already know how to do that.

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