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I have a an equation $x(-x^{2} + \frac{3}{2}x + \frac{z-1}{2}) = 0$.

I want to solve for the range of values of $z$ which keep |$x$| < 1.

I have to show the values of z which keep the spectral radius of a matrix less than 1. This is to guarantee convergence of an iterative method of solving matrix equations.

Solving the positive case using the quadratic formula is easy enough, but I am confused on how to solve for the negative case, as the negative sign cancels when squaring both sides to get rid of the square root, leaving the same answer as the positive root, namely from 0 < $z$ < 6.

This can't be right. For example, plugging in $z=1$ results in |x|<1 for the positive root and |x|>1 for the negative, meaning that the spectral radius is greater than one.

Graphing this equation, I can see that the appropriate range is $-\frac{1}{8} < z < 0$, but I don't know the algebra to show this on paper.

How do I show the correct range algebraically?

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1 Answer 1

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Using the quadratic formula tells us that the nonzero roots of the equation are $x = \frac14 (3 \pm \sqrt{ 8 z+1})$. (We see that there are no nonzero roots when $z<-\frac18$.) Notice that the sum of these two roots is $\frac32$ independent of $z$ (we could see that from the original polynomial's coefficients); so the smaller root cannot be greater than $1$, and if it's less than $-1$ then the larger root is definitely greater than $1$. So it suffices to find when the larger root is less than $1$, which should give you your range $-\frac18\le z<0$.

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  • $\begingroup$ Evening Sir, I'm afraid that I got something different under the radical, namely $\sqrt{\frac{1}{4}+2z}$. I see that by factoring out a $\frac{1}{4}$ you can get to $\frac{1}{4}(8z+1)$ under the radical, but how did you move the $\frac{1}{4}$ outside the square root? Would you mind expounding? I hate to ask you to show a college student how to do algebra, but I would really like to understand your process. Thanks in advance! $\endgroup$
    – rocksNwaves
    Commented Nov 28, 2016 at 4:11
  • $\begingroup$ I did a little more work trying to replicate what you show above, and the closest I could get to what you arrived at by my own manipulation is $\frac{1}{4}(-3 \pm 2 \sqrt{\frac{1}{4}(8z+1)})$ $\endgroup$
    – rocksNwaves
    Commented Nov 28, 2016 at 4:21
  • $\begingroup$ Note that $2\sqrt{\frac14\cdot\text{something}}$ is the same as $\sqrt{\text{something}}$...! That should get you almost all the way there. The only remaining issue is the negative sign in front of the $3$; I think that must be a typo in your work (either you missed the negative sign in $-x^2$, or else you forgot the negative sign in the $-b$ part of the quadratic formula). $\endgroup$
    – Greg Martin
    Commented Nov 28, 2016 at 4:57
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    $\begingroup$ That indeed got me there. Sometimes I question my decision to go back to school for math almost 10 years after undergrad, and without a math or engineering back ground. Thanks to people like you who generously help me fill in the gaps, I am finding my way. $\endgroup$
    – rocksNwaves
    Commented Nov 28, 2016 at 15:09

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