I would appreciate it if anyone could tell me if what I am doing is correct The question is:

Which is greater $1$ or $|x|$ if $x^2+2y^2=2$ and $y=\frac{1}{2} $

Here is what I go after simplifying $x = \pm \sqrt{\frac{3}{2}}$ = $\pm1.2250$

Thus $|x|=1.2250$ which is greater than $1$.

However the text says that $1$ is greater than $|x|$ , should I consider that a misprint ?

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    $\begingroup$ Your computation looks right to me. $\endgroup$ – hmakholm left over Monica Jul 12 '12 at 19:45
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    $\begingroup$ The fact that $x^2 > 1$ immediately gives that $|x| > 1$. To see this, $x^2 = |x|^2$, and is increasing for positive $x$. Yeah though your work seems to pan out. $\endgroup$ – Eugene Shvarts Jul 12 '12 at 19:45

Yes, it’s an error in the book. Your work is basically fine, though it isn’t really correct to write $$\pm\sqrt\frac32=\pm 1.2250\;:$$ the equals sign should be reserved for things that are actually equal to each other. It would be much better to write

$$\pm\sqrt\frac32\approx\pm 1.2250\;.$$

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    $\begingroup$ +1, if nothing else for correcting the misuse of $=$, yet another peeve of mine... $\endgroup$ – Arturo Magidin Jul 12 '12 at 23:45
  • $\begingroup$ @Arturo: I routinely warned my students that I would always penalize them for misusing $=$, though generally not very much. $\endgroup$ – Brian M. Scott Jul 13 '12 at 6:07

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