Question 235591
<font face="Garamond" size="+2">


Since *[tex \Large a^2\ -\ b^2\ =\ 0], we can say *[tex \Large a^2\ =\ b^2]


Hence, *[tex \Large a\ =\ \pm b]


Since *[tex \Large 2ab\ =\ 1], we can say *[tex \Large ab\ =\ \frac{1}{2}]


Since *[tex \Large \frac{1}{2}\ >\ 0], we know that *[tex \Large a] and *[tex \Large b] must have the same sign.  Therefore we can now say that *[tex \Large a\ =\ b]


But since *[tex \Large a\ =\ b], we now can say *[tex \Large a^2\ =\ \frac{1}{2}] and therefore *[tex \Large a\ =\ \pm \sqrt{\frac{1}{2}}\ =\ \pm\frac{1}{\sqrt{2}}]


Rationalizing the denominator:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ \pm \frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}}{\sqrt{2}}\right)\ =\ \pm\frac{\sqrt{2}}{2}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>