Question 237819
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Use a substitution:


Let *[tex \Large u\ =\ x^2\ -\ 7]


Now the problem becomes


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u^2\ +\ 3u\ -\ 12\ =\ 0]


Which, more's the pity, does not factor.


Use the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ u = \frac{-(3) \pm sqrt{(3)^2 - 4(1)(-12)}}{2(1)}\ =\ \frac{-3\ \pm\ 3\sqrt{5}}{2}]


(Verification left as an exercise for the student)


But now we have to go back and unwind the substitution:


Since *[tex \Large u\ =\ x^2\ -\ 7],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 7\ =\ \frac{-3\ \pm\ 3\sqrt{5}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ =\ \frac{11\ \pm\ 3\sqrt{5}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \pm\sqrt{\frac{11\ \pm\ 3\sqrt{5}}{2}}]


Yes, that is uglier than a mud fence.  Notice that you actually have four roots. This is, of course, to be expected because this equation began life as a quartic not a quadratic.


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