Question 191926
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ s -4\sqrt{s} - 1 = 0]


The trick to this one is a substitution.  Let *[tex \LARGE x^2 = s], then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ x^2 -4x - 1 = 0]


Using the quadratic formula:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{-(-4) \pm \sqrt{(-4)^2 - (4)(1)(-1)}}{2(1)}]


You can do your own arithmetic to check that


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


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ x^2 = 9 \pm 4\sqrt{5}]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ s = 9 \pm 4\sqrt{5}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ \sqrt{s} = 2 \pm \sqrt{5}]


Check:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 9 + 4\sqrt{5} - 4(2 + \sqrt{5}) - 1 = 0]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \ \ 9 - 4\sqrt{5} - 4(2 - \sqrt{5}) - 1 = 0]



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