Question 188987
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So, you know the answer is *[tex \large sqrt{2}].  Well, let's just find out if that is correct.


If


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  sqrt{b^2 + 16 + 2b} = 5b]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  b = sqrt{2}]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  sqrt{( \sqrt{2} )^2 + 16 + 2( \sqrt{2})} = 5 sqrt{2}]


Well, root 2 squared is just 2 plus 16 is 18 plus 2 times root 2 which is a little less than 3, so total somewhere between 20 and 21, take the root of that and you get maybe 4 and a half.  But 5 times root 2 is a little less than 7 and a half.  < <i><b>Buzz</b></i> > Thank you for playing, we have some lovely parting gifts.


Once you have


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0= 2(12b^2 - b - 8)]


You know that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12b^2 - b - 8 = 0]


Just use the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b_{1,2} = \frac{-\beta \pm sqrt{\beta^2 - 4\alpha\gamma}}{2\alpha}]


Where *[tex \large \alpha = 12], *[tex \large \beta = -1], and *[tex \large \gamma = -8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b_{1,2} = \frac{-1 \pm sqrt{(-1)^2 - 4(12)(-8)}}{2(12)}]


You get to do your own arithmetic.


You will get two roots to the quadratic, but in the process of squaring the variable you quite often get an extraneous root.  One of the roots to this equation will be extraneous, so you have to check both answers in the original equation to determine which one to throw out.


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