Question 179112
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*[tex \Large x^2+18x+14=(x+a)^2+b]


*[tex \Large (x + a)^2 = x^2 + 2ax + a^2] so


*[tex \Large x^2+18x+14= x^2 + 2ax + a^2+b]


Clearly, *[tex \Large 2ax = 18x] and *[tex \Large a^2 + b = 14]


*[tex \Large 2ax = 18x \rightarrow 2a = 18 \rightarrow a = 9]


Since we know that *[tex \Large a = 9] we can substitute this value into:


*[tex \Large (9)^2 + b = 14 \rightarrow 81 + b = 14 \rightarrow b = 14 - 81 \rightarrow b = -67 ]


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