Question 262307
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Take the square root of both sides, remembering to consider both the positive and negative root.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(x\ +\ 4\right)^2\ =\ 36]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 4\ =\ \pm 6]


Hence,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 4\ =\ 6\ \Rightarrow\ x\ =\ 2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ 4\ =\ -6\ \Rightarrow\ x\ =\ -10]


Check


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left((2)\ +\ 4\right)^2\ =^?\ 36]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6^2\ =\ 36\ \ \ \ ]True


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left((-10)\ +\ 4\right)^2\ =^?\ 36]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(-6\right)^2\ =\ 36\ \ \ \ ]True



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