Question 1015213
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(y\ +\ 4\right)^{\frac{2}{3}}\ =\ -2]


Raise both sides to the third power


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(y\ +\ 4\right)^2\ =\ -8]


Expand the squared binomial


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ 8y\ +\ 16\ =\ -8]


Put the quadratic in standard form


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y^2\ +\ 8y\ +\ 24\ =\ 0]


but the discriminant, *[tex \Large b^2\ -\ 4ac] is negative


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 64\ -\ (4)(24)\ <\ 0]


So the set of real number solutions to



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(y\ +\ 4\right)^{\frac{2}{3}}\ =\ -2]


is the empty set.


This result squares with an intuitive analysis of the original problem.  Whenever a fractional exponent has an even numerator, the result of raising to that power must be positive.  Hence, the LHS of the original equation can never be negative, much less -2.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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