Question 407493
# 1


{{{2(x-7)^(1/3)^"" +5 = 7}}}



{{{2(x-7)^(1/3)^"" = 7-5}}}



{{{2(x-7)^(1/3)^"" = 2}}}



{{{(x-7)^(1/3)^"" = 2/2}}}



{{{(x-7)^(1/3)^"" = 1}}}



{{{root(3,x-7) = 1}}}



{{{x-7 = 1^3}}}



{{{x-7 = 1}}}



{{{x = 1+7}}}



{{{x = 8}}}



So the solution is {{{x = 8}}}


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# 2


{{{2x^(1/2)^""  +8 = 0 }}}



{{{2x^(1/2)^""  = -8 }}}



{{{x^(1/2)^""  = -4 }}}



{{{sqrt(x) = -4 }}}



Recall that the range of the square root function is nonnegative. So the last equation is NOT possible. So there are NO solutions to the last equation.



Because the set of equations above are all equivalent, this means that there are NO solutions to the equation {{{2x^(1/2) +8 = 0 }}}



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