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anonymous
 5 years ago
is there a value for h that makes it possible for the equation √(x+h)+5=0 to have any real number solution? explain
anonymous
 5 years ago
is there a value for h that makes it possible for the equation √(x+h)+5=0 to have any real number solution? explain

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is x+h+5 all under the square root? or just x+h?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok.. so you want to find a value for h that makes the expression \[\sqrt{x+h}+5=0 \implies \sqrt{x+h}=5\] but the square root can't have a negative value. so, such h doesn't exist.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, the sqroot of a number can have a negative value, the SQUARE of a number can't have a real value

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0square both sides and you get x+h = 25

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's not right mindedone, you're suggesting that h=25x is a solution.. check by plugging back into the equation and see what you get.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Keep in mind though that if you plug h=25x into the original equation you get +sqrt(25) = 5, and so the negative square root does satisfy the equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you get sqrt(25) = 5
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