anonymous
  • anonymous
is there a value for h that makes it possible for the equation √(x+h)+5=0 to have any real number solution? explain
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
is x+h+5 all under the square root? or just x+h?
anonymous
  • anonymous
just x+h
anonymous
  • anonymous
ok.. 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.

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anonymous
  • anonymous
wait, the sqroot of a number can have a negative value, the SQUARE of a number can't have a real value
anonymous
  • anonymous
square both sides and you get x+h = 25
anonymous
  • anonymous
that's not right mindedone, you're suggesting that h=25-x is a solution.. check by plugging back into the equation and see what you get.
anonymous
  • anonymous
Keep in mind though that if you plug h=25-x into the original equation you get +-sqrt(25) = -5, and so the negative square root does satisfy the equation.
anonymous
  • anonymous
you get sqrt(25) = -5
anonymous
  • anonymous
sqrt(25) = +-5

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