2180.11 – Real-valued Radical


For how many real numbers xx is (x+1)2\sqrt{-(x+1)^2} a real number?

  1. none

  2. one

  3. two

  4. a finite number greater than two

  5. infinitely many


Solution

The quantity (x+1)2\sqrt{-(x+1)^2} is real if (x+1)20–(x + 1)^2 \geq 0, i.e., (x+1)20.(x + 1)^2 \leq 0. This means that (x+1)2=0 or (x+1)=0.(x + 1)^2 = 0 \text{ or } (x + 1) = 0.

So x=1x = -1 and (b) is the correct answer.