Question

solve the following:

x^2=-16
x(x+2)=0
(x-3)(x-3)=0
do they have one solution, two solutions, or no solution? explain

Answer 1

Assume the usual real-number domain unless stated otherwise.

1) x^2 = −16
- Over the real numbers: no solution, because a real square is ≥ 0 and cannot equal −16.
- Over the complex numbers: two solutions, x = 4i and x = −4i.

2) x(x + 2) = 0
- Use the zero-product property: x = 0 or x + 2 = 0 → x = −2.
- Two distinct real solutions: x = 0 and x = −2.

3) (x − 3)(x − 3) = 0 (i.e. (x − 3)^2 = 0)
- x − 3 = 0 → x = 3.
- One distinct real solution (x = 3). Note: as a quadratic it has a double root (multiplicity 2), but only one distinct solution.