The equation x^3 = -27 can be written as x^3 + 27 = 0. By factoring the sum of cubes formula, we have (x + 3)(x^2 - 3x + 9) = 0.
To find the real solutions, we set each factor equal to zero:
x + 3 = 0 ➝ x = -3
x^2 - 3x + 9 = 0
Using the quadratic formula:
x = (-(-3) ± √((-3)^2 - 4(1)(9))) / (2(1))
x = (3 ± √(9 - 36)) / 2
x = (3 ± √(-27)) / 2
Since square root of negative numbers are not real, there are no additional real solutions to the equation x^3 = -27. Therefore, the only real solution is x = -3.
what are the real solutions to the equation x^3 = -27?
1 answer