Question
What are the real and complex solutions of the polynomial equation? x3−64=0 3 − 64 = 0 (1 point) Responses 4, 1+2i3–√, 1+2i3–√ 4 , 1 + 2 3 , 1 + 2 3 4, 1+2i3–√, 1+2i3–√ 4 , 1 + 2 3 , 1 + 2 3 4,−1+2i3, −−−√−1+2i3–√ 4 , − 1 + 2 3 , − 1 + 2 3 4,−1+2i3, −−−√−1+2i3–√ 4 , − 1 + 2 3 , − 1 + 2 3 4, −2+2i3–√, −2−2i3–√ 4 , − 2 + 2 3 , − 2 − 2 3 4, −2+2i3–√, −2−2i3–√ 4 , − 2 + 2 3 , − 2 − 2 3 4, −2+2i3–√, 2+2i3–√ 4 , − 2 + 2 3 , 2 + 2 3 4, −2+2i3–√, 2+2i3–√
Answers
GPT 3.5
The polynomial equation is x^3 - 64 = 0.
To find the solutions, we can factor the equation as (x - 4)(x^2 + 4x + 16) = 0.
Setting each factor equal to zero, we have two real solutions:
x - 4 = 0, which gives x = 4.
x^2 + 4x + 16 = 0 does not have any real solutions. Instead, it has complex solutions. We can use the quadratic formula to find the complex solutions:
x = (-4 ± √(4^2 - 4(1)(16))) / (2(1))
x = (-4 ± √(-48)) / 2
x = (-4 ± 4i√3) / 2
x = -2 ± 2i√3
Therefore, the real solution is x = 4, and the complex solutions are x = -2 + 2i√3 and x = -2 - 2i√3.
To find the solutions, we can factor the equation as (x - 4)(x^2 + 4x + 16) = 0.
Setting each factor equal to zero, we have two real solutions:
x - 4 = 0, which gives x = 4.
x^2 + 4x + 16 = 0 does not have any real solutions. Instead, it has complex solutions. We can use the quadratic formula to find the complex solutions:
x = (-4 ± √(4^2 - 4(1)(16))) / (2(1))
x = (-4 ± √(-48)) / 2
x = (-4 ± 4i√3) / 2
x = -2 ± 2i√3
Therefore, the real solution is x = 4, and the complex solutions are x = -2 + 2i√3 and x = -2 - 2i√3.