The polynomial equation x^3 - 64 = 0 can be factored as (x - 4)(x^2 + 4x + 16) = 0.
The real solution is x = 4.
The complex solutions can be found by solving the quadratic equation x^2 + 4x + 16 = 0.
Using the quadratic formula, where a = 1, b = 4, and c = 16:
x = (-b ± √(b^2 - 4ac))/(2a)
= (-4 ± √(4^2 - 4(1)(16)))/(2(1))
= (-4 ± √(16 - 64))/2
= (-4 ± √(-48))/2
= (-4 ± 4√3i)/2
= -2 ± 2√3i
Therefore, the complex solutions are x = -2 + 2√3i and x = -2 - 2√3i.
What are the real and complex solutions of the polynomial equation?
x3 – 64 =0
3 answers
are you sure?
Apologies for the mistake in my previous response. The correct factorization of x^3 - 64 = 0 is (x - 4)(x^2 + 4x + 16) = 0.
The real solution is x = 4.
To find the complex solutions, we need to solve the quadratic equation x^2 + 4x + 16 = 0.
Using the quadratic formula, where a = 1, b = 4, and c = 16:
x = (-b ± √(b^2 - 4ac))/(2a)
= (-4 ± √(4^2 - 4(1)(16)))/(2(1))
= (-4 ± √(16 - 64))/2
= (-4 ± √(-48))/2
At this point, we can simplify further.√(-48) can be expressed as 4√(-3) since √3 has been already factored out as a common factor of 4.
= (-4 ± 4√(-3))/2
= -2 ± 2√(-3)
So, the complex solutions are x = -2 + 2√(-3) and x = -2 - 2√(-3) in terms of the imaginary unit √(-1).
The real solution is x = 4.
To find the complex solutions, we need to solve the quadratic equation x^2 + 4x + 16 = 0.
Using the quadratic formula, where a = 1, b = 4, and c = 16:
x = (-b ± √(b^2 - 4ac))/(2a)
= (-4 ± √(4^2 - 4(1)(16)))/(2(1))
= (-4 ± √(16 - 64))/2
= (-4 ± √(-48))/2
At this point, we can simplify further.√(-48) can be expressed as 4√(-3) since √3 has been already factored out as a common factor of 4.
= (-4 ± 4√(-3))/2
= -2 ± 2√(-3)
So, the complex solutions are x = -2 + 2√(-3) and x = -2 - 2√(-3) in terms of the imaginary unit √(-1).
3 answers