What are the real and complex solutions of the polynomial equation?
27x^3 + 125 = 0
I already got the real solution x = -5/3, I just need to figure out how to find the complex solution. Any help?
you have the sum of two cubes, so it factors into
(3x+5)((3x)^2-(3x)(5)+5^2)
= (3x+5)(9x^2-15x+25)
It's easy to check that the quadratic has no real solutions. Just use the quadratic formula, learned long ago in Algebra I...
To find the complex solutions of the polynomial equation 27x^3 + 125 = 0, we can use the concept of complex numbers and the fact that every polynomial of degree n has exactly n complex solutions.
Step 1: Rewrite the equation in factored form
We start by factoring the polynomial equation:
27x^3 + 125 = 0
Since 27x^3 = (3x)^3 and 125 = 5^3, we can write the equation as:
(3x)^3 + 5^3 = 0
This can be further factored using the sum of cubes formula:
(a^3 + b^3) = (a + b)(a^2 - ab + b^2)
Applying this formula, we get:
(3x + 5)((3x)^2 - (3x)(5) + 5^2) = 0
Simplifying:
(3x + 5)(9x^2 - 15x + 25) = 0
Now we have two factors: (3x + 5) and (9x^2 - 15x + 25). To find the complex solutions, we set each factor equal to zero and solve for x.
Step 2: Solve the first factor (3x + 5) = 0
Setting 3x + 5 = 0 and solving for x, we get:
3x = -5
x = -5/3
As you mentioned, this is the real solution.
Step 3: Solve the second factor (9x^2 - 15x + 25) = 0
To solve the quadratic factor, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, the quadratic equation is:
9x^2 - 15x + 25 = 0
where a = 9, b = -15, and c = 25.
Plugging the values into the quadratic formula:
x = (-(-15) ± √((-15)^2 - 4(9)(25))) / (2(9))
Simplifying:
x = (15 ± √(225 - 900)) / 18
x = (15 ± √(-675)) / 18
This is where we encounter complex numbers. The square root of a negative number is not possible in the realm of real numbers, but in the complex number system, we can represent it as the imaginary unit, denoted by "i."
Step 4: Simplify the square root of -675
√(-675) = √(675)√(-1) = √(225 * 3)√(-1) = 15√(3) i
Substituting this result back into the quadratic formula:
x = (15 ± 15√(3) i) / 18
We can simplify this further by factoring out a common factor of 3:
x = (5 ± 5√(3) i) / 6
These are the complex solutions to the polynomial equation 27x^3 + 125 = 0:
x = -5/3 (real solution)
x = (5 ± 5√(3) i) / 6 (complex solutions)
To find the complex solutions of the polynomial equation, we can use the fact that complex numbers are of the form a + bi, where a is the real part and b is the imaginary part.
Let's start by rewriting the equation: 27x^3 + 125 = 0
We can factor out the common factor from the first term:
27(x^3 + 125/27) = 0
We can see that the second term is actually (5/3)^3, so we can rewrite the equation as:
27(x + 5/3)(x^2 - (5/3)x + (25/9)) = 0
Now we have a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = -(5/3), and c = (25/9).
To solve this quadratic equation, we can use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / 2a
Substituting the values of a, b, and c, we get:
x = [-(5/3) ± √((5/3)^2 - 4(1)(25/9))] / (2(1))
Simplifying this equation, we have:
x = [-5/3 ± √(25/9 - 100/9)] / 2
x = [-5/3 ± √(-75/9)] / 2
Now, √(-75/9) is not a real number, which means we have complex solutions. To simplify it, we can express √(-75/9) in terms of the imaginary unit 'i', where i^2 = -1.
√(-75/9) = √(-25/3) = √(-1 × 25/3) = √(-1) × √(25/3) = i √(25/3)
Now, we can rewrite the solutions:
x = (-5/3 ± i √(25/3)) / 2
Therefore, the two complex solutions are:
x = (-5/3 + i √(25/3)) / 2
x = (-5/3 - i √(25/3)) / 2