what are the real solutions to the equation x^3 = -27?
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.
To find the real solutions to the equation x^3 = -27, follow these steps:
Step 1: Start with the given equation: x^3 = -27.
Step 2: Take the cube root of both sides of the equation to eliminate the exponent. The cube root of -27 is -3, since -3 * -3 * -3 = -27.
∛(x^3) = ∛(-27)
x = -3
Therefore, the real solution to the equation x^3 = -27 is x = -3.
To find the real solutions to the equation x^3 = -27, we need to find the values of x that satisfy this equation. Let's break down the process step by step:
Step 1: Rewrite the equation
x^3 = -27 can be rewritten as x^3 + 27 = 0.
Step 2: Use factoring techniques
We can leverage the difference of cubes formula to factor the equation further:
(x + 3)(x^2 - 3x + 9) = 0.
Step 3: Solve for x
To find the solutions, we set each factor equal to zero and solve for x.
a) x + 3 = 0
By solving this linear equation, we find that x = -3.
b) x^2 - 3x + 9 = 0
To find the roots of this quadratic equation, we can either solve it by factoring or by using the quadratic formula.
Using the quadratic formula, where a = 1, b = -3, and c = 9, we substitute the values:
x = (-(-3) ± √((-3)^2 - 4(1)(9))) / (2(1)).
This simplifies to: x = (3 ± √(9 - 36)) / 2.
Further simplification gives us: x = (3 ± √(-27)) / 2.
Since √(-27) produces complex roots (imaginary roots involving the square root of a negative number), there are no real solutions for the quadratic equation.
Therefore, the real solution to the equation x^3 = -27 is x = -3.