Find all real solutions of the equation. (Enter your answers as a comma-separated list. If there is no real solution, enter NO REAL SOLUTION.)
x4 + 27x = 0
x^4 + 27x = 0
x(x^3+27) = 0
recall your sum-of-cubes factoring to get
x(x+3)(x^2-3x+9)
two real roots, two complex roots
To find the real solutions of the equation x^4 + 27x = 0, we need to factorize it. Notice that both terms have a common factor of x:
x(x^3 + 27) = 0
Now, we can see that x = 0 satisfies the equation.
Next, let's focus on the expression inside the parentheses, x^3 + 27. This can be further factored using the sum of cubes formula: a^3 + b^3 = (a + b)(a^2 - ab + b^2). In this case, a = x and b = 3:
x^3 + 27 = (x + 3)(x^2 - 3x + 9)
Setting each factor equal to zero:
x + 3 = 0 --> x = -3
x^2 - 3x + 9 = 0
To solve for x in the quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. In this case, a = 1, b = -3, and c = 9. Plugging in these values:
x = (-(-3) ± √((-3)^2 - 4(1)(9))) / (2(1))
= (3 ± √(9 - 36)) / 2
= (3 ± √(-27)) / 2
Since the square root of a negative number is not a real number, there are no real solutions for the equation x^2 - 3x + 9 = 0.
Therefore, the real solutions of the equation x^4 + 27x = 0 are:
x = 0, -3
To find the real solutions of the equation x^4 + 27x = 0, we need to solve for x.
Step 1: Factor out the common factor.
x(x^3 + 27) = 0
Step 2: Set each factor equal to zero and solve for x.
x = 0
To find the solutions of the second factor x^3 + 27 = 0, we can rewrite it as (x + 3)(x^2 - 3x + 9) = 0 and solve for x.
Setting each factor equal to zero:
x + 3 = 0 --> x = -3
x^2 - 3x + 9 = 0
Since the quadratic equation does not factor further, we can use the quadratic formula to find its solutions:
x = (-(-3) ± √((-3)^2 - 4(1)(9))) / (2(1))
x = (3 ± √(9 - 36)) / 2
x = (3 ± √(-27)) / 2
Since the discriminant is negative (√(-27)), there are no real solutions for this quadratic equation. Therefore, the only real solutions for the given equation x^4 + 27x = 0 are x = 0 and x = -3.
Final Answer: 0, -3