Certainly! Let's solve the equation [tex]\( x^4 - 27x = 0 \)[/tex] step by step.
1. Factor out the common term:
First, we can factor out [tex]\( x \)[/tex] from the equation:
[tex]\[
x^4 - 27x = 0 \implies x(x^3 - 27) = 0
\][/tex]
2. Solve for the first solution:
Now we have a product of factors equal to zero. According to the zero-product property, if the product of two factors is zero, at least one of the factors must be zero.
So, we set each factor equal to zero and solve:
[tex]\[
x = 0
\][/tex]
3. Solve the cubic equation:
Next, we solve the cubic equation [tex]\( x^3 - 27 = 0 \)[/tex]:
[tex]\[
x^3 - 27 = 0
\][/tex]
4. Express the cubic equation in standard form:
We recognize that [tex]\( 27 \)[/tex] is a perfect cube, [tex]\( 27 = 3^3 \)[/tex], so our equation can be rewritten as:
[tex]\[
x^3 - 3^3 = 0
\][/tex]
This is a difference of cubes, which can be factored using the formula [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex]:
[tex]\[
x^3 - 3^3 = (x - 3)(x^2 + 3x + 9) = 0
\][/tex]
5. Solve for the real root:
We now set each factor of the factored form to zero and solve:
[tex]\[
x - 3 = 0 \implies x = 3
\][/tex]
6. Solve the quadratic equation:
For the quadratic part:
[tex]\[
x^2 + 3x + 9 = 0
\][/tex]
We use the quadratic formula to solve it, where [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 9\)[/tex]:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substituting the values:
[tex]\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 36}}{2} = \frac{-3 \pm \sqrt{-27}}{2} = \frac{-3 \pm 3i\sqrt{3}}{2}
\][/tex]
Thus, we get two complex solutions:
[tex]\[
x = \frac{-3 + 3i\sqrt{3}}{2} \quad \text{and} \quad x = \frac{-3 - 3i\sqrt{3}}{2}
\][/tex]
These can be further simplified to:
[tex]\[
x = -\frac{3}{2} + \frac{3i\sqrt{3}}{2} \quad \text{and} \quad x = -\frac{3}{2} - \frac{3i\sqrt{3}}{2}
\][/tex]
7. Compile all solutions:
Therefore, the solutions to the equation [tex]\( x^4 - 27x = 0 \)[/tex] are:
[tex]\[
x = 0, \quad x = 3, \quad x = -\frac{3}{2} + \frac{3i\sqrt{3}}{2}, \quad x = -\frac{3}{2} - \frac{3i\sqrt{3}}{2}
\][/tex]
Hence, the roots of the equation are:
[tex]\[
\boxed{0, 3, -\frac{3}{2} - \frac{3\sqrt{3}i}{2}, -\frac{3}{2} + \frac{3\sqrt{3}i}{2}}
\][/tex]
