How to solve x ^ 3 + 27 = 0

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve x ^ 3 + 27 = 0 | Equatix

Question

$$x^{3}+27=0$$

Solve for x (complex solution)

$x=\frac{-3\sqrt{3}i+3}{2}\approx 1.5-2.598076211i$
$x=-3$
$x=\frac{3+3\sqrt{3}i}{2}\approx 1.5+2.598076211i$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $27$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±27,±9,±3,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-3$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}+27$ by $x+3$ to get $x^{2}-3x+9$. Solve the equation where the result equals to $0$.

$$x^{2}-3x+9=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $-3$ for $b$, and $9$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-27}}{2}$$

Solve the equation $x^{2}-3x+9=0$ when $±$ is plus and when $±$ is minus.

$$x=\frac{-3i\sqrt{3}+3}{2}$$ $$x=\frac{3+3i\sqrt{3}}{2}$$

List all found solutions.

$$x=-3$$ $$x=\frac{-3i\sqrt{3}+3}{2}$$ $$x=\frac{3+3i\sqrt{3}}{2}$$

Show Solution

Solve for x

$x=-3$

Solution Steps

$$±27,±9,±3,±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=-3$$

By Factor theorem, $x-k$ is a factor of the polynomial for each root $k$. Divide $x^{3}+27$ by $x+3$ to get $x^{2}-3x+9$. Solve the equation where the result equals to $0$.

$$x^{2}-3x+9=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $-3$ for $b$, and $9$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\times 9}}{2}$$

Do the calculations.

$$x=\frac{3±\sqrt{-27}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=-3$$

Show Solution

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