Find the Roots (Zeros) 9x^4-18x^3+27x^2=0

$9 x^{4} - 18 x^{3} + 27 x^{2} = 0$

Factor $9 x^{2}$ out of $9 x^{4} - 18 x^{3} + 27 x^{2}$ .

Factor $9 x^{2}$ out of $9 x^{4}$ .

$9 x^{2} (x^{2}) - 18 x^{3} + 27 x^{2} = 0$

Factor $9 x^{2}$ out of $- 18 x^{3}$ .

$9 x^{2} (x^{2}) + 9 x^{2} (- 2 x) + 27 x^{2} = 0$

Factor $9 x^{2}$ out of $27 x^{2}$ .

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

Factor $9 x^{2}$ out of $9 x^{2} (x^{2}) + 9 x^{2} (- 2 x)$ .

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

Factor $9 x^{2}$ out of $9 x^{2} (x^{2} - 2 x) + 9 x^{2} (3)$ .

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

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x^{2} = 0$

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

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $x^{2} - 2 x + 3$ equal to $0$ and solve for $x$ .

Set $x^{2} - 2 x + 3$ equal to $0$ .

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

Solve $x^{2} - 2 x + 3 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 2$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot 3)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 2$ to the power of $2$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot 3}}{2 \cdot 1}$

Multiply $- 4$ by $3$ .

$x = \frac{2 \pm \sqrt{4 - 12}}{2 \cdot 1}$

Subtract $12$ from $4$ .

$x = \frac{2 \pm \sqrt{- 8}}{2 \cdot 1}$

Rewrite $- 8$ as $- 1 (8)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 8}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{8}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{2 \pm i \cdot \sqrt{8}}{2 \cdot 1}$

Rewrite $8$ as $2^{2} \cdot 2$ .

Factor $4$ out of $8$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (2)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 2}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm i \cdot (2 \sqrt{2})}{2 \cdot 1}$

Move $2$ to the left of $i$ .

$x = \frac{2 \pm 2 i \sqrt{2}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 i \sqrt{2}}{2}$

Simplify $\frac{2 \pm 2 i \sqrt{2}}{2}$ .

$x = 1 \pm i \sqrt{2}$

The final answer is the combination of both solutions.

$x = 1 + i \sqrt{2}, 1 - i \sqrt{2}$

The final solution is all the values that make $9 x^{2} (x^{2} - 2 x + 3) = 0$ true.

$x = 0, 1 + i \sqrt{2}, 1 - i \sqrt{2}$

Find the Roots (Zeros) 9x^4-18x^3+27x^2=0

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