Solve left(3xright)^2-4x+1=0

$${ \left(3x \right) }^{ 2 } -4x+1=0$$

Solve for x (complex solution)

$$x=\frac{2+\sqrt{5}i}{9}\approx 0.222222222+0.248451997i$$

$$x=\frac{-\sqrt{5}i+2}{9}\approx 0.222222222-0.248451997i$$

Steps Using the Quadratic Formula

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-4x+1=0$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-4x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -4 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 9}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-4\right)±\sqrt{16-36}}{2\times 9}$$

Add 16 to -36.

$$x=\frac{-\left(-4\right)±\sqrt{-20}}{2\times 9}$$

Take the square root of -20.

$$x=\frac{-\left(-4\right)±2\sqrt{5}i}{2\times 9}$$

The opposite of -4 is 4.

$$x=\frac{4±2\sqrt{5}i}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{4±2\sqrt{5}i}{18}$$

Now solve the equation x=\frac{4±2\sqrt{5}i}{18} when ± is plus. Add 4 to 2i\sqrt{5}.

$$x=\frac{4+2\sqrt{5}i}{18}$$

Divide 4+2i\sqrt{5} by 18.

$$x=\frac{2+\sqrt{5}i}{9}$$

Now solve the equation x=\frac{4±2\sqrt{5}i}{18} when ± is minus. Subtract 2i\sqrt{5} from 4.

$$x=\frac{-2\sqrt{5}i+4}{18}$$

Divide 4-2i\sqrt{5} by 18.

$$x=\frac{-\sqrt{5}i+2}{9}$$

The equation is now solved.

$$x=\frac{2+\sqrt{5}i}{9} x=\frac{-\sqrt{5}i+2}{9}$$

Steps for Completing the Square

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-4x+1=0$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-4x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-4x=-1$$

Divide both sides by 9.

$$\frac{9x^{2}-4x}{9}=-\frac{1}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{4}{9}x=-\frac{1}{9}$$

Divide -\frac{4}{9}, the coefficient of the x term, by 2 to get -\frac{2}{9}. Then add the square of -\frac{2}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{9}x+\left(-\frac{2}{9}\right)^{2}=-\frac{1}{9}+\left(-\frac{2}{9}\right)^{2}$$

Square -\frac{2}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{9}x+\frac{4}{81}=-\frac{1}{9}+\frac{4}{81}$$

Add -\frac{1}{9} to \frac{4}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{9}x+\frac{4}{81}=-\frac{5}{81}$$

Factor x^{2}-\frac{4}{9}x+\frac{4}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{9}\right)^{2}=-\frac{5}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{9}\right)^{2}}=\sqrt{-\frac{5}{81}}$$

Simplify.

$$x-\frac{2}{9}=\frac{\sqrt{5}i}{9} x-\frac{2}{9}=-\frac{\sqrt{5}i}{9}$$

Add \frac{2}{9} to both sides of the equation.

$$x=\frac{2+\sqrt{5}i}{9} x=\frac{-\sqrt{5}i+2}{9}$$