Solve x-9x^2+2x-4=0

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

Solve for x (complex solution)

$$x=\frac{-\sqrt{15}i+1}{6}\approx 0.166666667-0.645497224i$$

$$x=\frac{1+\sqrt{15}i}{6}\approx 0.166666667+0.645497224i$$

Steps Using the Quadratic Formula

Combine x and 2x to get 3x.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 3 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 3.

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

Multiply -4 times -9.

$$x=\frac{-3±\sqrt{9+36\left(-4\right)}}{2\left(-9\right)}$$

Multiply 36 times -4.

$$x=\frac{-3±\sqrt{9-144}}{2\left(-9\right)}$$

Add 9 to -144.

$$x=\frac{-3±\sqrt{-135}}{2\left(-9\right)}$$

Take the square root of -135.

$$x=\frac{-3±3\sqrt{15}i}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-3±3\sqrt{15}i}{-18}$$

Now solve the equation x=\frac{-3±3\sqrt{15}i}{-18} when ± is plus. Add -3 to 3i\sqrt{15}.

$$x=\frac{-3+3\sqrt{15}i}{-18}$$

Divide -3+3i\sqrt{15} by -18.

$$x=\frac{-\sqrt{15}i+1}{6}$$

Now solve the equation x=\frac{-3±3\sqrt{15}i}{-18} when ± is minus. Subtract 3i\sqrt{15} from -3.

$$x=\frac{-3\sqrt{15}i-3}{-18}$$

Divide -3-3i\sqrt{15} by -18.

$$x=\frac{1+\sqrt{15}i}{6}$$

The equation is now solved.

$$x=\frac{-\sqrt{15}i+1}{6} x=\frac{1+\sqrt{15}i}{6}$$

Steps for Completing the Square

Combine x and 2x to get 3x.

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

Add 4 to both sides. Anything plus zero gives itself.

$$3x-9x^{2}=4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -9.

$$\frac{-9x^{2}+3x}{-9}=\frac{4}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{3}{-9}x=\frac{4}{-9}$$

Reduce the fraction \frac{3}{-9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{1}{3}x=\frac{4}{-9}$$

Divide 4 by -9.

$$x^{2}-\frac{1}{3}x=-\frac{4}{9}$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=-\frac{4}{9}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=-\frac{4}{9}+\frac{1}{36}$$

Add -\frac{4}{9} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=-\frac{5}{12}$$

Factor x^{2}-\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{6}\right)^{2}=-\frac{5}{12}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{-\frac{5}{12}}$$

Simplify.

$$x-\frac{1}{6}=\frac{\sqrt{15}i}{6} x-\frac{1}{6}=-\frac{\sqrt{15}i}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{1+\sqrt{15}i}{6} x=\frac{-\sqrt{15}i+1}{6}$$