Solve (x+1/3)^2-9x^2-(1-2x)^2=1/9

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

Solve for x (complex solution)

$$x=\frac{7+\sqrt{59}i}{36}\approx 0.194444444+0.21336516i$$

$$x=\frac{-\sqrt{59}i+7}{36}\approx 0.194444444-0.21336516i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{1}{3}\right)^{2}.

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

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}+\frac{2}{3}x+\frac{1}{9}-\left(1-2x\right)^{2}=\frac{1}{9}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-2x\right)^{2}.

$$-8x^{2}+\frac{2}{3}x+\frac{1}{9}-\left(1-4x+4x^{2}\right)=\frac{1}{9}$$

To find the opposite of 1-4x+4x^{2}, find the opposite of each term.

$$-8x^{2}+\frac{2}{3}x+\frac{1}{9}-1+4x-4x^{2}=\frac{1}{9}$$

Subtract 1 from \frac{1}{9} to get -\frac{8}{9}.

$$-8x^{2}+\frac{2}{3}x-\frac{8}{9}+4x-4x^{2}=\frac{1}{9}$$

Combine \frac{2}{3}x and 4x to get \frac{14}{3}x.

$$-8x^{2}+\frac{14}{3}x-\frac{8}{9}-4x^{2}=\frac{1}{9}$$

Combine -8x^{2} and -4x^{2} to get -12x^{2}.

$$-12x^{2}+\frac{14}{3}x-\frac{8}{9}=\frac{1}{9}$$

Subtract \frac{1}{9} from both sides.

$$-12x^{2}+\frac{14}{3}x-\frac{8}{9}-\frac{1}{9}=0$$

Subtract \frac{1}{9} from -\frac{8}{9} to get -1.

$$-12x^{2}+\frac{14}{3}x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -12 for a, \frac{14}{3} for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{14}{3}±\sqrt{\left(\frac{14}{3}\right)^{2}-4\left(-12\right)\left(-1\right)}}{2\left(-12\right)}$$

Square \frac{14}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{14}{3}±\sqrt{\frac{196}{9}-4\left(-12\right)\left(-1\right)}}{2\left(-12\right)}$$

Multiply -4 times -12.

$$x=\frac{-\frac{14}{3}±\sqrt{\frac{196}{9}+48\left(-1\right)}}{2\left(-12\right)}$$

Multiply 48 times -1.

$$x=\frac{-\frac{14}{3}±\sqrt{\frac{196}{9}-48}}{2\left(-12\right)}$$

Add \frac{196}{9} to -48.

$$x=\frac{-\frac{14}{3}±\sqrt{-\frac{236}{9}}}{2\left(-12\right)}$$

Take the square root of -\frac{236}{9}.

$$x=\frac{-\frac{14}{3}±\frac{2\sqrt{59}i}{3}}{2\left(-12\right)}$$

Multiply 2 times -12.

$$x=\frac{-\frac{14}{3}±\frac{2\sqrt{59}i}{3}}{-24}$$

Now solve the equation x=\frac{-\frac{14}{3}±\frac{2\sqrt{59}i}{3}}{-24} when ± is plus. Add -\frac{14}{3} to \frac{2i\sqrt{59}}{3}.

$$x=\frac{-14+2\sqrt{59}i}{-24\times 3}$$

Divide \frac{-14+2i\sqrt{59}}{3} by -24.

$$x=\frac{-\sqrt{59}i+7}{36}$$

Now solve the equation x=\frac{-\frac{14}{3}±\frac{2\sqrt{59}i}{3}}{-24} when ± is minus. Subtract \frac{2i\sqrt{59}}{3} from -\frac{14}{3}.

$$x=\frac{-2\sqrt{59}i-14}{-24\times 3}$$

Divide \frac{-14-2i\sqrt{59}}{3} by -24.

$$x=\frac{7+\sqrt{59}i}{36}$$

The equation is now solved.

$$x=\frac{-\sqrt{59}i+7}{36} x=\frac{7+\sqrt{59}i}{36}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+\frac{1}{3}\right)^{2}.

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

Combine x^{2} and -9x^{2} to get -8x^{2}.

$$-8x^{2}+\frac{2}{3}x+\frac{1}{9}-\left(1-2x\right)^{2}=\frac{1}{9}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-2x\right)^{2}.

$$-8x^{2}+\frac{2}{3}x+\frac{1}{9}-\left(1-4x+4x^{2}\right)=\frac{1}{9}$$

To find the opposite of 1-4x+4x^{2}, find the opposite of each term.

$$-8x^{2}+\frac{2}{3}x+\frac{1}{9}-1+4x-4x^{2}=\frac{1}{9}$$

Subtract 1 from \frac{1}{9} to get -\frac{8}{9}.

$$-8x^{2}+\frac{2}{3}x-\frac{8}{9}+4x-4x^{2}=\frac{1}{9}$$

Combine \frac{2}{3}x and 4x to get \frac{14}{3}x.

$$-8x^{2}+\frac{14}{3}x-\frac{8}{9}-4x^{2}=\frac{1}{9}$$

Combine -8x^{2} and -4x^{2} to get -12x^{2}.

$$-12x^{2}+\frac{14}{3}x-\frac{8}{9}=\frac{1}{9}$$

Add \frac{8}{9} to both sides.

$$-12x^{2}+\frac{14}{3}x=\frac{1}{9}+\frac{8}{9}$$

Add \frac{1}{9} and \frac{8}{9} to get 1.

$$-12x^{2}+\frac{14}{3}x=1$$

Divide both sides by -12.

$$\frac{-12x^{2}+\frac{14}{3}x}{-12}=\frac{1}{-12}$$

Dividing by -12 undoes the multiplication by -12.

$$x^{2}+\frac{\frac{14}{3}}{-12}x=\frac{1}{-12}$$

Divide \frac{14}{3} by -12.

$$x^{2}-\frac{7}{18}x=\frac{1}{-12}$$

Divide 1 by -12.

$$x^{2}-\frac{7}{18}x=-\frac{1}{12}$$

Divide -\frac{7}{18}, the coefficient of the x term, by 2 to get -\frac{7}{36}. Then add the square of -\frac{7}{36} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{18}x+\left(-\frac{7}{36}\right)^{2}=-\frac{1}{12}+\left(-\frac{7}{36}\right)^{2}$$

Square -\frac{7}{36} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{18}x+\frac{49}{1296}=-\frac{1}{12}+\frac{49}{1296}$$

Add -\frac{1}{12} to \frac{49}{1296} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{18}x+\frac{49}{1296}=-\frac{59}{1296}$$

Factor x^{2}-\frac{7}{18}x+\frac{49}{1296}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{7}{36}\right)^{2}=-\frac{59}{1296}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{36}\right)^{2}}=\sqrt{-\frac{59}{1296}}$$

Simplify.

$$x-\frac{7}{36}=\frac{\sqrt{59}i}{36} x-\frac{7}{36}=-\frac{\sqrt{59}i}{36}$$

Add \frac{7}{36} to both sides of the equation.

$$x=\frac{7+\sqrt{59}i}{36} x=\frac{-\sqrt{59}i+7}{36}$$