Solve x+(1-x)^2=19/27

$$x + ( 1 - x ) ^ { 2 } = \frac { 19 } { 27 }$$

Solve for x (complex solution)

$$x=\frac{\sqrt{15}i}{18}+\frac{1}{2}\approx 0.5+0.215165741i$$

$$x=-\frac{\sqrt{15}i}{18}+\frac{1}{2}\approx 0.5-0.215165741i$$

Steps Using the Quadratic Formula

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

$$x+1-2x+x^{2}=\frac{19}{27}$$

Combine x and -2x to get -x.

$$-x+1+x^{2}=\frac{19}{27}$$

Subtract \frac{19}{27} from both sides.

$$-x+1+x^{2}-\frac{19}{27}=0$$

Subtract \frac{19}{27} from 1 to get \frac{8}{27}.

$$-x+\frac{8}{27}+x^{2}=0$$

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.

$$x^{2}-x+\frac{8}{27}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -1 for b, and \frac{8}{27} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{8}{27}}}{2}$$

Multiply -4 times \frac{8}{27}.

$$x=\frac{-\left(-1\right)±\sqrt{1-\frac{32}{27}}}{2}$$

Add 1 to -\frac{32}{27}.

$$x=\frac{-\left(-1\right)±\sqrt{-\frac{5}{27}}}{2}$$

Take the square root of -\frac{5}{27}.

$$x=\frac{-\left(-1\right)±\frac{\sqrt{15}i}{9}}{2}$$

The opposite of -1 is 1.

$$x=\frac{1±\frac{\sqrt{15}i}{9}}{2}$$

Now solve the equation x=\frac{1±\frac{\sqrt{15}i}{9}}{2} when ± is plus. Add 1 to \frac{i\sqrt{15}}{9}.

$$x=\frac{\frac{\sqrt{15}i}{9}+1}{2}$$

Divide 1+\frac{i\sqrt{15}}{9} by 2.

$$x=\frac{\sqrt{15}i}{18}+\frac{1}{2}$$

Now solve the equation x=\frac{1±\frac{\sqrt{15}i}{9}}{2} when ± is minus. Subtract \frac{i\sqrt{15}}{9} from 1.

$$x=\frac{-\frac{\sqrt{15}i}{9}+1}{2}$$

Divide 1-\frac{i\sqrt{15}}{9} by 2.

$$x=-\frac{\sqrt{15}i}{18}+\frac{1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{15}i}{18}+\frac{1}{2} x=-\frac{\sqrt{15}i}{18}+\frac{1}{2}$$

Steps for Completing the Square

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

$$x+1-2x+x^{2}=\frac{19}{27}$$

Combine x and -2x to get -x.

$$-x+1+x^{2}=\frac{19}{27}$$

Subtract 1 from both sides.

$$-x+x^{2}=\frac{19}{27}-1$$

Subtract 1 from \frac{19}{27} to get -\frac{8}{27}.

$$-x+x^{2}=-\frac{8}{27}$$

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.

$$x^{2}-x=-\frac{8}{27}$$

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

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=-\frac{8}{27}+\left(-\frac{1}{2}\right)^{2}$$

Square -\frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-x+\frac{1}{4}=-\frac{8}{27}+\frac{1}{4}$$

Add -\frac{8}{27} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-x+\frac{1}{4}=-\frac{5}{108}$$

Factor x^{2}-x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{2}\right)^{2}=-\frac{5}{108}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{-\frac{5}{108}}$$

Simplify.

$$x-\frac{1}{2}=\frac{\sqrt{15}i}{18} x-\frac{1}{2}=-\frac{\sqrt{15}i}{18}$$

Add \frac{1}{2} to both sides of the equation.

$$x=\frac{\sqrt{15}i}{18}+\frac{1}{2} x=-\frac{\sqrt{15}i}{18}+\frac{1}{2}$$