Solve +x=4/9(x^2-2x+1)

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

Solve for x

$$x=4$$

$$x=\frac{1}{4}=0.25$$

Steps Using the Quadratic Formula

Use the distributive property to multiply \frac{4}{9} by x^{2}-2x+1.

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

Subtract \frac{4}{9}x^{2} from both sides.

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

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

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

Combine x and \frac{8}{9}x to get \frac{17}{9}x.

$$\frac{17}{9}x-\frac{4}{9}x^{2}=\frac{4}{9}$$

Subtract \frac{4}{9} from both sides.

$$\frac{17}{9}x-\frac{4}{9}x^{2}-\frac{4}{9}=0$$

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.

$$-\frac{4}{9}x^{2}+\frac{17}{9}x-\frac{4}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{4}{9} for a, \frac{17}{9} for b, and -\frac{4}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{17}{9}±\sqrt{\left(\frac{17}{9}\right)^{2}-4\left(-\frac{4}{9}\right)\left(-\frac{4}{9}\right)}}{2\left(-\frac{4}{9}\right)}$$

Square \frac{17}{9} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{17}{9}±\sqrt{\frac{289}{81}-4\left(-\frac{4}{9}\right)\left(-\frac{4}{9}\right)}}{2\left(-\frac{4}{9}\right)}$$

Multiply -4 times -\frac{4}{9}.

$$x=\frac{-\frac{17}{9}±\sqrt{\frac{289}{81}+\frac{16}{9}\left(-\frac{4}{9}\right)}}{2\left(-\frac{4}{9}\right)}$$

Multiply \frac{16}{9} times -\frac{4}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{17}{9}±\sqrt{\frac{289-64}{81}}}{2\left(-\frac{4}{9}\right)}$$

Add \frac{289}{81} to -\frac{64}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{17}{9}±\sqrt{\frac{25}{9}}}{2\left(-\frac{4}{9}\right)}$$

Take the square root of \frac{25}{9}.

$$x=\frac{-\frac{17}{9}±\frac{5}{3}}{2\left(-\frac{4}{9}\right)}$$

Multiply 2 times -\frac{4}{9}.

$$x=\frac{-\frac{17}{9}±\frac{5}{3}}{-\frac{8}{9}}$$

Now solve the equation x=\frac{-\frac{17}{9}±\frac{5}{3}}{-\frac{8}{9}} when ± is plus. Add -\frac{17}{9} to \frac{5}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{2}{9}}{-\frac{8}{9}}$$

Divide -\frac{2}{9} by -\frac{8}{9} by multiplying -\frac{2}{9} by the reciprocal of -\frac{8}{9}.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-\frac{17}{9}±\frac{5}{3}}{-\frac{8}{9}} when ± is minus. Subtract \frac{5}{3} from -\frac{17}{9} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{32}{9}}{-\frac{8}{9}}$$

Divide -\frac{32}{9} by -\frac{8}{9} by multiplying -\frac{32}{9} by the reciprocal of -\frac{8}{9}.

$$x=4$$

The equation is now solved.

$$x=\frac{1}{4} x=4$$

Steps for Completing the Square

Use the distributive property to multiply \frac{4}{9} by x^{2}-2x+1.

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

Subtract \frac{4}{9}x^{2} from both sides.

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

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

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

Combine x and \frac{8}{9}x to get \frac{17}{9}x.

$$\frac{17}{9}x-\frac{4}{9}x^{2}=\frac{4}{9}$$

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.

$$-\frac{4}{9}x^{2}+\frac{17}{9}x=\frac{4}{9}$$

Divide both sides of the equation by -\frac{4}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{4}{9}x^{2}+\frac{17}{9}x}{-\frac{4}{9}}=\frac{\frac{4}{9}}{-\frac{4}{9}}$$

Dividing by -\frac{4}{9} undoes the multiplication by -\frac{4}{9}.

$$x^{2}+\frac{\frac{17}{9}}{-\frac{4}{9}}x=\frac{\frac{4}{9}}{-\frac{4}{9}}$$

Divide \frac{17}{9} by -\frac{4}{9} by multiplying \frac{17}{9} by the reciprocal of -\frac{4}{9}.

$$x^{2}-\frac{17}{4}x=\frac{\frac{4}{9}}{-\frac{4}{9}}$$

Divide \frac{4}{9} by -\frac{4}{9} by multiplying \frac{4}{9} by the reciprocal of -\frac{4}{9}.

$$x^{2}-\frac{17}{4}x=-1$$

Divide -\frac{17}{4}, the coefficient of the x term, by 2 to get -\frac{17}{8}. Then add the square of -\frac{17}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{4}x+\left(-\frac{17}{8}\right)^{2}=-1+\left(-\frac{17}{8}\right)^{2}$$

Square -\frac{17}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{4}x+\frac{289}{64}=-1+\frac{289}{64}$$

Add -1 to \frac{289}{64}.

$$x^{2}-\frac{17}{4}x+\frac{289}{64}=\frac{225}{64}$$

Factor x^{2}-\frac{17}{4}x+\frac{289}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{17}{8}\right)^{2}=\frac{225}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{8}\right)^{2}}=\sqrt{\frac{225}{64}}$$

Simplify.

$$x-\frac{17}{8}=\frac{15}{8} x-\frac{17}{8}=-\frac{15}{8}$$

Add \frac{17}{8} to both sides of the equation.

$$x=4 x=\frac{1}{4}$$