Solve (3x-2)^2=0

$$( 3 x - 2 ) ^ { 2 } = 0$$

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

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

$$9x^{2}-12x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-12 ab=9\times 4=36$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -12.

$$a=-6 b=-6$$

Rewrite 9x^{2}-12x+4 as \left(9x^{2}-6x\right)+\left(-6x+4\right).

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

Factor out 3x in the first and -2 in the second group.

$$3x\left(3x-2\right)-2\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(3x-2\right)$$

Rewrite as a binomial square.

$$\left(3x-2\right)^{2}$$

To find equation solution, solve 3x-2=0.

$$x=\frac{2}{3}$$

Steps Using the Quadratic Formula

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

$$9x^{2}-12x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -12 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\times 4}}{2\times 9}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 9\times 4}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-12\right)±\sqrt{144-36\times 4}}{2\times 9}$$

Multiply -36 times 4.

$$x=\frac{-\left(-12\right)±\sqrt{144-144}}{2\times 9}$$

Add 144 to -144.

$$x=\frac{-\left(-12\right)±\sqrt{0}}{2\times 9}$$

Take the square root of 0.

$$x=-\frac{-12}{2\times 9}$$

The opposite of -12 is 12.

$$x=\frac{12}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{12}{18}$$

Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{2}{3}$$

Steps for Completing the Square

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

$$9x^{2}-12x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-12x=-4$$

Divide both sides by 9.

$$\frac{9x^{2}-12x}{9}=-\frac{4}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{12}{9}\right)x=-\frac{4}{9}$$

Reduce the fraction \frac{-12}{9} to lowest terms by extracting and canceling out 3.

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

Divide -\frac{4}{3}, the coefficient of the x term, by 2 to get -\frac{2}{3}. Then add the square of -\frac{2}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

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

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

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

Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{2}{3}=0 x-\frac{2}{3}=0$$

Add \frac{2}{3} to both sides of the equation.

$$x=\frac{2}{3} x=\frac{2}{3}$$

The equation is now solved. Solutions are the same.

$$x=\frac{2}{3}$$