Solve (2x-4)^2=0

$$( 2 x - 4 ) ^ { 2 } = 0$$

Solve for x

$$x=2$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$4x^{2}-16x+16=0$$

Divide both sides by 4.

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

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

$$a+b=-4 ab=1\times 4=4$$

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 4.

$$-1,-4 -2,-2$$

Calculate the sum for each pair.

$$-1-4=-5 -2-2=-4$$

The solution is the pair that gives sum -4.

$$a=-2 b=-2$$

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

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

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

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

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

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

Rewrite as a binomial square.

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

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

$$x=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$4x^{2}-16x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -16 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\times 16}}{2\times 4}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 4\times 16}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-16\right)±\sqrt{256-16\times 16}}{2\times 4}$$

Multiply -16 times 16.

$$x=\frac{-\left(-16\right)±\sqrt{256-256}}{2\times 4}$$

Add 256 to -256.

$$x=\frac{-\left(-16\right)±\sqrt{0}}{2\times 4}$$

Take the square root of 0.

$$x=-\frac{-16}{2\times 4}$$

The opposite of -16 is 16.

$$x=\frac{16}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{16}{8}$$

Divide 16 by 8.

$$x=2$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-4\right)^{2}.

$$4x^{2}-16x+16=0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$4x^{2}-16x=-16$$

Divide both sides by 4.

$$\frac{4x^{2}-16x}{4}=-\frac{16}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{16}{4}\right)x=-\frac{16}{4}$$

Divide -16 by 4.

$$x^{2}-4x=-\frac{16}{4}$$

Divide -16 by 4.

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

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

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

Square -2.

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

Add -4 to 4.

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

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=0 x-2=0$$

Add 2 to both sides of the equation.

$$x=2 x=2$$

The equation is now solved. Solutions are the same.

$$x=2$$