Solve (4x+4)^2=64

$$( 4 x + 4 ) ^ { 2 } = 64$$

Solve for x

$$x=1$$

$$x=-3$$

Steps Using Factoring By Grouping

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

$$16x^{2}+32x+16=64$$

Subtract 64 from both sides.

$$16x^{2}+32x+16-64=0$$

Subtract 64 from 16 to get -48.

$$16x^{2}+32x-48=0$$

Divide both sides by 16.

$$x^{2}+2x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=2 ab=1\left(-3\right)=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=-1 b=3$$

Rewrite x^{2}+2x-3 as \left(x^{2}-x\right)+\left(3x-3\right).

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

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

$$x\left(x-1\right)+3\left(x-1\right)$$

Factor out common term x-1 by using distributive property.

$$\left(x-1\right)\left(x+3\right)$$

To find equation solutions, solve x-1=0 and x+3=0.

$$x=1 x=-3$$

Steps Using the Quadratic Formula

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

$$16x^{2}+32x+16=64$$

Subtract 64 from both sides.

$$16x^{2}+32x+16-64=0$$

Subtract 64 from 16 to get -48.

$$16x^{2}+32x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 32 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-32±\sqrt{32^{2}-4\times 16\left(-48\right)}}{2\times 16}$$

Square 32.

$$x=\frac{-32±\sqrt{1024-4\times 16\left(-48\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-32±\sqrt{1024-64\left(-48\right)}}{2\times 16}$$

Multiply -64 times -48.

$$x=\frac{-32±\sqrt{1024+3072}}{2\times 16}$$

Add 1024 to 3072.

$$x=\frac{-32±\sqrt{4096}}{2\times 16}$$

Take the square root of 4096.

$$x=\frac{-32±64}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-32±64}{32}$$

Now solve the equation x=\frac{-32±64}{32} when ± is plus. Add -32 to 64.

$$x=\frac{32}{32}$$

Divide 32 by 32.

$$x=1$$

Now solve the equation x=\frac{-32±64}{32} when ± is minus. Subtract 64 from -32.

$$x=-\frac{96}{32}$$

Divide -96 by 32.

$$x=-3$$

The equation is now solved.

$$x=1 x=-3$$

Steps for Completing the Square

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

$$16x^{2}+32x+16=64$$

Subtract 16 from both sides.

$$16x^{2}+32x=64-16$$

Subtract 16 from 64 to get 48.

$$16x^{2}+32x=48$$

Divide both sides by 16.

$$\frac{16x^{2}+32x}{16}=\frac{48}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{32}{16}x=\frac{48}{16}$$

Divide 32 by 16.

$$x^{2}+2x=\frac{48}{16}$$

Divide 48 by 16.

$$x^{2}+2x=3$$

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

$$x^{2}+2x+1^{2}=3+1^{2}$$

Square 1.

$$x^{2}+2x+1=3+1$$

Add 3 to 1.

$$x^{2}+2x+1=4$$

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

$$\left(x+1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x+1=2 x+1=-2$$

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$