Solve 4x^2+16x+1=4x(x+4)+1

$$4 x ^ { 2 } + 16 x + 1 = 4 x ( x + 4 ) + 1$$

Solve for x (complex solution)

$$x\in \mathrm{C}$$

Solution Steps

Use the distributive property to multiply 4x by x+4.

$$4x^{2}+16x+1=4x^{2}+16x+1$$

Subtract 4x^{2} from both sides.

$$4x^{2}+16x+1-4x^{2}=16x+1$$

Combine 4x^{2} and -4x^{2} to get 0.

$$16x+1=16x+1$$

Subtract 16x from both sides.

$$16x+1-16x=1$$

Combine 16x and -16x to get 0.

$$1=1$$

Compare 1 and 1.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{C}$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

Use the distributive property to multiply 4x by x+4.

$$4x^{2}+16x+1=4x^{2}+16x+1$$

Subtract 4x^{2} from both sides.

$$4x^{2}+16x+1-4x^{2}=16x+1$$

Combine 4x^{2} and -4x^{2} to get 0.

$$16x+1=16x+1$$

Subtract 16x from both sides.

$$16x+1-16x=1$$

Combine 16x and -16x to get 0.

$$1=1$$

Compare 1 and 1.

$$\text{true}$$

This is true for any x.

$$x\in \mathrm{R}$$