Solve left(4x+1right)^2=left(4xright)^2+left(x+1right)^2

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

Solve for x

$$x=6$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=\left(4x\right)^{2}+\left(x+1\right)^{2}$$

Expand \left(4x\right)^{2}.

$$16x^{2}+8x+1=4^{2}x^{2}+\left(x+1\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+8x+1=16x^{2}+\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$16x^{2}+8x+1=16x^{2}+x^{2}+2x+1$$

Combine 16x^{2} and x^{2} to get 17x^{2}.

$$16x^{2}+8x+1=17x^{2}+2x+1$$

Subtract 17x^{2} from both sides.

$$16x^{2}+8x+1-17x^{2}=2x+1$$

Combine 16x^{2} and -17x^{2} to get -x^{2}.

$$-x^{2}+8x+1=2x+1$$

Subtract 2x from both sides.

$$-x^{2}+8x+1-2x=1$$

Combine 8x and -2x to get 6x.

$$-x^{2}+6x+1=1$$

Subtract 1 from both sides.

$$-x^{2}+6x+1-1=0$$

Subtract 1 from 1 to get 0.

$$-x^{2}+6x=0$$

Factor out x.

$$x\left(-x+6\right)=0$$

To find equation solutions, solve x=0 and -x+6=0.

$$x=0 x=6$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=\left(4x\right)^{2}+\left(x+1\right)^{2}$$

Expand \left(4x\right)^{2}.

$$16x^{2}+8x+1=4^{2}x^{2}+\left(x+1\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+8x+1=16x^{2}+\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$16x^{2}+8x+1=16x^{2}+x^{2}+2x+1$$

Combine 16x^{2} and x^{2} to get 17x^{2}.

$$16x^{2}+8x+1=17x^{2}+2x+1$$

Subtract 17x^{2} from both sides.

$$16x^{2}+8x+1-17x^{2}=2x+1$$

Combine 16x^{2} and -17x^{2} to get -x^{2}.

$$-x^{2}+8x+1=2x+1$$

Subtract 2x from both sides.

$$-x^{2}+8x+1-2x=1$$

Combine 8x and -2x to get 6x.

$$-x^{2}+6x+1=1$$

Subtract 1 from both sides.

$$-x^{2}+6x+1-1=0$$

Subtract 1 from 1 to get 0.

$$-x^{2}+6x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}}}{2\left(-1\right)}$$

Take the square root of 6^{2}.

$$x=\frac{-6±6}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-6±6}{-2}$$

Now solve the equation x=\frac{-6±6}{-2} when ± is plus. Add -6 to 6.

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

Divide 0 by -2.

$$x=0$$

Now solve the equation x=\frac{-6±6}{-2} when ± is minus. Subtract 6 from -6.

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

Divide -12 by -2.

$$x=6$$

The equation is now solved.

$$x=0 x=6$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+1\right)^{2}.

$$16x^{2}+8x+1=\left(4x\right)^{2}+\left(x+1\right)^{2}$$

Expand \left(4x\right)^{2}.

$$16x^{2}+8x+1=4^{2}x^{2}+\left(x+1\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}+8x+1=16x^{2}+\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$16x^{2}+8x+1=16x^{2}+x^{2}+2x+1$$

Combine 16x^{2} and x^{2} to get 17x^{2}.

$$16x^{2}+8x+1=17x^{2}+2x+1$$

Subtract 17x^{2} from both sides.

$$16x^{2}+8x+1-17x^{2}=2x+1$$

Combine 16x^{2} and -17x^{2} to get -x^{2}.

$$-x^{2}+8x+1=2x+1$$

Subtract 2x from both sides.

$$-x^{2}+8x+1-2x=1$$

Combine 8x and -2x to get 6x.

$$-x^{2}+6x+1=1$$

Subtract 1 from both sides.

$$-x^{2}+6x=1-1$$

Subtract 1 from 1 to get 0.

$$-x^{2}+6x=0$$

Divide both sides by -1.

$$\frac{-x^{2}+6x}{-1}=\frac{0}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{6}{-1}x=\frac{0}{-1}$$

Divide 6 by -1.

$$x^{2}-6x=\frac{0}{-1}$$

Divide 0 by -1.

$$x^{2}-6x=0$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-6x+\left(-3\right)^{2}=\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=9$$

Factor x^{2}-6x+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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{9}$$

Simplify.

$$x-3=3 x-3=-3$$

Add 3 to both sides of the equation.

$$x=6 x=0$$