Solve 1/2(x-4)^2=8

$$\frac { 1 } { 2 } ( x - 4 ) ^ { 2 } = 8$$

Solve for x

$$x=8$$

$$x=0$$

Steps Using Factoring

Multiply both sides by 2, the reciprocal of \frac{1}{2}.

$$\left(x-4\right)^{2}=8\times 2$$

Multiply 8 and 2 to get 16.

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

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

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

Subtract 16 from both sides.

$$x^{2}-8x+16-16=0$$

Subtract 16 from 16 to get 0.

$$x^{2}-8x=0$$

Factor out x.

$$x\left(x-8\right)=0$$

To find equation solutions, solve x=0 and x-8=0.

$$x=0 x=8$$

Steps Using the Quadratic Formula

Multiply both sides by 2, the reciprocal of \frac{1}{2}.

$$\left(x-4\right)^{2}=8\times 2$$

Multiply 8 and 2 to get 16.

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

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

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

Subtract 16 from both sides.

$$x^{2}-8x+16-16=0$$

Subtract 16 from 16 to get 0.

$$x^{2}-8x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2}$$

Take the square root of \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±8}{2}$$

Now solve the equation x=\frac{8±8}{2} when ± is plus. Add 8 to 8.

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

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{8±8}{2} when ± is minus. Subtract 8 from 8.

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

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=8 x=0$$

Steps for Completing the Square

Multiply both sides by 2, the reciprocal of \frac{1}{2}.

$$\left(x-4\right)^{2}=8\times 2$$

Multiply 8 and 2 to get 16.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-4=4 x-4=-4$$

Add 4 to both sides of the equation.

$$x=8 x=0$$