Solve (2x-1)^2=8

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

Solve for x

$$x=\sqrt{2}+\frac{1}{2}\approx 1.914213562$$

$$x=\frac{1}{2}-\sqrt{2}\approx -0.914213562$$

Steps by Finding Square Root

Take the square root of both sides of the equation.

$$2x-1=2\sqrt{2} 2x-1=-2\sqrt{2}$$

Add 1 to both sides of the equation.

$$2x-1-\left(-1\right)=2\sqrt{2}-\left(-1\right) 2x-1-\left(-1\right)=-2\sqrt{2}-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$2x=2\sqrt{2}-\left(-1\right) 2x=-2\sqrt{2}-\left(-1\right)$$

Subtract -1 from 2\sqrt{2}.

$$2x=2\sqrt{2}+1$$

Subtract -1 from -2\sqrt{2}.

$$2x=1-2\sqrt{2}$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{2\sqrt{2}+1}{2} \frac{2x}{2}=\frac{1-2\sqrt{2}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{2\sqrt{2}+1}{2} x=\frac{1-2\sqrt{2}}{2}$$

Divide 2\sqrt{2}+1 by 2.

$$x=\sqrt{2}+\frac{1}{2}$$

Divide -2\sqrt{2}+1 by 2.

$$x=\frac{1}{2}-\sqrt{2}$$

The equation is now solved.

$$x=\sqrt{2}+\frac{1}{2} x=\frac{1}{2}-\sqrt{2}$$