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

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

Solve for x

$$x=0$$

Steps by Finding Square Root

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

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

Add 4x to both sides.

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

Combine -4x and 4x to get 0.

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

$$4x^{2}=0$$

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

Take the square root of both sides of the equation.

$$x=0 x=0$$

The equation is now solved. Solutions are the same.

$$x=0$$

Steps Using the Quadratic Formula

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

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

Subtract 1 from both sides.

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

Subtract 1 from 1 to get 0.

$$4x^{2}-4x=-4x$$

Add 4x to both sides.

$$4x^{2}-4x+4x=0$$

Combine -4x and 4x to get 0.

$$4x^{2}=0$$

Divide both sides by 4. Zero divided by any non-zero number gives zero.

$$x^{2}=0$$

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

$$x=\frac{0±\sqrt{0^{2}}}{2}$$

Take the square root of 0^{2}.

$$x=\frac{0±0}{2}$$

Divide 0 by 2.

$$x=0$$