Solve left(x-2right)^2=left(1-2xright)^2

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

Solve for x

$$x=1$$

$$x=-1$$

Steps by Finding Square Root

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

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

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

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

Add 4x to both sides.

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

Combine -4x and 4x to get 0.

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

Subtract 4x^{2} from both sides.

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

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+4=1$$

Subtract 4 from both sides.

$$-3x^{2}=1-4$$

Subtract 4 from 1 to get -3.

$$-3x^{2}=-3$$

Divide both sides by -3.

$$x^{2}=\frac{-3}{-3}$$

Divide -3 by -3 to get 1.

$$x^{2}=1$$

Take the square root of both sides of the equation.

$$x=1 x=-1$$

Steps Using the Quadratic Formula

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

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

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

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

Subtract 1 from both sides.

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

Subtract 1 from 4 to get 3.

$$x^{2}-4x+3=-4x+4x^{2}$$

Add 4x to both sides.

$$x^{2}-4x+3+4x=4x^{2}$$

Combine -4x and 4x to get 0.

$$x^{2}+3=4x^{2}$$

Subtract 4x^{2} from both sides.

$$x^{2}+3-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$-3x^{2}+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 0 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-3\right)\times 3}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{0±\sqrt{12\times 3}}{2\left(-3\right)}$$

Multiply 12 times 3.

$$x=\frac{0±\sqrt{36}}{2\left(-3\right)}$$

Take the square root of 36.

$$x=\frac{0±6}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{0±6}{-6}$$

Now solve the equation x=\frac{0±6}{-6} when ± is plus. Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{0±6}{-6} when ± is minus. Divide -6 by -6.

$$x=1$$

The equation is now solved.

$$x=-1 x=1$$