Solve sqrt{x^2-2x+1}=2x

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

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x^{2}-2x+1} to the power of 2 and get x^{2}-2x+1.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

Subtract 4x^{2} from both sides.

$$x^{2}-2x+1-4x^{2}=0$$

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

$$-3x^{2}-2x+1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+1. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-3$$

Rewrite -3x^{2}-2x+1 as \left(-3x^{2}+x\right)+\left(-3x+1\right).

$$\left(-3x^{2}+x\right)+\left(-3x+1\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(3x-1\right)-\left(3x-1\right)$$

Factor out common term 3x-1 by using distributive property.

$$\left(3x-1\right)\left(-x-1\right)$$

To find equation solutions, solve 3x-1=0 and -x-1=0.

$$x=\frac{1}{3} x=-1$$

Substitute \frac{1}{3} for x in the equation \sqrt{x^{2}-2x+1}=2x.

$$\sqrt{\left(\frac{1}{3}\right)^{2}-2\times \frac{1}{3}+1}=2\times \frac{1}{3}$$

Simplify. The value x=\frac{1}{3} satisfies the equation.

$$\frac{2}{3}=\frac{2}{3}$$

Substitute -1 for x in the equation \sqrt{x^{2}-2x+1}=2x.

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

Simplify. The value x=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

Equation \sqrt{x^{2}-2x+1}=2x has a unique solution.

$$x=\frac{1}{3}$$