Solve sqrt{4x^2-8x+1}=x-1

$$\sqrt { 4 x ^ { 2 } - 8 x + 1 } = x - 1$$

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{4x^{2}-8x+1}\right)^{2}=\left(x-1\right)^{2}$$

Calculate \sqrt{4x^{2}-8x+1} to the power of 2 and get 4x^{2}-8x+1.

$$4x^{2}-8x+1=\left(x-1\right)^{2}$$

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

$$4x^{2}-8x+1=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$4x^{2}-8x+1-x^{2}=-2x+1$$

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

$$3x^{2}-8x+1=-2x+1$$

Add 2x to both sides.

$$3x^{2}-8x+1+2x=1$$

Combine -8x and 2x to get -6x.

$$3x^{2}-6x+1=1$$

Subtract 1 from both sides.

$$3x^{2}-6x+1-1=0$$

Subtract 1 from 1 to get 0.

$$3x^{2}-6x=0$$

Factor out x.

$$x\left(3x-6\right)=0$$

To find equation solutions, solve x=0 and 3x-6=0.

$$x=0 x=2$$

Substitute 0 for x in the equation \sqrt{4x^{2}-8x+1}=x-1.

$$\sqrt{4\times 0^{2}-8\times 0+1}=0-1$$

Simplify. The value x=0 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

Substitute 2 for x in the equation \sqrt{4x^{2}-8x+1}=x-1.

$$\sqrt{4\times 2^{2}-8\times 2+1}=2-1$$

Simplify. The value x=2 satisfies the equation.

$$1=1$$

Equation \sqrt{4x^{2}-8x+1}=x-1 has a unique solution.

$$x=2$$