Solve (2x-2)^2/x^2+1=1

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

Solve for x

$$x = \frac{\sqrt{7} + 4}{3} \approx 2.215250437$$

$$x=\frac{4-\sqrt{7}}{3}\approx 0.45141623$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by x^{2}+1.

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

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

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

Subtract x^{2} from both sides.

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

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

$$3x^{2}-8x+4=1$$

Subtract 1 from both sides.

$$3x^{2}-8x+4-1=0$$

Subtract 1 from 4 to get 3.

$$3x^{2}-8x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -8 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3\times 3}}{2\times 3}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12\times 3}}{2\times 3}$$

Multiply -12 times 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-36}}{2\times 3}$$

Add 64 to -36.

$$x=\frac{-\left(-8\right)±\sqrt{28}}{2\times 3}$$

Take the square root of 28.

$$x=\frac{-\left(-8\right)±2\sqrt{7}}{2\times 3}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{7}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{8±2\sqrt{7}}{6}$$

Now solve the equation x=\frac{8±2\sqrt{7}}{6} when ± is plus. Add 8 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}+8}{6}$$

Divide 8+2\sqrt{7} by 6.

$$x=\frac{\sqrt{7}+4}{3}$$

Now solve the equation x=\frac{8±2\sqrt{7}}{6} when ± is minus. Subtract 2\sqrt{7} from 8.

$$x=\frac{8-2\sqrt{7}}{6}$$

Divide 8-2\sqrt{7} by 6.

$$x=\frac{4-\sqrt{7}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{7}+4}{3} x=\frac{4-\sqrt{7}}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by x^{2}+1.

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

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

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

Subtract x^{2} from both sides.

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

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

$$3x^{2}-8x+4=1$$

Subtract 4 from both sides.

$$3x^{2}-8x=1-4$$

Subtract 4 from 1 to get -3.

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

Divide both sides by 3.

$$\frac{3x^{2}-8x}{3}=-\frac{3}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{8}{3}x=-\frac{3}{3}$$

Divide -3 by 3.

$$x^{2}-\frac{8}{3}x=-1$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=-1+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=-1+\frac{16}{9}$$

Add -1 to \frac{16}{9}.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{7}{9}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{7}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{7}{9}}$$

Simplify.

$$x-\frac{4}{3}=\frac{\sqrt{7}}{3} x-\frac{4}{3}=-\frac{\sqrt{7}}{3}$$

Add \frac{4}{3} to both sides of the equation.

$$x=\frac{\sqrt{7}+4}{3} x=\frac{4-\sqrt{7}}{3}$$