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

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

Solve for x

$$x=1$$

$$x=-3$$

Steps Using the Quadratic Formula

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

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

Use the distributive property to multiply \frac{1}{2} by x^{2}-2x+1.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract x^{2} from both sides.

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

Combine \frac{1}{2}x^{2} and -x^{2} to get -\frac{1}{2}x^{2}.

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

Add 1 to both sides.

$$-\frac{1}{2}x^{2}-x+\frac{1}{2}+1=0$$

Add \frac{1}{2} and 1 to get \frac{3}{2}.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, -1 for b, and \frac{3}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-\frac{1}{2}\right)\times \frac{3}{2}}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

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

Multiply 2 times \frac{3}{2}.

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

Add 1 to 3.

$$x=\frac{-\left(-1\right)±\sqrt{4}}{2\left(-\frac{1}{2}\right)}$$

Take the square root of 4.

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

The opposite of -1 is 1.

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

Multiply 2 times -\frac{1}{2}.

$$x=\frac{1±2}{-1}$$

Now solve the equation x=\frac{1±2}{-1} when ± is plus. Add 1 to 2.

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

Divide 3 by -1.

$$x=-3$$

Now solve the equation x=\frac{1±2}{-1} when ± is minus. Subtract 2 from 1.

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

Divide -1 by -1.

$$x=1$$

The equation is now solved.

$$x=-3 x=1$$

Steps for Completing the Square

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

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

Use the distributive property to multiply \frac{1}{2} by x^{2}-2x+1.

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

Consider \left(x-1\right)\left(x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

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

Subtract x^{2} from both sides.

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

Combine \frac{1}{2}x^{2} and -x^{2} to get -\frac{1}{2}x^{2}.

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

Subtract \frac{1}{2} from both sides.

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

Subtract \frac{1}{2} from -1 to get -\frac{3}{2}.

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

Multiply both sides by -2.

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

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

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

Divide -1 by -\frac{1}{2} by multiplying -1 by the reciprocal of -\frac{1}{2}.

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

Divide -\frac{3}{2} by -\frac{1}{2} by multiplying -\frac{3}{2} by the reciprocal of -\frac{1}{2}.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 1.

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

Add 3 to 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x+1=2 x+1=-2$$

Subtract 1 from both sides of the equation.

$$x=1 x=-3$$