Solve (x-1)(x+1)=(1/2x+x-2/2)(2x-3/2-x-1/2)-frac{x^2}{2}

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

Solve for x

$$x=\frac{\sqrt{41}-3}{4}\approx 0.850781059$$

$$x=\frac{-\sqrt{41}-3}{4}\approx -2.350781059$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply 2 by x-1.

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

Use the distributive property to multiply 2x-2 by x+1 and combine like terms.

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

Since \frac{2x-3}{2} and \frac{x-1}{2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2x-3-\left(x-1\right).

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

Combine like terms in 2x-3-x+1.

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

Express 2\times \frac{x-2}{2} as a single fraction.

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

Cancel out 2 and 2.

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

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

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

Express x\times \frac{x-2}{2} as a single fraction.

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

Express -2\times \frac{x-2}{2} as a single fraction.

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

Cancel out 2 and 2.

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

To find the opposite of x-2, find the opposite of each term.

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

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

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

Use the distributive property to multiply x by x-2.

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

Divide each term of x^{2}-2x by 2 to get \frac{1}{2}x^{2}-x.

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

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

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

Combine -x and -2x to get -3x.

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

Subtract x^{2} from both sides.

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

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

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

Add 3x to both sides.

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

Subtract 2 from both sides.

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

Subtract 2 from -2 to get -4.

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

Add x^{2} to both sides.

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

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

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

Square 3.

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

Multiply -4 times 2.

$$x=\frac{-3±\sqrt{9-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$x=\frac{-3±\sqrt{9+32}}{2\times 2}$$

Add 9 to 32.

$$x=\frac{-3±\sqrt{41}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-3±\sqrt{41}}{4}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{4} when ± is plus. Add -3 to \sqrt{41}.

$$x=\frac{\sqrt{41}-3}{4}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{4} when ± is minus. Subtract \sqrt{41} from -3.

$$x=\frac{-\sqrt{41}-3}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Use the distributive property to multiply 2 by x-1.

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

Use the distributive property to multiply 2x-2 by x+1 and combine like terms.

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

Since \frac{2x-3}{2} and \frac{x-1}{2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 2x-3-\left(x-1\right).

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

Combine like terms in 2x-3-x+1.

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

Express 2\times \frac{x-2}{2} as a single fraction.

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

Cancel out 2 and 2.

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

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

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

Express x\times \frac{x-2}{2} as a single fraction.

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

Express -2\times \frac{x-2}{2} as a single fraction.

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

Cancel out 2 and 2.

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

To find the opposite of x-2, find the opposite of each term.

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

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

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

Use the distributive property to multiply x by x-2.

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

Divide each term of x^{2}-2x by 2 to get \frac{1}{2}x^{2}-x.

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

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

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

Combine -x and -2x to get -3x.

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

Subtract x^{2} from both sides.

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

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

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

Add 3x to both sides.

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

Add x^{2} to both sides.

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Add 2 to both sides.

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

Add 2 and 2 to get 4.

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

Divide both sides by 2.

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

Dividing by 2 undoes the multiplication by 2.

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

Divide 4 by 2.

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

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

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

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

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

Add 2 to \frac{9}{16}.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{41}{16}$$

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

$$\left(x+\frac{3}{4}\right)^{2}=\frac{41}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{41}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{41}}{4} x+\frac{3}{4}=-\frac{\sqrt{41}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{41}-3}{4} x=\frac{-\sqrt{41}-3}{4}$$