Solve 1/2left(x-1right)^2=left(x-1right)left(x+1right)+15/8

$$\frac{ 1 }{ 2 } { \left(x-1 \right) }^{ 2 } = \left( x-1 \right) \left( x+1 \right) + \frac{ 15 }{ 8 }$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=-\frac{1}{2}=-0.5$$

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}.

$$\frac{1}{2}\left(x^{2}-2x+1\right)=\left(x-1\right)\left(x+1\right)+\frac{15}{8}$$

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

$$\frac{1}{2}x^{2}-x+\frac{1}{2}=\left(x-1\right)\left(x+1\right)+\frac{15}{8}$$

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.

$$\frac{1}{2}x^{2}-x+\frac{1}{2}=x^{2}-1+\frac{15}{8}$$

Add -1 and \frac{15}{8} to get \frac{7}{8}.

$$\frac{1}{2}x^{2}-x+\frac{1}{2}=x^{2}+\frac{7}{8}$$

Subtract x^{2} from both sides.

$$\frac{1}{2}x^{2}-x+\frac{1}{2}-x^{2}=\frac{7}{8}$$

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

$$-\frac{1}{2}x^{2}-x+\frac{1}{2}=\frac{7}{8}$$

Subtract \frac{7}{8} from both sides.

$$-\frac{1}{2}x^{2}-x+\frac{1}{2}-\frac{7}{8}=0$$

Subtract \frac{7}{8} from \frac{1}{2} to get -\frac{3}{8}.

$$-\frac{1}{2}x^{2}-x-\frac{3}{8}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, -1 for b, and -\frac{3}{8} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

Multiply 2 times -\frac{3}{8}.

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

Add 1 to -\frac{3}{4}.

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

Take the square root of \frac{1}{4}.

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

The opposite of -1 is 1.

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

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

$$x=\frac{1±\frac{1}{2}}{-1}$$

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

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

Divide \frac{3}{2} by -1.

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

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

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

Divide \frac{1}{2} by -1.

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

The equation is now solved.

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

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}.

$$\frac{1}{2}\left(x^{2}-2x+1\right)=\left(x-1\right)\left(x+1\right)+\frac{15}{8}$$

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

$$\frac{1}{2}x^{2}-x+\frac{1}{2}=\left(x-1\right)\left(x+1\right)+\frac{15}{8}$$

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.

$$\frac{1}{2}x^{2}-x+\frac{1}{2}=x^{2}-1+\frac{15}{8}$$

Add -1 and \frac{15}{8} to get \frac{7}{8}.

$$\frac{1}{2}x^{2}-x+\frac{1}{2}=x^{2}+\frac{7}{8}$$

Subtract x^{2} from both sides.

$$\frac{1}{2}x^{2}-x+\frac{1}{2}-x^{2}=\frac{7}{8}$$

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

$$-\frac{1}{2}x^{2}-x+\frac{1}{2}=\frac{7}{8}$$

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

$$-\frac{1}{2}x^{2}-x=\frac{7}{8}-\frac{1}{2}$$

Subtract \frac{1}{2} from \frac{7}{8} to get \frac{3}{8}.

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

Multiply both sides by -2.

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

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

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

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

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

Divide \frac{3}{8} by -\frac{1}{2} by multiplying \frac{3}{8} by the reciprocal of -\frac{1}{2}.

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

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.

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

Square 1.

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

Add -\frac{3}{4} to 1.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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