Solve 1/2(x+1/2)^2-1/8(x+1/2)-frac{3}{16}=0

$$\frac { 1 } { 2 } ( x + \frac { 1 } { 2 } ) ^ { 2 } - \frac { 1 } { 8 } ( x + \frac { 1 } { 2 } ) - \frac { 3 } { 16 } = 0$$

Solve for x

$$x=-1$$

$$x=\frac{1}{4}=0.25$$

Steps Using the Quadratic Formula

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

$$\frac{1}{2}\left(x^{2}+x+\frac{1}{4}\right)-\frac{1}{8}\left(x+\frac{1}{2}\right)-\frac{3}{16}=0$$

Use the distributive property to multiply \frac{1}{2} by x^{2}+x+\frac{1}{4}.

$$\frac{1}{2}x^{2}+\frac{1}{2}x+\frac{1}{8}-\frac{1}{8}\left(x+\frac{1}{2}\right)-\frac{3}{16}=0$$

Use the distributive property to multiply -\frac{1}{8} by x+\frac{1}{2}.

$$\frac{1}{2}x^{2}+\frac{1}{2}x+\frac{1}{8}-\frac{1}{8}x-\frac{1}{16}-\frac{3}{16}=0$$

Combine \frac{1}{2}x and -\frac{1}{8}x to get \frac{3}{8}x.

$$\frac{1}{2}x^{2}+\frac{3}{8}x+\frac{1}{8}-\frac{1}{16}-\frac{3}{16}=0$$

Subtract \frac{1}{16} from \frac{1}{8} to get \frac{1}{16}.

$$\frac{1}{2}x^{2}+\frac{3}{8}x+\frac{1}{16}-\frac{3}{16}=0$$

Subtract \frac{3}{16} from \frac{1}{16} to get -\frac{1}{8}.

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

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

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

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{3}{8}±\sqrt{\frac{9}{64}-4\times \frac{1}{2}\left(-\frac{1}{8}\right)}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-\frac{3}{8}±\sqrt{\frac{9}{64}-2\left(-\frac{1}{8}\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -\frac{1}{8}.

$$x=\frac{-\frac{3}{8}±\sqrt{\frac{9}{64}+\frac{1}{4}}}{2\times \frac{1}{2}}$$

Add \frac{9}{64} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{3}{8}±\sqrt{\frac{25}{64}}}{2\times \frac{1}{2}}$$

Take the square root of \frac{25}{64}.

$$x=\frac{-\frac{3}{8}±\frac{5}{8}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-\frac{3}{8}±\frac{5}{8}}{1}$$

Now solve the equation x=\frac{-\frac{3}{8}±\frac{5}{8}}{1} when ± is plus. Add -\frac{3}{8} to \frac{5}{8} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{\frac{1}{4}}{1}$$

Divide \frac{1}{4} by 1.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-\frac{3}{8}±\frac{5}{8}}{1} when ± is minus. Subtract \frac{5}{8} from -\frac{3}{8} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

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

Divide -1 by 1.

$$x=-1$$

The equation is now solved.

$$x=\frac{1}{4} x=-1$$

Steps for Completing the Square

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

$$\frac{1}{2}\left(x^{2}+x+\frac{1}{4}\right)-\frac{1}{8}\left(x+\frac{1}{2}\right)-\frac{3}{16}=0$$

Use the distributive property to multiply \frac{1}{2} by x^{2}+x+\frac{1}{4}.

$$\frac{1}{2}x^{2}+\frac{1}{2}x+\frac{1}{8}-\frac{1}{8}\left(x+\frac{1}{2}\right)-\frac{3}{16}=0$$

Use the distributive property to multiply -\frac{1}{8} by x+\frac{1}{2}.

$$\frac{1}{2}x^{2}+\frac{1}{2}x+\frac{1}{8}-\frac{1}{8}x-\frac{1}{16}-\frac{3}{16}=0$$

Combine \frac{1}{2}x and -\frac{1}{8}x to get \frac{3}{8}x.

$$\frac{1}{2}x^{2}+\frac{3}{8}x+\frac{1}{8}-\frac{1}{16}-\frac{3}{16}=0$$

Subtract \frac{1}{16} from \frac{1}{8} to get \frac{1}{16}.

$$\frac{1}{2}x^{2}+\frac{3}{8}x+\frac{1}{16}-\frac{3}{16}=0$$

Subtract \frac{3}{16} from \frac{1}{16} to get -\frac{1}{8}.

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

Add \frac{1}{8} to both sides. Anything plus zero gives itself.

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

Multiply both sides by 2.

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

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

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

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

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

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

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

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

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

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{1}{4}+\frac{9}{64}$$

Add \frac{1}{4} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{25}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{25}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{25}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{5}{8} x+\frac{3}{8}=-\frac{5}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$x=\frac{1}{4} x=-1$$