Solve 1/2x-7/14+12/16x^2

$$\frac { 1 } { 2 x } - \frac { 7 } { 14 } + \frac { 12 } { 16 x ^ { 2 } }$$

Evaluate

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

Short Solution Steps

Reduce the fraction \frac{7}{14} to lowest terms by extracting and canceling out 7.

$$\frac{1}{2x}-\frac{1}{2}+\frac{12}{16x^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x and 2 is 2x. Multiply \frac{1}{2} times \frac{x}{x}.

$$\frac{1}{2x}-\frac{x}{2x}+\frac{12}{16x^{2}}$$

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

$$\frac{1-x}{2x}+\frac{12}{16x^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2x and 16x^{2} is 16x^{2}. Multiply \frac{1-x}{2x} times \frac{8x}{8x}.

$$\frac{\left(1-x\right)\times 8x}{16x^{2}}+\frac{12}{16x^{2}}$$

Since \frac{\left(1-x\right)\times 8x}{16x^{2}} and \frac{12}{16x^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(1-x\right)\times 8x+12}{16x^{2}}$$

Do the multiplications in \left(1-x\right)\times 8x+12.

$$\frac{8x-8x^{2}+12}{16x^{2}}$$

Factor the expressions that are not already factored in \frac{8x-8x^{2}+12}{16x^{2}}.

$$\frac{-2\times 4\left(x-\left(-\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)}{16x^{2}}$$

Cancel out 2\times 4 in both numerator and denominator.

$$\frac{-\left(x-\left(-\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)}{2x^{2}}$$

Cancel out -1 in both numerator and denominator.

$$\frac{\left(x-\left(-\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)}{-2x^{2}}$$

To find the opposite of -\frac{1}{2}\sqrt{7}+\frac{1}{2}, find the opposite of each term.

$$\frac{\left(x+\frac{1}{2}\sqrt{7}-\frac{1}{2}\right)\left(x-\left(\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)}{-2x^{2}}$$

To find the opposite of \frac{1}{2}\sqrt{7}+\frac{1}{2}, find the opposite of each term.

$$\frac{\left(x+\frac{1}{2}\sqrt{7}-\frac{1}{2}\right)\left(x-\frac{1}{2}\sqrt{7}-\frac{1}{2}\right)}{-2x^{2}}$$

Use the distributive property to multiply x+\frac{1}{2}\sqrt{7}-\frac{1}{2} by x-\frac{1}{2}\sqrt{7}-\frac{1}{2} and combine like terms.

$$\frac{x^{2}-x-\frac{1}{4}\left(\sqrt{7}\right)^{2}+\frac{1}{4}}{-2x^{2}}$$

The square of \sqrt{7} is 7.

$$\frac{x^{2}-x-\frac{1}{4}\times 7+\frac{1}{4}}{-2x^{2}}$$

Multiply -\frac{1}{4} and 7 to get -\frac{7}{4}.

$$\frac{x^{2}-x-\frac{7}{4}+\frac{1}{4}}{-2x^{2}}$$

Add -\frac{7}{4} and \frac{1}{4} to get -\frac{3}{2}.

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

Factor the expressions that are not already factored.

$$\frac{\frac{1}{2}\times 2\left(x-\left(-\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)}{-2x^{2}}$$

Cancel out 2 in both numerator and denominator.

$$\frac{\frac{1}{2}\left(x-\left(-\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)\left(x-\left(\frac{1}{2}\sqrt{7}+\frac{1}{2}\right)\right)}{-x^{2}}$$

Expand the expression.

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

Factor

$$-\frac{\frac{1}{2}\left(x-\frac{1-\sqrt{7}}{2}\right)\left(x-\frac{\sqrt{7}+1}{2}\right)}{x^{2}}$$