Solve f(x)=(x-1)(x+1)-3/2x^2

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

Evaluate

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

Short Solution Steps

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.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-1 times \frac{2x^{2}}{2x^{2}}.

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

Since \frac{\left(x^{2}-1\right)\times 2x^{2}}{2x^{2}} and \frac{3}{2x^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(x^{2}-1\right)\times 2x^{2}-3.

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

Factor the expressions that are not already factored in \frac{2x^{4}-2x^{2}-3}{2x^{2}}.

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

Cancel out 2 in both numerator and denominator.

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

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

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

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

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

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

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

The square of \sqrt{7} is 7.

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

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

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

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

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

Expand

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

Short Solution Steps

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.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{2}-1 times \frac{2x^{2}}{2x^{2}}.

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

Since \frac{\left(x^{2}-1\right)\times 2x^{2}}{2x^{2}} and \frac{3}{2x^{2}} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(x^{2}-1\right)\times 2x^{2}-3.

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

Factor the expressions that are not already factored in \frac{2x^{4}-2x^{2}-3}{2x^{2}}.

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

Cancel out 2 in both numerator and denominator.

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

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

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

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

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

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

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

The square of \sqrt{7} is 7.

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

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

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

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

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