Solve frac{(sqrt{5}-3)(sqrt{5}-3)}{(sqrt{5}+3)(sqrt{5}-3)}

$$\frac { ( \sqrt { 5 } - 3 ) ( \sqrt { 5 } - 3 ) } { ( \sqrt { 5 } + 3 ) ( \sqrt { 5 } - 3 ) }$$

Evaluate

$$\frac{3\sqrt{5}-7}{2}\approx -0.145898034$$

Solution Steps

Consider \left(\sqrt{5}+3\right)\left(\sqrt{5}-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{5}-3\right)\left(\sqrt{5}-3\right)}{\left(\sqrt{5}\right)^{2}-3^{2}}$$

Square \sqrt{5}. Square 3.

$$\frac{\left(\sqrt{5}-3\right)\left(\sqrt{5}-3\right)}{5-9}$$

Subtract 9 from 5 to get -4.

$$\frac{\left(\sqrt{5}-3\right)\left(\sqrt{5}-3\right)}{-4}$$

Multiply \sqrt{5}-3 and \sqrt{5}-3 to get \left(\sqrt{5}-3\right)^{2}.

$$\frac{\left(\sqrt{5}-3\right)^{2}}{-4}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-3\right)^{2}.

$$\frac{\left(\sqrt{5}\right)^{2}-6\sqrt{5}+9}{-4}$$

The square of \sqrt{5} is 5.

$$\frac{5-6\sqrt{5}+9}{-4}$$

Add 5 and 9 to get 14.

$$\frac{14-6\sqrt{5}}{-4}$$

Factor

$$\frac{3 \sqrt{5} - 7}{2} = -0.1458980337503153$$