Solve frac{(3+sqrt{5})^2-(2-sqrt{5})(2+sqrt{5})}{(sqrt{7}-2)(sqrt{7}+2)}

$$\frac { ( 3 + \sqrt { 5 } ) ^ { 2 } - ( 2 - \sqrt { 5 } ) ( 2 + \sqrt { 5 } ) } { ( \sqrt { 7 } - 2 ) ( \sqrt { 7 } + 2 ) }$$

Evaluate

$$2\sqrt{5}+5\approx 9.472135955$$

Solution Steps

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

$$\frac{9+6\sqrt{5}+\left(\sqrt{5}\right)^{2}-\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}$$

The square of \sqrt{5} is 5.

$$\frac{9+6\sqrt{5}+5-\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}$$

Add 9 and 5 to get 14.

$$\frac{14+6\sqrt{5}-\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}$$

Consider \left(2-\sqrt{5}\right)\left(2+\sqrt{5}\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 2.

$$\frac{14+6\sqrt{5}-\left(4-\left(\sqrt{5}\right)^{2}\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}$$

The square of \sqrt{5} is 5.

$$\frac{14+6\sqrt{5}-\left(4-5\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}$$

Subtract 5 from 4 to get -1.

$$\frac{14+6\sqrt{5}-\left(-1\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}$$

The opposite of -1 is 1.

$$\frac{14+6\sqrt{5}+1}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}$$

Consider \left(\sqrt{7}-2\right)\left(\sqrt{7}+2\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 2.

$$\frac{14+6\sqrt{5}+1}{\left(\sqrt{7}\right)^{2}-4}$$

The square of \sqrt{7} is 7.

$$\frac{14+6\sqrt{5}+1}{7-4}$$

Subtract 4 from 7 to get 3.

$$\frac{14+6\sqrt{5}+1}{3}$$

Add 14 and 1 to get 15.

$$\frac{15+6\sqrt{5}}{3}$$

Divide each term of 15+6\sqrt{5} by 3 to get 5+2\sqrt{5}.

$$5+2\sqrt{5}$$