Solve [(sqrt{5}-2)+(sqrt{3}+2)]^2-(sqrt{3}-2)(sqrt{3}+2)

$$[ ( \sqrt { 5 } - 2 ) + ( \sqrt { 3 } + 2 ) ] ^ { 2 } - ( \sqrt { 3 } - 2 ) ( \sqrt { 3 } + 2 )$$

Evaluate

$$2\sqrt{15}+9\approx 16.745966692$$

Solution Steps

Add -2 and 2 to get 0.

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

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

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

The square of \sqrt{5} is 5.

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

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$5+2\sqrt{15}+\left(\sqrt{3}\right)^{2}-\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)$$

The square of \sqrt{3} is 3.

$$5+2\sqrt{15}+3-\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)$$

Add 5 and 3 to get 8.

$$8+2\sqrt{15}-\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)$$

Consider \left(\sqrt{3}-2\right)\left(\sqrt{3}+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.

$$8+2\sqrt{15}-\left(\left(\sqrt{3}\right)^{2}-4\right)$$

The square of \sqrt{3} is 3.

$$8+2\sqrt{15}-\left(3-4\right)$$

Subtract 4 from 3 to get -1.

$$8+2\sqrt{15}-\left(-1\right)$$

The opposite of -1 is 1.

$$8+2\sqrt{15}+1$$

Add 8 and 1 to get 9.

$$9+2\sqrt{15}$$