Solve (sqrt{5}-3)^2+(sqrt{11}+3)(sqrt{11}-3)

$$( \sqrt { 5 } - 3 ) ^ { 2 } + ( \sqrt { 11 } + 3 ) ( \sqrt { 11 } - 3 )$$

Evaluate

$$16-6\sqrt{5}\approx 2.583592135$$

Solution Steps

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

$$\left(\sqrt{5}\right)^{2}-6\sqrt{5}+9+\left(\sqrt{11}+3\right)\left(\sqrt{11}-3\right)$$

The square of \sqrt{5} is 5.

$$5-6\sqrt{5}+9+\left(\sqrt{11}+3\right)\left(\sqrt{11}-3\right)$$

Add 5 and 9 to get 14.

$$14-6\sqrt{5}+\left(\sqrt{11}+3\right)\left(\sqrt{11}-3\right)$$

Consider \left(\sqrt{11}+3\right)\left(\sqrt{11}-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}. Square 3.

$$14-6\sqrt{5}+\left(\sqrt{11}\right)^{2}-9$$

The square of \sqrt{11} is 11.

$$14-6\sqrt{5}+11-9$$

Subtract 9 from 11 to get 2.

$$14-6\sqrt{5}+2$$

Add 14 and 2 to get 16.

$$16-6\sqrt{5}$$