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

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

Evaluate

$$1-2\sqrt{6}\approx -3.898979486$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

The square of \sqrt{2} is 2.

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

Add 3 and 2 to get 5.

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

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}. Square 3.

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

The square of \sqrt{5} is 5.

$$5-2\sqrt{6}+5-9$$

Subtract 9 from 5 to get -4.

$$5-2\sqrt{6}-4$$

Subtract 4 from 5 to get 1.

$$1-2\sqrt{6}$$