Solve (sqrt{7}+sqrt{2})^2-(sqrt{7}+sqrt{2})(sqrt{7}-sqrt{2})

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

Evaluate

$$2\left(\sqrt{14}+2\right)\approx 11.483314774$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7}+\sqrt{2}\right)^{2}.

$$\left(\sqrt{7}\right)^{2}+2\sqrt{7}\sqrt{2}+\left(\sqrt{2}\right)^{2}-\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)$$

The square of \sqrt{7} is 7.

$$7+2\sqrt{7}\sqrt{2}+\left(\sqrt{2}\right)^{2}-\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)$$

To multiply \sqrt{7} and \sqrt{2}, multiply the numbers under the square root.

$$7+2\sqrt{14}+\left(\sqrt{2}\right)^{2}-\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)$$

The square of \sqrt{2} is 2.

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

Add 7 and 2 to get 9.

$$9+2\sqrt{14}-\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)$$

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

$$9+2\sqrt{14}-\left(\left(\sqrt{7}\right)^{2}-\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{7} is 7.

$$9+2\sqrt{14}-\left(7-\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$9+2\sqrt{14}-\left(7-2\right)$$

Subtract 2 from 7 to get 5.

$$9+2\sqrt{14}-5$$

Subtract 5 from 9 to get 4.

$$4+2\sqrt{14}$$

Factor

$$2 {(\sqrt{14} + 2)} = 11.483314774$$