Solve (3+2sqrt{2})(3-2sqrt{2})-sqrt{54}+sqrt{6}

$$( 3 + 2 \sqrt { 2 } ) ( 3 - 2 \sqrt { 2 } ) - \sqrt { 54 } + \sqrt { 6 }$$

Evaluate

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

Solution Steps

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

$$3^{2}-\left(2\sqrt{2}\right)^{2}-\sqrt{54}+\sqrt{6}$$

Calculate 3 to the power of 2 and get 9.

$$9-\left(2\sqrt{2}\right)^{2}-\sqrt{54}+\sqrt{6}$$

Expand \left(2\sqrt{2}\right)^{2}.

$$9-2^{2}\left(\sqrt{2}\right)^{2}-\sqrt{54}+\sqrt{6}$$

Calculate 2 to the power of 2 and get 4.

$$9-4\left(\sqrt{2}\right)^{2}-\sqrt{54}+\sqrt{6}$$

The square of \sqrt{2} is 2.

$$9-4\times 2-\sqrt{54}+\sqrt{6}$$

Multiply 4 and 2 to get 8.

$$9-8-\sqrt{54}+\sqrt{6}$$

Subtract 8 from 9 to get 1.

$$1-\sqrt{54}+\sqrt{6}$$

Factor 54=3^{2}\times 6. Rewrite the square root of the product \sqrt{3^{2}\times 6} as the product of square roots \sqrt{3^{2}}\sqrt{6}. Take the square root of 3^{2}.

$$1-3\sqrt{6}+\sqrt{6}$$

Combine -3\sqrt{6} and \sqrt{6} to get -2\sqrt{6}.

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