Solve (2+sqrt{2})^2-sqrt{18}

$$( 2 + \sqrt { 2 } ) ^ { 2 } - \sqrt { 18 }$$

Evaluate

$$\sqrt{2}+6\approx 7.414213562$$

Solution Steps

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

$$4+4\sqrt{2}+\left(\sqrt{2}\right)^{2}-\sqrt{18}$$

The square of \sqrt{2} is 2.

$$4+4\sqrt{2}+2-\sqrt{18}$$

Add 4 and 2 to get 6.

$$6+4\sqrt{2}-\sqrt{18}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$6+4\sqrt{2}-3\sqrt{2}$$

Combine 4\sqrt{2} and -3\sqrt{2} to get \sqrt{2}.

$$6+\sqrt{2}$$