Solve sqrt{18}+{left(sqrt{2}-1right)}^2+sqrt{5}*sqrt{10}

$$\sqrt{ 18 } + { \left( \sqrt{ 2 } -1 \right) }^{ 2 } + \sqrt{ 5 } \times \sqrt{ 10 }$$

Evaluate

$$6\sqrt{2}+3\approx 11.485281374$$

Solution Steps

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}.

$$3\sqrt{2}+\left(\sqrt{2}-1\right)^{2}+\sqrt{5}\sqrt{10}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}-1\right)^{2}.

$$3\sqrt{2}+\left(\sqrt{2}\right)^{2}-2\sqrt{2}+1+\sqrt{5}\sqrt{10}$$

The square of \sqrt{2} is 2.

$$3\sqrt{2}+2-2\sqrt{2}+1+\sqrt{5}\sqrt{10}$$

Add 2 and 1 to get 3.

$$3\sqrt{2}+3-2\sqrt{2}+\sqrt{5}\sqrt{10}$$

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

$$\sqrt{2}+3+\sqrt{5}\sqrt{10}$$

Factor 10=5\times 2. Rewrite the square root of the product \sqrt{5\times 2} as the product of square roots \sqrt{5}\sqrt{2}.

$$\sqrt{2}+3+\sqrt{5}\sqrt{5}\sqrt{2}$$

Multiply \sqrt{5} and \sqrt{5} to get 5.

$$\sqrt{2}+3+5\sqrt{2}$$

Combine \sqrt{2} and 5\sqrt{2} to get 6\sqrt{2}.

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