Solve frac{sqrt{24}}{sqrt{3}}-sqrt{2}*sqrt{18}+{left(sqrt{2}+1right)}^2

$$\frac{ \sqrt{ 24 } }{ \sqrt{ 3 } } - \sqrt{ 2 } \times \sqrt{ 18 } + { \left( \sqrt{ 2 } +1 \right) }^{ 2 }$$

Evaluate

$$4\sqrt{2}-3\approx 2.656854249$$

Solution Steps

Rewrite the division of square roots \frac{\sqrt{24}}{\sqrt{3}} as the square root of the division \sqrt{\frac{24}{3}} and perform the division.

$$\sqrt{8}-\sqrt{2}\sqrt{18}+\left(\sqrt{2}+1\right)^{2}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

Factor 18=2\times 9. Rewrite the square root of the product \sqrt{2\times 9} as the product of square roots \sqrt{2}\sqrt{9}.

$$2\sqrt{2}-\sqrt{2}\sqrt{2}\sqrt{9}+\left(\sqrt{2}+1\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$2\sqrt{2}-2\sqrt{9}+\left(\sqrt{2}+1\right)^{2}$$

Calculate the square root of 9 and get 3.

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

Multiply 2 and 3 to get 6.

$$2\sqrt{2}-6+\left(\sqrt{2}+1\right)^{2}$$

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

$$2\sqrt{2}-6+\left(\sqrt{2}\right)^{2}+2\sqrt{2}+1$$

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

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

Add -6 and 3 to get -3.

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

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

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