Solve (sqrt{48}-sqrt{27})divsqrt{3}+(2sqrt{2}-1)^2

$$( \sqrt { 48 } - \sqrt { 27 } ) \div \sqrt { 3 } + ( 2 \sqrt { 2 } - 1 ) ^ { 2 }$$

Evaluate

$$10-4\sqrt{2}\approx 4.343145751$$

Solution Steps

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{4\sqrt{3}-\sqrt{27}}{\sqrt{3}}+\left(2\sqrt{2}-1\right)^{2}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$\frac{4\sqrt{3}-3\sqrt{3}}{\sqrt{3}}+\left(2\sqrt{2}-1\right)^{2}$$

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

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

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

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

Calculate the square root of 1 and get 1.

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

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

$$1+4\left(\sqrt{2}\right)^{2}-4\sqrt{2}+1$$

The square of \sqrt{2} is 2.

$$1+4\times 2-4\sqrt{2}+1$$

Multiply 4 and 2 to get 8.

$$1+8-4\sqrt{2}+1$$

Add 8 and 1 to get 9.

$$1+9-4\sqrt{2}$$

Add 1 and 9 to get 10.

$$10-4\sqrt{2}$$