Solve sqrt{3}(sqrt{6}-sqrt{3})+(sqrt{2}+1)^2

$$\sqrt { 3 } ( \sqrt { 6 } - \sqrt { 3 } ) + ( \sqrt { 2 } + 1 ) ^ { 2 }$$

Evaluate

$$5\sqrt{2}\approx 7.071067812$$

Solution Steps

Use the distributive property to multiply \sqrt{3} by \sqrt{6}-\sqrt{3}.

$$\sqrt{3}\sqrt{6}-\left(\sqrt{3}\right)^{2}+\left(\sqrt{2}+1\right)^{2}$$

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

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

The square of \sqrt{3} is 3.

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

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

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

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

Add -3 and 3 to get 0.

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

Combine 3\sqrt{2} and 2\sqrt{2} to get 5\sqrt{2}.

$$5\sqrt{2}$$