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

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

Evaluate

$$2\sqrt{2}+7\approx 9.828427125$$

Solution Steps

Use the distributive property to multiply -2\sqrt{2} by \sqrt{3}-1.

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{6}+1\right)^{2}.

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

The square of \sqrt{6} is 6.

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

Add 6 and 1 to get 7.

$$-2\sqrt{6}+2\sqrt{2}+7+2\sqrt{6}$$

Combine -2\sqrt{6} and 2\sqrt{6} to get 0.

$$2\sqrt{2}+7$$