Solve (sqrt{3}+2)^2-sqrt{48}+sqrt{8}*sqrt{1/2}

$$( \sqrt { 3 } + 2 ) ^ { 2 } - \sqrt { 48 } + \sqrt { 8 } \times \sqrt { \frac { 1 } { 2 } }$$

Evaluate

$$9$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+2\right)^{2}.

$$\left(\sqrt{3}\right)^{2}+4\sqrt{3}+4-\sqrt{48}+\sqrt{8}\sqrt{\frac{1}{2}}$$

The square of \sqrt{3} is 3.

$$3+4\sqrt{3}+4-\sqrt{48}+\sqrt{8}\sqrt{\frac{1}{2}}$$

Add 3 and 4 to get 7.

$$7+4\sqrt{3}-\sqrt{48}+\sqrt{8}\sqrt{\frac{1}{2}}$$

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}.

$$7+4\sqrt{3}-4\sqrt{3}+\sqrt{8}\sqrt{\frac{1}{2}}$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$7+\sqrt{8}\sqrt{\frac{1}{2}}$$

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}.

$$7+2\sqrt{2}\sqrt{\frac{1}{2}}$$

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

$$7+2\sqrt{2}\times \frac{\sqrt{1}}{\sqrt{2}}$$

Calculate the square root of 1 and get 1.

$$7+2\sqrt{2}\times \frac{1}{\sqrt{2}}$$

Rationalize the denominator of \frac{1}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$7+2\sqrt{2}\times \frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

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

Cancel out 2 and 2.

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

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

$$7+2$$

Add 7 and 2 to get 9.

$$9$$

Factor

$$3^{2}$$