Solve 4+(frac{sqrt{2}}{2}-1)-sqrt{3}+sqrt{12}

$$4 + ( \frac { \sqrt { 2 } } { 2 } - 1 ) - \sqrt { 3 } + \sqrt { 12 }$$

Evaluate

$$\frac{\sqrt{2}}{2}+\sqrt{3}+3\approx 5.439157589$$

Short Solution Steps

Subtract 1 from 4 to get 3.

$$3+\frac{\sqrt{2}}{2}-\sqrt{3}+\sqrt{12}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 3+\sqrt{3} times \frac{2}{2}.

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

Since \frac{2\left(3+\sqrt{3}\right)}{2} and \frac{\sqrt{2}}{2} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2\left(3+\sqrt{3}\right)+\sqrt{2}.

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

Factor

$$\frac{\sqrt{2} + 2 \sqrt{3} + 6}{2} = 5.439157588755425$$