Solve frac{sqrt{12}}{sqrt{2}}+sqrt{18}+2^0-|-3|+(-1/2)^-1

$$\frac { \sqrt { 12 } } { \sqrt { 2 } } + \sqrt { 18 } + 2 ^ { 0 } - | - 3 | + ( - \frac { 1 } { 2 } ) ^ { - 1 }$$

Evaluate

$$\sqrt{6}+3\sqrt{2}-4\approx 2.69213043$$

Solution Steps

Rewrite the division of square roots \frac{\sqrt{12}}{\sqrt{2}} as the square root of the division \sqrt{\frac{12}{2}} and perform the division.

$$\sqrt{6}+\sqrt{18}+2^{0}-|-3|+\left(-\frac{1}{2}\right)^{-1}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$\sqrt{6}+3\sqrt{2}+2^{0}-|-3|+\left(-\frac{1}{2}\right)^{-1}$$

Calculate 2 to the power of 0 and get 1.

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

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -3 is 3.

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

Subtract 3 from 1 to get -2.

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

Calculate -\frac{1}{2} to the power of -1 and get -2.

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

Subtract 2 from -2 to get -4.

$$\sqrt{6}+3\sqrt{2}-4$$