Solve 6/sqrt{2}-sqrt{18}-(1/2)^0quad(sqrt{48}+1/4sqrt{6})

$$\frac { 6 } { \sqrt { 2 } } - \sqrt { 18 } - ( \frac { 1 } { 2 } ) ^ { 0 } \quad ( \sqrt { 48 } + \frac { 1 } { 4 } \sqrt { 6 } )$$

Evaluate

$$-\frac{\sqrt{6}}{4}-4\sqrt{3}\approx -7.540575666$$

Solution Steps

Rationalize the denominator of \frac{6}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{6\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-\sqrt{18}-\left(\frac{1}{2}\right)^{0}\left(\sqrt{48}+\frac{1}{4}\sqrt{6}\right)$$

The square of \sqrt{2} is 2.

$$\frac{6\sqrt{2}}{2}-\sqrt{18}-\left(\frac{1}{2}\right)^{0}\left(\sqrt{48}+\frac{1}{4}\sqrt{6}\right)$$

Divide 6\sqrt{2} by 2 to get 3\sqrt{2}.

$$3\sqrt{2}-\sqrt{18}-\left(\frac{1}{2}\right)^{0}\left(\sqrt{48}+\frac{1}{4}\sqrt{6}\right)$$

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}.

$$3\sqrt{2}-3\sqrt{2}-\left(\frac{1}{2}\right)^{0}\left(\sqrt{48}+\frac{1}{4}\sqrt{6}\right)$$

Combine 3\sqrt{2} and -3\sqrt{2} to get 0.

$$0-\left(\frac{1}{2}\right)^{0}\left(\sqrt{48}+\frac{1}{4}\sqrt{6}\right)$$

Calculate \frac{1}{2} to the power of 0 and get 1.

$$0-1\left(\sqrt{48}+\frac{1}{4}\sqrt{6}\right)$$

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}.

$$0-1\left(4\sqrt{3}+\frac{1}{4}\sqrt{6}\right)$$

Multiply -1 and 1 to get -1.

$$0-\left(4\sqrt{3}+\frac{1}{4}\sqrt{6}\right)$$

Anything plus zero gives itself.

$$-\left(4\sqrt{3}+\frac{1}{4}\sqrt{6}\right)$$

To find the opposite of 4\sqrt{3}+\frac{1}{4}\sqrt{6}, find the opposite of each term.

$$-4\sqrt{3}-\frac{1}{4}\sqrt{6}$$

Factor

$$\frac{-\sqrt{6} - 16 \sqrt{3}}{4} = -7.5405756659713035$$