Solve (sqrt{48}+1/4sqrt{8})-sqrt{27}

$$( \sqrt { 48 } + \frac { 1 } { 4 } \sqrt { 8 } ) - \sqrt { 27 }$$

Evaluate

$$\frac{\sqrt{2}\left(\sqrt{6}+1\right)}{2}\approx 2.439157589$$

Solution Steps

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}.

$$4\sqrt{3}+\frac{1}{4}\sqrt{8}-\sqrt{27}$$

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}.

$$4\sqrt{3}+\frac{1}{4}\times 2\sqrt{2}-\sqrt{27}$$

Multiply \frac{1}{4} and 2 to get \frac{2}{4}.

$$4\sqrt{3}+\frac{2}{4}\sqrt{2}-\sqrt{27}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$4\sqrt{3}+\frac{1}{2}\sqrt{2}-\sqrt{27}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

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

$$\sqrt{3}+\frac{1}{2}\sqrt{2}$$

Factor

$$\frac{\sqrt{2} {(\sqrt{2} \sqrt{3} + 1)}}{2} = 2.439157588755425$$