Solve sqrt{8}+sqrt{2}+sqrt{72}quadtext{(f)}sqrt{300}

$$\sqrt { 8 } + \sqrt { 2 } + \sqrt { 72 } \quad \text { (f) } \sqrt { 300 }$$

Evaluate

$$3\left(20\sqrt{6}f+\sqrt{2}\right)$$

Solution Steps

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}.

$$2\sqrt{2}+\sqrt{2}+\sqrt{72}f\sqrt{300}$$

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

$$3\sqrt{2}+\sqrt{72}f\sqrt{300}$$

Factor 72=6^{2}\times 2. Rewrite the square root of the product \sqrt{6^{2}\times 2} as the product of square roots \sqrt{6^{2}}\sqrt{2}. Take the square root of 6^{2}.

$$3\sqrt{2}+6\sqrt{2}f\sqrt{300}$$

Factor 300=10^{2}\times 3. Rewrite the square root of the product \sqrt{10^{2}\times 3} as the product of square roots \sqrt{10^{2}}\sqrt{3}. Take the square root of 10^{2}.

$$3\sqrt{2}+6\sqrt{2}f\times 10\sqrt{3}$$

Multiply 6 and 10 to get 60.

$$3\sqrt{2}+60\sqrt{2}f\sqrt{3}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$3\sqrt{2}+60\sqrt{6}f$$

Differentiate w.r.t. f

$$60 \sqrt{6} = 146.969384567$$