Solve sqrt{7+1/2+1+40}

$$\sqrt { \frac { 7 + 1 } { 2 + 1 } + 40 }$$

Evaluate

$$\frac{8\sqrt{6}}{3}\approx 6.531972647$$

Solution Steps

Add 7 and 1 to get 8.

$$\sqrt{\frac{8}{2+1}+40}$$

Add 2 and 1 to get 3.

$$\sqrt{\frac{8}{3}+40}$$

Convert 40 to fraction \frac{120}{3}.

$$\sqrt{\frac{8}{3}+\frac{120}{3}}$$

Since \frac{8}{3} and \frac{120}{3} have the same denominator, add them by adding their numerators.

$$\sqrt{\frac{8+120}{3}}$$

Add 8 and 120 to get 128.

$$\sqrt{\frac{128}{3}}$$

Rewrite the square root of the division \sqrt{\frac{128}{3}} as the division of square roots \frac{\sqrt{128}}{\sqrt{3}}.

$$\frac{\sqrt{128}}{\sqrt{3}}$$

Factor 128=8^{2}\times 2. Rewrite the square root of the product \sqrt{8^{2}\times 2} as the product of square roots \sqrt{8^{2}}\sqrt{2}. Take the square root of 8^{2}.

$$\frac{8\sqrt{2}}{\sqrt{3}}$$

Rationalize the denominator of \frac{8\sqrt{2}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{8\sqrt{2}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{8\sqrt{2}\sqrt{3}}{3}$$

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

$$\frac{8\sqrt{6}}{3}$$