Solve sqrt{25+400/9}+sqrt{1+16/9}

$$\sqrt { 25 + \frac { 400 } { 9 } } + \sqrt { 1 + \frac { 16 } { 9 } }$$

Evaluate

$$10$$

Solution Steps

Convert 25 to fraction \frac{225}{9}.

$$\sqrt{\frac{225}{9}+\frac{400}{9}}+\sqrt{1+\frac{16}{9}}$$

Since \frac{225}{9} and \frac{400}{9} have the same denominator, add them by adding their numerators.

$$\sqrt{\frac{225+400}{9}}+\sqrt{1+\frac{16}{9}}$$

Add 225 and 400 to get 625.

$$\sqrt{\frac{625}{9}}+\sqrt{1+\frac{16}{9}}$$

Rewrite the square root of the division \frac{625}{9} as the division of square roots \frac{\sqrt{625}}{\sqrt{9}}. Take the square root of both numerator and denominator.

$$\frac{25}{3}+\sqrt{1+\frac{16}{9}}$$

Convert 1 to fraction \frac{9}{9}.

$$\frac{25}{3}+\sqrt{\frac{9}{9}+\frac{16}{9}}$$

Since \frac{9}{9} and \frac{16}{9} have the same denominator, add them by adding their numerators.

$$\frac{25}{3}+\sqrt{\frac{9+16}{9}}$$

Add 9 and 16 to get 25.

$$\frac{25}{3}+\sqrt{\frac{25}{9}}$$

Rewrite the square root of the division \frac{25}{9} as the division of square roots \frac{\sqrt{25}}{\sqrt{9}}. Take the square root of both numerator and denominator.

$$\frac{25}{3}+\frac{5}{3}$$

Since \frac{25}{3} and \frac{5}{3} have the same denominator, add them by adding their numerators.

$$\frac{25+5}{3}$$

Add 25 and 5 to get 30.

$$\frac{30}{3}$$

Divide 30 by 3 to get 10.

$$10$$

Factor

$$2\times 5$$