Solve sqrt{598/7.35}+1

$$\sqrt { \frac { 598 } { 7.35 } } + 1$$

Evaluate

$$\frac{2\sqrt{8970}}{21}+1\approx 10.020007994$$

Short Solution Steps

Expand \frac{598}{7.35} by multiplying both numerator and the denominator by 100.

$$\sqrt{\frac{59800}{735}}+1$$

Reduce the fraction \frac{59800}{735} to lowest terms by extracting and canceling out 5.

$$\sqrt{\frac{11960}{147}}+1$$

Rewrite the square root of the division \sqrt{\frac{11960}{147}} as the division of square roots \frac{\sqrt{11960}}{\sqrt{147}}.

$$\frac{\sqrt{11960}}{\sqrt{147}}+1$$

Factor 11960=2^{2}\times 2990. Rewrite the square root of the product \sqrt{2^{2}\times 2990} as the product of square roots \sqrt{2^{2}}\sqrt{2990}. Take the square root of 2^{2}.

$$\frac{2\sqrt{2990}}{\sqrt{147}}+1$$

Factor 147=7^{2}\times 3. Rewrite the square root of the product \sqrt{7^{2}\times 3} as the product of square roots \sqrt{7^{2}}\sqrt{3}. Take the square root of 7^{2}.

$$\frac{2\sqrt{2990}}{7\sqrt{3}}+1$$

Rationalize the denominator of \frac{2\sqrt{2990}}{7\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{2\sqrt{2990}\sqrt{3}}{7\left(\sqrt{3}\right)^{2}}+1$$

The square of \sqrt{3} is 3.

$$\frac{2\sqrt{2990}\sqrt{3}}{7\times 3}+1$$

To multiply \sqrt{2990} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{2\sqrt{8970}}{7\times 3}+1$$

Multiply 7 and 3 to get 21.

$$\frac{2\sqrt{8970}}{21}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{21}{21}.

$$\frac{2\sqrt{8970}}{21}+\frac{21}{21}$$

Since \frac{2\sqrt{8970}}{21} and \frac{21}{21} have the same denominator, add them by adding their numerators.

$$\frac{2\sqrt{8970}+21}{21}$$