Solve sqrt{{left(3/2right)}^2+6^2}

$$\sqrt{ { \left( \frac{ 3 }{ 2 } \right) }^{ 2 } + { 6 }^{ 2 } }$$

Evaluate

$$\frac{3\sqrt{17}}{2}\approx 6.184658438$$

Solution Steps

Calculate \frac{3}{2} to the power of 2 and get \frac{9}{4}.

$$\sqrt{\frac{9}{4}+6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\sqrt{\frac{9}{4}+36}$$

Convert 36 to fraction \frac{144}{4}.

$$\sqrt{\frac{9}{4}+\frac{144}{4}}$$

Since \frac{9}{4} and \frac{144}{4} have the same denominator, add them by adding their numerators.

$$\sqrt{\frac{9+144}{4}}$$

Add 9 and 144 to get 153.

$$\sqrt{\frac{153}{4}}$$

Rewrite the square root of the division \sqrt{\frac{153}{4}} as the division of square roots \frac{\sqrt{153}}{\sqrt{4}}.

$$\frac{\sqrt{153}}{\sqrt{4}}$$

Factor 153=3^{2}\times 17. Rewrite the square root of the product \sqrt{3^{2}\times 17} as the product of square roots \sqrt{3^{2}}\sqrt{17}. Take the square root of 3^{2}.

$$\frac{3\sqrt{17}}{\sqrt{4}}$$

Calculate the square root of 4 and get 2.

$$\frac{3\sqrt{17}}{2}$$