Solve sqrt{19/2^2+3^2+9^2}

$$\sqrt{ \frac{ 19 }{ { 2 }^{ 2 } + { 3 }^{ 2 } + { 9 }^{ 2 } } }$$

Evaluate

$$\frac{\sqrt{1786}}{94}\approx 0.449586098$$

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$\sqrt{\frac{19}{4+3^{2}+9^{2}}}$$

Calculate 3 to the power of 2 and get 9.

$$\sqrt{\frac{19}{4+9+9^{2}}}$$

Add 4 and 9 to get 13.

$$\sqrt{\frac{19}{13+9^{2}}}$$

Calculate 9 to the power of 2 and get 81.

$$\sqrt{\frac{19}{13+81}}$$

Add 13 and 81 to get 94.

$$\sqrt{\frac{19}{94}}$$

Rewrite the square root of the division \sqrt{\frac{19}{94}} as the division of square roots \frac{\sqrt{19}}{\sqrt{94}}.

$$\frac{\sqrt{19}}{\sqrt{94}}$$

Rationalize the denominator of \frac{\sqrt{19}}{\sqrt{94}} by multiplying numerator and denominator by \sqrt{94}.

$$\frac{\sqrt{19}\sqrt{94}}{\left(\sqrt{94}\right)^{2}}$$

The square of \sqrt{94} is 94.

$$\frac{\sqrt{19}\sqrt{94}}{94}$$

To multiply \sqrt{19} and \sqrt{94}, multiply the numbers under the square root.

$$\frac{\sqrt{1786}}{94}$$