Solve sqrt{1^2+(4-7/6)^2}

$$\sqrt { 1 ^ { 2 } + ( 4 - \frac { 7 } { 6 } ) ^ { 2 } }$$

Evaluate

$$\frac{5\sqrt{13}}{6}\approx 3.004626063$$

Solution Steps

Calculate 1 to the power of 2 and get 1.

$$\sqrt{1+\left(4-\frac{7}{6}\right)^{2}}$$

Subtract \frac{7}{6} from 4 to get \frac{17}{6}.

$$\sqrt{1+\left(\frac{17}{6}\right)^{2}}$$

Calculate \frac{17}{6} to the power of 2 and get \frac{289}{36}.

$$\sqrt{1+\frac{289}{36}}$$

Add 1 and \frac{289}{36} to get \frac{325}{36}.

$$\sqrt{\frac{325}{36}}$$

Rewrite the square root of the division \sqrt{\frac{325}{36}} as the division of square roots \frac{\sqrt{325}}{\sqrt{36}}.

$$\frac{\sqrt{325}}{\sqrt{36}}$$

Factor 325=5^{2}\times 13. Rewrite the square root of the product \sqrt{5^{2}\times 13} as the product of square roots \sqrt{5^{2}}\sqrt{13}. Take the square root of 5^{2}.

$$\frac{5\sqrt{13}}{\sqrt{36}}$$

Calculate the square root of 36 and get 6.

$$\frac{5\sqrt{13}}{6}$$