Solve sqrt{(1+6^2)[(11/3)^2-4*121/36]}

$$\sqrt { ( 1 + 6 ^ { 2 } ) [ ( \frac { 11 } { 3 } ) ^ { 2 } - 4 \times \frac { 121 } { 36 } ] }$$

Evaluate

$$0$$

Solution Steps

Calculate 6 to the power of 2 and get 36.

$$\sqrt{\left(1+36\right)\left(\left(\frac{11}{3}\right)^{2}-4\times \frac{121}{36}\right)}$$

Add 1 and 36 to get 37.

$$\sqrt{37\left(\left(\frac{11}{3}\right)^{2}-4\times \frac{121}{36}\right)}$$

Calculate \frac{11}{3} to the power of 2 and get \frac{121}{9}.

$$\sqrt{37\left(\frac{121}{9}-4\times \frac{121}{36}\right)}$$

Express 4\times \frac{121}{36} as a single fraction.

$$\sqrt{37\left(\frac{121}{9}-\frac{4\times 121}{36}\right)}$$

Multiply 4 and 121 to get 484.

$$\sqrt{37\left(\frac{121}{9}-\frac{484}{36}\right)}$$

Reduce the fraction \frac{484}{36} to lowest terms by extracting and canceling out 4.

$$\sqrt{37\left(\frac{121}{9}-\frac{121}{9}\right)}$$

Subtract \frac{121}{9} from \frac{121}{9} to get 0.

$$\sqrt{37\times 0}$$

Multiply 37 and 0 to get 0.

$$\sqrt{0}$$

Calculate the square root of 0 and get 0.

$$0$$

Factor

$$0$$