Solve 9sqrt{75}+1/3sqrt{147}

$$9 \sqrt { 75 } + \frac { 1 } { 3 } \sqrt { 147 }$$

Evaluate

$$\frac{142\sqrt{3}}{3}\approx 81.983738225$$

Solution Steps

Factor 75=5^{2}\times 3. Rewrite the square root of the product \sqrt{5^{2}\times 3} as the product of square roots \sqrt{5^{2}}\sqrt{3}. Take the square root of 5^{2}.

$$9\times 5\sqrt{3}+\frac{1}{3}\sqrt{147}$$

Multiply 9 and 5 to get 45.

$$45\sqrt{3}+\frac{1}{3}\sqrt{147}$$

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}.

$$45\sqrt{3}+\frac{1}{3}\times 7\sqrt{3}$$

Multiply \frac{1}{3} and 7 to get \frac{7}{3}.

$$45\sqrt{3}+\frac{7}{3}\sqrt{3}$$

Combine 45\sqrt{3} and \frac{7}{3}\sqrt{3} to get \frac{142}{3}\sqrt{3}.

$$\frac{142}{3}\sqrt{3}$$