Solve frac{2sqrt{98}+3sqrt{72}}{5sqrt{75}-3sqrt{48}}

$$\frac{ 2 \sqrt{ 98 } +3 \sqrt{ 72 } }{ 5 \sqrt{ 75 } -3 \sqrt{ 48 } }$$

Evaluate

$$\frac{32\sqrt{6}}{39}\approx 2.009837738$$

Solution Steps

Factor 98=7^{2}\times 2. Rewrite the square root of the product \sqrt{7^{2}\times 2} as the product of square roots \sqrt{7^{2}}\sqrt{2}. Take the square root of 7^{2}.

$$\frac{2\times 7\sqrt{2}+3\sqrt{72}}{5\sqrt{75}-3\sqrt{48}}$$

Multiply 2 and 7 to get 14.

$$\frac{14\sqrt{2}+3\sqrt{72}}{5\sqrt{75}-3\sqrt{48}}$$

Factor 72=6^{2}\times 2. Rewrite the square root of the product \sqrt{6^{2}\times 2} as the product of square roots \sqrt{6^{2}}\sqrt{2}. Take the square root of 6^{2}.

$$\frac{14\sqrt{2}+3\times 6\sqrt{2}}{5\sqrt{75}-3\sqrt{48}}$$

Multiply 3 and 6 to get 18.

$$\frac{14\sqrt{2}+18\sqrt{2}}{5\sqrt{75}-3\sqrt{48}}$$

Combine 14\sqrt{2} and 18\sqrt{2} to get 32\sqrt{2}.

$$\frac{32\sqrt{2}}{5\sqrt{75}-3\sqrt{48}}$$

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}.

$$\frac{32\sqrt{2}}{5\times 5\sqrt{3}-3\sqrt{48}}$$

Multiply 5 and 5 to get 25.

$$\frac{32\sqrt{2}}{25\sqrt{3}-3\sqrt{48}}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{32\sqrt{2}}{25\sqrt{3}-3\times 4\sqrt{3}}$$

Multiply -3 and 4 to get -12.

$$\frac{32\sqrt{2}}{25\sqrt{3}-12\sqrt{3}}$$

Combine 25\sqrt{3} and -12\sqrt{3} to get 13\sqrt{3}.

$$\frac{32\sqrt{2}}{13\sqrt{3}}$$

Rationalize the denominator of \frac{32\sqrt{2}}{13\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{32\sqrt{2}\sqrt{3}}{13\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{32\sqrt{2}\sqrt{3}}{13\times 3}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{32\sqrt{6}}{13\times 3}$$

Multiply 13 and 3 to get 39.

$$\frac{32\sqrt{6}}{39}$$