Solve 7sqrt{50}-2sqrt{75}+2sqrt{18}-3sqrt{720}+3sqrt{48}

$$7 \sqrt{ 50 } -2 \sqrt{ 75 } +2 \sqrt{ 18 } -3 \sqrt{ 720 } +3 \sqrt{ 48 }$$

Evaluate

$$2\sqrt{3}+41\sqrt{2}-36\sqrt{5}\approx -19.051589518$$

Solution Steps

Factor 50=5^{2}\times 2. Rewrite the square root of the product \sqrt{5^{2}\times 2} as the product of square roots \sqrt{5^{2}}\sqrt{2}. Take the square root of 5^{2}.

$$7\times 5\sqrt{2}-2\sqrt{75}+2\sqrt{18}-3\sqrt{720}+3\sqrt{48}$$

Multiply 7 and 5 to get 35.

$$35\sqrt{2}-2\sqrt{75}+2\sqrt{18}-3\sqrt{720}+3\sqrt{48}$$

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}.

$$35\sqrt{2}-2\times 5\sqrt{3}+2\sqrt{18}-3\sqrt{720}+3\sqrt{48}$$

Multiply -2 and 5 to get -10.

$$35\sqrt{2}-10\sqrt{3}+2\sqrt{18}-3\sqrt{720}+3\sqrt{48}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$35\sqrt{2}-10\sqrt{3}+2\times 3\sqrt{2}-3\sqrt{720}+3\sqrt{48}$$

Multiply 2 and 3 to get 6.

$$35\sqrt{2}-10\sqrt{3}+6\sqrt{2}-3\sqrt{720}+3\sqrt{48}$$

Combine 35\sqrt{2} and 6\sqrt{2} to get 41\sqrt{2}.

$$41\sqrt{2}-10\sqrt{3}-3\sqrt{720}+3\sqrt{48}$$

Factor 720=12^{2}\times 5. Rewrite the square root of the product \sqrt{12^{2}\times 5} as the product of square roots \sqrt{12^{2}}\sqrt{5}. Take the square root of 12^{2}.

$$41\sqrt{2}-10\sqrt{3}-3\times 12\sqrt{5}+3\sqrt{48}$$

Multiply -3 and 12 to get -36.

$$41\sqrt{2}-10\sqrt{3}-36\sqrt{5}+3\sqrt{48}$$

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}.

$$41\sqrt{2}-10\sqrt{3}-36\sqrt{5}+3\times 4\sqrt{3}$$

Multiply 3 and 4 to get 12.

$$41\sqrt{2}-10\sqrt{3}-36\sqrt{5}+12\sqrt{3}$$

Combine -10\sqrt{3} and 12\sqrt{3} to get 2\sqrt{3}.

$$41\sqrt{2}+2\sqrt{3}-36\sqrt{5}$$