Solve (sqrt{8}+sqrt{50})(sqrt{24}+sqrt{54})

$$( \sqrt { 8 } + \sqrt { 50 } ) ( \sqrt { 24 } + \sqrt { 54 } )$$

Evaluate

$$70\sqrt{3}\approx 121.24355653$$

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\left(2\sqrt{2}+\sqrt{50}\right)\left(\sqrt{24}+\sqrt{54}\right)$$

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}.

$$\left(2\sqrt{2}+5\sqrt{2}\right)\left(\sqrt{24}+\sqrt{54}\right)$$

Combine 2\sqrt{2} and 5\sqrt{2} to get 7\sqrt{2}.

$$7\sqrt{2}\left(\sqrt{24}+\sqrt{54}\right)$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$7\sqrt{2}\left(2\sqrt{6}+\sqrt{54}\right)$$

Factor 54=3^{2}\times 6. Rewrite the square root of the product \sqrt{3^{2}\times 6} as the product of square roots \sqrt{3^{2}}\sqrt{6}. Take the square root of 3^{2}.

$$7\sqrt{2}\left(2\sqrt{6}+3\sqrt{6}\right)$$

Combine 2\sqrt{6} and 3\sqrt{6} to get 5\sqrt{6}.

$$7\sqrt{2}\times 5\sqrt{6}$$

Multiply 7 and 5 to get 35.

$$35\sqrt{2}\sqrt{6}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$35\sqrt{2}\sqrt{2}\sqrt{3}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$35\times 2\sqrt{3}$$

Multiply 35 and 2 to get 70.

$$70\sqrt{3}$$