Solve (2)(sqrt{48}+sqrt{24})(sqrt{12}+sqrt{18})

$$( 2 ) ( \sqrt { 48 } + \sqrt { 24 } ) ( \sqrt { 12 } + \sqrt { 18 } )$$

Evaluate

$$24\left(\sqrt{2}+\sqrt{3}+\sqrt{6}+2\right)\approx 182.298098705$$

Solution Steps

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}.

$$2\left(4\sqrt{3}+\sqrt{24}\right)\left(\sqrt{12}+\sqrt{18}\right)$$

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}.

$$2\left(4\sqrt{3}+2\sqrt{6}\right)\left(\sqrt{12}+\sqrt{18}\right)$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$2\left(4\sqrt{3}+2\sqrt{6}\right)\left(2\sqrt{3}+\sqrt{18}\right)$$

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}.

$$2\left(4\sqrt{3}+2\sqrt{6}\right)\left(2\sqrt{3}+3\sqrt{2}\right)$$

Use the distributive property to multiply 2 by 4\sqrt{3}+2\sqrt{6}.

$$\left(8\sqrt{3}+4\sqrt{6}\right)\left(2\sqrt{3}+3\sqrt{2}\right)$$

Apply the distributive property by multiplying each term of 8\sqrt{3}+4\sqrt{6} by each term of 2\sqrt{3}+3\sqrt{2}.

$$16\left(\sqrt{3}\right)^{2}+24\sqrt{3}\sqrt{2}+8\sqrt{3}\sqrt{6}+12\sqrt{6}\sqrt{2}$$

The square of \sqrt{3} is 3.

$$16\times 3+24\sqrt{3}\sqrt{2}+8\sqrt{3}\sqrt{6}+12\sqrt{6}\sqrt{2}$$

Multiply 16 and 3 to get 48.

$$48+24\sqrt{3}\sqrt{2}+8\sqrt{3}\sqrt{6}+12\sqrt{6}\sqrt{2}$$

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

$$48+24\sqrt{6}+8\sqrt{3}\sqrt{6}+12\sqrt{6}\sqrt{2}$$

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

$$48+24\sqrt{6}+8\sqrt{3}\sqrt{3}\sqrt{2}+12\sqrt{6}\sqrt{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$48+24\sqrt{6}+8\times 3\sqrt{2}+12\sqrt{6}\sqrt{2}$$

Multiply 8 and 3 to get 24.

$$48+24\sqrt{6}+24\sqrt{2}+12\sqrt{6}\sqrt{2}$$

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}.

$$48+24\sqrt{6}+24\sqrt{2}+12\sqrt{2}\sqrt{3}\sqrt{2}$$

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

$$48+24\sqrt{6}+24\sqrt{2}+12\times 2\sqrt{3}$$

Multiply 12 and 2 to get 24.

$$48+24\sqrt{6}+24\sqrt{2}+24\sqrt{3}$$