Solve (sqrt{2}-sqrt{12})*(sqrt{18}+sqrt{48})

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

Evaluate

$$-2\sqrt{6}-18\approx -22.898979486$$

Solution Steps

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}.

$$\left(\sqrt{2}-2\sqrt{3}\right)\left(\sqrt{18}+\sqrt{48}\right)$$

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}.

$$\left(\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{48}\right)$$

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}.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

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

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

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

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

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

Combine 4\sqrt{6} and -6\sqrt{6} to get -2\sqrt{6}.

$$6-2\sqrt{6}-8\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$6-2\sqrt{6}-8\times 3$$

Multiply -8 and 3 to get -24.

$$6-2\sqrt{6}-24$$

Subtract 24 from 6 to get -18.

$$-18-2\sqrt{6}$$