Solve (sqrt{12}-sqrt{16})+sqrt{13}(sqrt{3}+1)(sqrt{3}-1)

$$( \sqrt { 12 } - \sqrt { 16 } ) + \sqrt { 13 } ( \sqrt { 3 } + 1 ) ( \sqrt { 3 } - 1 )$$

Evaluate

$$2\left(\sqrt{3}+\sqrt{13}-2\right)\approx 6.675204166$$

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}.

$$2\sqrt{3}-\sqrt{16}+\sqrt{13}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)$$

Calculate the square root of 16 and get 4.

$$2\sqrt{3}-4+\sqrt{13}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)$$

Use the distributive property to multiply \sqrt{13} by \sqrt{3}+1.

$$2\sqrt{3}-4+\left(\sqrt{13}\sqrt{3}+\sqrt{13}\right)\left(\sqrt{3}-1\right)$$

To multiply \sqrt{13} and \sqrt{3}, multiply the numbers under the square root.

$$2\sqrt{3}-4+\left(\sqrt{39}+\sqrt{13}\right)\left(\sqrt{3}-1\right)$$

Apply the distributive property by multiplying each term of \sqrt{39}+\sqrt{13} by each term of \sqrt{3}-1.

$$2\sqrt{3}-4+\sqrt{39}\sqrt{3}-\sqrt{39}+\sqrt{13}\sqrt{3}-\sqrt{13}$$

Factor 39=3\times 13. Rewrite the square root of the product \sqrt{3\times 13} as the product of square roots \sqrt{3}\sqrt{13}.

$$2\sqrt{3}-4+\sqrt{3}\sqrt{13}\sqrt{3}-\sqrt{39}+\sqrt{13}\sqrt{3}-\sqrt{13}$$

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

$$2\sqrt{3}-4+3\sqrt{13}-\sqrt{39}+\sqrt{13}\sqrt{3}-\sqrt{13}$$

To multiply \sqrt{13} and \sqrt{3}, multiply the numbers under the square root.

$$2\sqrt{3}-4+3\sqrt{13}-\sqrt{39}+\sqrt{39}-\sqrt{13}$$

Combine -\sqrt{39} and \sqrt{39} to get 0.

$$2\sqrt{3}-4+3\sqrt{13}-\sqrt{13}$$

Combine 3\sqrt{13} and -\sqrt{13} to get 2\sqrt{13}.

$$2\sqrt{3}-4+2\sqrt{13}$$