Solve sqrt{98}-sqrt{12}-(sqrt{32}-sqrt{108})

$$\sqrt { 98 } - \sqrt { 12 } - ( \sqrt { 32 } - \sqrt { 108 } )$$

Evaluate

$$\sqrt{3}\left(\sqrt{6}+4\right)\approx 11.170843917$$

Solution Steps

Factor 98=7^{2}\times 2. Rewrite the square root of the product \sqrt{7^{2}\times 2} as the product of square roots \sqrt{7^{2}}\sqrt{2}. Take the square root of 7^{2}.

$$7\sqrt{2}-\sqrt{12}-\left(\sqrt{32}-\sqrt{108}\right)$$

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}.

$$7\sqrt{2}-2\sqrt{3}-\left(\sqrt{32}-\sqrt{108}\right)$$

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$7\sqrt{2}-2\sqrt{3}-\left(4\sqrt{2}-\sqrt{108}\right)$$

Factor 108=6^{2}\times 3. Rewrite the square root of the product \sqrt{6^{2}\times 3} as the product of square roots \sqrt{6^{2}}\sqrt{3}. Take the square root of 6^{2}.

$$7\sqrt{2}-2\sqrt{3}-\left(4\sqrt{2}-6\sqrt{3}\right)$$

To find the opposite of 4\sqrt{2}-6\sqrt{3}, find the opposite of each term.

$$7\sqrt{2}-2\sqrt{3}-4\sqrt{2}-\left(-6\sqrt{3}\right)$$

Combine 7\sqrt{2} and -4\sqrt{2} to get 3\sqrt{2}.

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

The opposite of -6\sqrt{3} is 6\sqrt{3}.

$$3\sqrt{2}-2\sqrt{3}+6\sqrt{3}$$

Combine -2\sqrt{3} and 6\sqrt{3} to get 4\sqrt{3}.

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