Solve sqrt{18}-(sqrt{98}-2sqrt{75}+sqrt{27})

$$\sqrt{ 18 } -( \sqrt{ 98 } -2 \sqrt{ 75 } + \sqrt{ 27 } )$$

Evaluate

$$7\sqrt{3}-4\sqrt{2}\approx 6.467501403$$

Solution Steps

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}.

$$3\sqrt{2}-\left(\sqrt{98}-2\sqrt{75}+\sqrt{27}\right)$$

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}.

$$3\sqrt{2}-\left(7\sqrt{2}-2\sqrt{75}+\sqrt{27}\right)$$

Factor 75=5^{2}\times 3. Rewrite the square root of the product \sqrt{5^{2}\times 3} as the product of square roots \sqrt{5^{2}}\sqrt{3}. Take the square root of 5^{2}.

$$3\sqrt{2}-\left(7\sqrt{2}-2\times 5\sqrt{3}+\sqrt{27}\right)$$

Multiply -2 and 5 to get -10.

$$3\sqrt{2}-\left(7\sqrt{2}-10\sqrt{3}+\sqrt{27}\right)$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$3\sqrt{2}-\left(7\sqrt{2}-10\sqrt{3}+3\sqrt{3}\right)$$

Combine -10\sqrt{3} and 3\sqrt{3} to get -7\sqrt{3}.

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

To find the opposite of 7\sqrt{2}-7\sqrt{3}, find the opposite of each term.

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

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

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

The opposite of -7\sqrt{3} is 7\sqrt{3}.

$$-4\sqrt{2}+7\sqrt{3}$$