Solve (6sqrt{3}-2sqrt{1/2})-(sqrt{75}-sqrt{32})

$$( 6 \sqrt { 3 } - 2 \sqrt { \frac { 1 } { 2 } } ) - ( \sqrt { 75 } - \sqrt { 32 } )$$

Evaluate

$$\sqrt{3}+3\sqrt{2}\approx 5.974691495$$

Solution Steps

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

$$6\sqrt{3}-2\times \frac{\sqrt{1}}{\sqrt{2}}-\left(\sqrt{75}-\sqrt{32}\right)$$

Calculate the square root of 1 and get 1.

$$6\sqrt{3}-2\times \frac{1}{\sqrt{2}}-\left(\sqrt{75}-\sqrt{32}\right)$$

Rationalize the denominator of \frac{1}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$6\sqrt{3}-2\times \frac{\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-\left(\sqrt{75}-\sqrt{32}\right)$$

The square of \sqrt{2} is 2.

$$6\sqrt{3}-2\times \frac{\sqrt{2}}{2}-\left(\sqrt{75}-\sqrt{32}\right)$$

Cancel out 2 and 2.

$$6\sqrt{3}-\sqrt{2}-\left(\sqrt{75}-\sqrt{32}\right)$$

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}.

$$6\sqrt{3}-\sqrt{2}-\left(5\sqrt{3}-\sqrt{32}\right)$$

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}.

$$6\sqrt{3}-\sqrt{2}-\left(5\sqrt{3}-4\sqrt{2}\right)$$

To find the opposite of 5\sqrt{3}-4\sqrt{2}, find the opposite of each term.

$$6\sqrt{3}-\sqrt{2}-5\sqrt{3}-\left(-4\sqrt{2}\right)$$

Combine 6\sqrt{3} and -5\sqrt{3} to get \sqrt{3}.

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

The opposite of -4\sqrt{2} is 4\sqrt{2}.

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

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

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