Solve (sqrt{12}+sqrt{18}-sqrt{32})divsqrt{2}

$$( \sqrt{ 12 } + \sqrt{ 18 } - \sqrt{ 32 } ) \div \sqrt{ 2 }$$

Evaluate

$$\sqrt{6}-1\approx 1.449489743$$

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}.

$$\frac{2\sqrt{3}+\sqrt{18}-\sqrt{32}}{\sqrt{2}}$$

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}.

$$\frac{2\sqrt{3}+3\sqrt{2}-\sqrt{32}}{\sqrt{2}}$$

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}.

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

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

$$\frac{2\sqrt{3}-\sqrt{2}}{\sqrt{2}}$$

Rationalize the denominator of \frac{2\sqrt{3}-\sqrt{2}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Use the distributive property to multiply 2\sqrt{3}-\sqrt{2} by \sqrt{2}.

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

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

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

The square of \sqrt{2} is 2.

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

Divide each term of 2\sqrt{6}-2 by 2 to get \sqrt{6}-1.

$$\sqrt{6}-1$$