Solve (sqrt{24}-sqrt{32})divsqrt{2}

$$( \sqrt { 24 } - \sqrt { 32 } ) \div \sqrt { 2 }$$

Evaluate

$$2\left(\sqrt{3}-2\right)\approx -0.535898385$$

Solution Steps

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$\frac{2\sqrt{6}-\sqrt{32}}{\sqrt{2}}$$

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}.

$$\frac{2\sqrt{6}-4\sqrt{2}}{\sqrt{2}}$$

Rationalize the denominator of \frac{2\sqrt{6}-4\sqrt{2}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\left(2\sqrt{6}-4\sqrt{2}\right)\sqrt{2}}{\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

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

Use the distributive property to multiply 2\sqrt{6}-4\sqrt{2} by \sqrt{2}.

$$\frac{2\sqrt{6}\sqrt{2}-4\left(\sqrt{2}\right)^{2}}{2}$$

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

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

Multiply \sqrt{2} and \sqrt{2} to get 2.

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

Multiply 2 and 2 to get 4.

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

The square of \sqrt{2} is 2.

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

Multiply -4 and 2 to get -8.

$$\frac{4\sqrt{3}-8}{2}$$