Solve frac{3sqrt{18}}{6sqrt{12}}=frac{sqrt{18}}{2sqrt{12}}=

$$\frac { 3 \sqrt { 18 } } { 6 \sqrt { 12 } } = \frac { \sqrt { 18 } } { 2 \sqrt { 12 } } =$$

Verify

Solution Steps

Cancel out 3 in both numerator and denominator.

$$\frac{\sqrt{18}}{2\sqrt{12}}=\frac{\sqrt{18}}{2\sqrt{12}}$$

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}.

$$\frac{3\sqrt{2}}{2\sqrt{12}}=\frac{\sqrt{18}}{2\sqrt{12}}$$

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}.

$$\frac{3\sqrt{2}}{2\times 2\sqrt{3}}=\frac{\sqrt{18}}{2\sqrt{12}}$$

Multiply 2 and 2 to get 4.

$$\frac{3\sqrt{2}}{4\sqrt{3}}=\frac{\sqrt{18}}{2\sqrt{12}}$$

Rationalize the denominator of \frac{3\sqrt{2}}{4\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{3\sqrt{2}\sqrt{3}}{4\left(\sqrt{3}\right)^{2}}=\frac{\sqrt{18}}{2\sqrt{12}}$$

The square of \sqrt{3} is 3.

$$\frac{3\sqrt{2}\sqrt{3}}{4\times 3}=\frac{\sqrt{18}}{2\sqrt{12}}$$

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

$$\frac{3\sqrt{6}}{4\times 3}=\frac{\sqrt{18}}{2\sqrt{12}}$$

Multiply 4 and 3 to get 12.

$$\frac{3\sqrt{6}}{12}=\frac{\sqrt{18}}{2\sqrt{12}}$$

Divide 3\sqrt{6} by 12 to get \frac{1}{4}\sqrt{6}.

$$\frac{1}{4}\sqrt{6}=\frac{\sqrt{18}}{2\sqrt{12}}$$

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}.

$$\frac{1}{4}\sqrt{6}=\frac{3\sqrt{2}}{2\sqrt{12}}$$

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}.

$$\frac{1}{4}\sqrt{6}=\frac{3\sqrt{2}}{2\times 2\sqrt{3}}$$

Multiply 2 and 2 to get 4.

$$\frac{1}{4}\sqrt{6}=\frac{3\sqrt{2}}{4\sqrt{3}}$$

Rationalize the denominator of \frac{3\sqrt{2}}{4\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{1}{4}\sqrt{6}=\frac{3\sqrt{2}\sqrt{3}}{4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{1}{4}\sqrt{6}=\frac{3\sqrt{2}\sqrt{3}}{4\times 3}$$

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

$$\frac{1}{4}\sqrt{6}=\frac{3\sqrt{6}}{4\times 3}$$

Multiply 4 and 3 to get 12.

$$\frac{1}{4}\sqrt{6}=\frac{3\sqrt{6}}{12}$$

Divide 3\sqrt{6} by 12 to get \frac{1}{4}\sqrt{6}.

$$\frac{1}{4}\sqrt{6}=\frac{1}{4}\sqrt{6}$$

Subtract \frac{1}{4}\sqrt{6} from both sides.

$$\frac{1}{4}\sqrt{6}-\frac{1}{4}\sqrt{6}=0$$

Combine \frac{1}{4}\sqrt{6} and -\frac{1}{4}\sqrt{6} to get 0.

$$0=0$$

Compare 0 and 0.

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