Solve frac{6sqrt{19}}{sqrt{38}}=3sqrt{2}

$$\frac { 6 \sqrt { 19 } } { \sqrt { 38 } } = 3 \sqrt { 2 }$$

Verify

Solution Steps

Rationalize the denominator of \frac{6\sqrt{19}}{\sqrt{38}} by multiplying numerator and denominator by \sqrt{38}.

$$\frac{6\sqrt{19}\sqrt{38}}{\left(\sqrt{38}\right)^{2}}=3\sqrt{2}$$

The square of \sqrt{38} is 38.

$$\frac{6\sqrt{19}\sqrt{38}}{38}=3\sqrt{2}$$

Factor 38=19\times 2. Rewrite the square root of the product \sqrt{19\times 2} as the product of square roots \sqrt{19}\sqrt{2}.

$$\frac{6\sqrt{19}\sqrt{19}\sqrt{2}}{38}=3\sqrt{2}$$

Multiply \sqrt{19} and \sqrt{19} to get 19.

$$\frac{6\times 19\sqrt{2}}{38}=3\sqrt{2}$$

Multiply 6 and 19 to get 114.

$$\frac{114\sqrt{2}}{38}=3\sqrt{2}$$

Divide 114\sqrt{2} by 38 to get 3\sqrt{2}.

$$3\sqrt{2}=3\sqrt{2}$$

Subtract 3\sqrt{2} from both sides.

$$3\sqrt{2}-3\sqrt{2}=0$$

Combine 3\sqrt{2} and -3\sqrt{2} to get 0.

$$0=0$$

Compare 0 and 0.

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