Solve sqrt{3}*sqrt{24}=6sqrt{2}

$$\sqrt { 3 } \cdot \sqrt { 24 } = 6 \sqrt { 2 }$$

Verify

Solution Steps

Factor 24=3\times 8. Rewrite the square root of the product \sqrt{3\times 8} as the product of square roots \sqrt{3}\sqrt{8}.

$$\sqrt{3}\sqrt{3}\sqrt{8}=6\sqrt{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$3\sqrt{8}=6\sqrt{2}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$3\times 2\sqrt{2}=6\sqrt{2}$$

Multiply 3 and 2 to get 6.

$$6\sqrt{2}=6\sqrt{2}$$

Subtract 6\sqrt{2} from both sides.

$$6\sqrt{2}-6\sqrt{2}=0$$

Combine 6\sqrt{2} and -6\sqrt{2} to get 0.

$$0=0$$

Compare 0 and 0.

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