1. Let's state the problem: We want to check if $\sqrt{3} + 2\sqrt{2}\sqrt{3} + \sqrt{2} = \sqrt{24}$ is correct. 2. Recall the properties of square roots: $\sqrt{a}\sqrt{b} = \sqrt{ab}$ and simplify expressions step-by-step. 3. Simplify the left side: - $\sqrt{3}$ stays as is. - $2\sqrt{2}\sqrt{3} = 2\sqrt{6}$ because $\sqrt{2}\sqrt{3} = \sqrt{6}$. - $\sqrt{2}$ stays as is. So the left side is $\sqrt{3} + 2\sqrt{6} + \sqrt{2}$. 4. The right side is $\sqrt{24}$. Simplify $\sqrt{24}$: $$\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}.$$ 5. Now compare both sides: Left side: $\sqrt{3} + 2\sqrt{6} + \sqrt{2}$ Right side: $2\sqrt{6}$ They are not equal because the left side has extra terms $\sqrt{3}$ and $\sqrt{2}$. 6. Therefore, the original equation is incorrect. 7. The left side expression $\sqrt{3} + 2\sqrt{6} + \sqrt{2}$ cannot be simplified further by combining terms because the radicals are different. Final answer: $\sqrt{3} + 2\sqrt{2}\sqrt{3} + \sqrt{2} \neq \sqrt{24}$ and the left side equals $\sqrt{3} + 2\sqrt{6} + \sqrt{2}$.