Question
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Simplio \[ \begin{array}{l}\sqrt{12}-\sqrt{20} \\ \sqrt{48}-\sqrt{80}\end{array} \]

Ask by Horton Colon. in South Africa
Jan 23,2025

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Tutor-Verified Answer

Answer

\[ \sqrt{12} - \sqrt{20} = 2(\sqrt{3} - \sqrt{5}) \] \[ \sqrt{48} - \sqrt{80} = 4(\sqrt{3} - \sqrt{5}) \]

Solution

Certainly! Let's simplify each of the given expressions step by step. ### 1. Simplifying \(\sqrt{12} - \sqrt{20}\) **Step 1: Factor each radicand into its prime factors.** \[ \sqrt{12} = \sqrt{4 \times 3} \\ \sqrt{20} = \sqrt{4 \times 5} \] **Step 2: Simplify the square roots by taking out the square factors.** \[ \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3} \\ \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5} \] **Step 3: Substitute back into the original expression.** \[ \sqrt{12} - \sqrt{20} = 2\sqrt{3} - 2\sqrt{5} \] **Step 4: Factor out the common term.** \[ 2\sqrt{3} - 2\sqrt{5} = 2(\sqrt{3} - \sqrt{5}) \] **Final Simplified Form:** \[ \sqrt{12} - \sqrt{20} = 2(\sqrt{3} - \sqrt{5}) \] --- ### 2. Simplifying \(\sqrt{48} - \sqrt{80}\) **Step 1: Factor each radicand into its prime factors.** \[ \sqrt{48} = \sqrt{16 \times 3} \\ \sqrt{80} = \sqrt{16 \times 5} \] **Step 2: Simplify the square roots by taking out the square factors.** \[ \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \\ \sqrt{16 \times 5} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5} \] **Step 3: Substitute back into the original expression.** \[ \sqrt{48} - \sqrt{80} = 4\sqrt{3} - 4\sqrt{5} \] **Step 4: Factor out the common term.** \[ 4\sqrt{3} - 4\sqrt{5} = 4(\sqrt{3} - \sqrt{5}) \] **Final Simplified Form:** \[ \sqrt{48} - \sqrt{80} = 4(\sqrt{3} - \sqrt{5}) \] --- ### Summary: \[ \begin{align*} \sqrt{12} - \sqrt{20} &= 2(\sqrt{3} - \sqrt{5}) \\ \sqrt{48} - \sqrt{80} &= 4(\sqrt{3} - \sqrt{5}) \end{align*} \] These simplified forms factor out the common terms, making the expressions more concise.

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Let's simplify each expression step by step! Starting with \(\sqrt{12}-\sqrt{20}\): 1. Break down the square roots: \[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \] \[ \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5} \] 2. Now, substitute back: \[ \sqrt{12}-\sqrt{20} = 2\sqrt{3} - 2\sqrt{5} \] 3. Factor out the common term: \[ = 2(\sqrt{3} - \sqrt{5}) \] Now for \(\sqrt{48}-\sqrt{80}\): 1. Break down the square roots: \[ \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \] \[ \sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5} \] 2. Now, substitute back: \[ \sqrt{48}-\sqrt{80} = 4\sqrt{3} - 4\sqrt{5} \] 3. Factor out the common term: \[ = 4(\sqrt{3} - \sqrt{5}) \] So in conclusion, the simplified forms are: - \(\sqrt{12} - \sqrt{20} = 2(\sqrt{3} - \sqrt{5})\) - \(\sqrt{48} - \sqrt{80} = 4(\sqrt{3} - \sqrt{5})\)

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