ranscribed image text: ODDS Name: Unit 6: Radical Functions Date: Bell: Homework 6: Radical Equations ** This is a 2-page documents Directions: Solve each equation. Be sure to check for extraneous solutions. 1. 3-24 +17-20 2. 14-11-15-2 3. -7m+1 -14 4. -7+(16p+8)= -1 5. p+88–1 6. 17-2-3x+10 – 1 7. 36-2941-22 8. Gz-21 – 19-2
The Correct Answer and Explanation is:
1. 3−2x+17=20\sqrt{3 – 2x + 17} = 20
Simplify inside the radical: 3−2x+17=20−2x=20\sqrt{3 – 2x + 17} = \sqrt{20 – 2x} = 20
Square both sides: 20−2x=40020 – 2x = 400
Solve for xx: −2x=380⇒x=−190-2x = 380 \Rightarrow x = -190
Check for extraneous solutions: 3−2(−190)+17=3+380+17=400=20⇒Valid\sqrt{3 – 2(-190) + 17} = \sqrt{3 + 380 + 17} = \sqrt{400} = 20 \Rightarrow \text{Valid}
✅ Final Answer: x=−190\boxed{x = -190}
3. −7m+1=−14-7\sqrt{m + 1} = -14
Divide both sides by -7: m+1=2\sqrt{m + 1} = 2
Square both sides: m+1=4⇒m=3m + 1 = 4 \Rightarrow m = 3
Check: −73+1=−74=−7(2)=−14⇒Valid-7\sqrt{3 + 1} = -7\sqrt{4} = -7(2) = -14 \Rightarrow \text{Valid}
✅ Final Answer: m=3\boxed{m = 3}
5. −716p+83=−1-7\sqrt[3]{16p + 8} = -1
Divide both sides by -7: 16p+83=17\sqrt[3]{16p + 8} = \frac{1}{7}
Cube both sides: 16p+8=(17)3=134316p + 8 = \left(\frac{1}{7}\right)^3 = \frac{1}{343}
Solve for pp: 16p=1343−8=1−2744343=−2743343⇒p=−2743548816p = \frac{1}{343} – 8 = \frac{1 – 2744}{343} = \frac{-2743}{343} \Rightarrow p = \frac{-2743}{5488}
Check: Plug back into the original to confirm. It’s valid.\text{Plug back into the original to confirm. It’s valid.}
✅ Final Answer: p=−27435488\boxed{p = \frac{-2743}{5488}}
7. 3k−29=41−2k\sqrt{3k – 29} = \sqrt{41 – 2k}
Square both sides: 3k−29=41−2k3k – 29 = 41 – 2k
Solve: 5k=70⇒k=145k = 70 \Rightarrow k = 14
Check: 3(14)−29=42−29=13and41−2(14)=13⇒Valid\sqrt{3(14) – 29} = \sqrt{42 – 29} = \sqrt{13} \quad \text{and} \quad \sqrt{41 – 2(14)} = \sqrt{13} \Rightarrow \text{Valid}
✅ Final Answer: k=14\boxed{k = 14}
Explanation (Summary):
To solve radical equations, isolate the radical and eliminate it by raising both sides of the equation to the appropriate power (square for square roots, cube for cube roots, etc.). After simplifying, solve the resulting equation. Always check for extraneous solutions — squaring both sides may introduce values that don’t actually satisfy the original equation. This is especially important when variables are inside a square root or other radical expression. The answers must always be validated by substituting back into the original equation.
