Sure! Below are ten random radical equations, followed by solutions and explanations for each step.
1. Solve \( \sqrt{2x + 1} = 5 \)
Step 1: Square both sides to eliminate the square root. \[ (\sqrt{2x + 1})^2 = 5^2 \ 2x + 1 = 25 \]
Step 2: Subtract 1 from both sides. \[ 2x = 24 \]
Step 3: Divide by 2. \[ x = 12 \]
Answer: \( x = 12 \)
2. Solve \( \sqrt{3x - 4} + 2 = 7 \)
Step 1: Subtract 2 from both sides. \[ \sqrt{3x - 4} = 5 \]
Step 2: Square both sides. \[ (\sqrt{3x - 4})^2 = 5^2 \ 3x - 4 = 25 \]
Step 3: Add 4 to both sides. \[ 3x = 29 \]
Step 4: Divide by 3. \[ x = \frac{29}{3} \]
Answer: \( x = \frac{29}{3} \)
3. Solve \( 2\sqrt{x + 1} = 10 \)
Step 1: Divide by 2 on both sides. \[ \sqrt{x + 1} = 5 \]
Step 2: Square both sides. \[ (\sqrt{x + 1})^2 = 5^2 \ x + 1 = 25 \]
Step 3: Subtract 1 from both sides. \[ x = 24 \]
Answer: \( x = 24 \)
4. Solve \( \sqrt{5x + 7} - 3 = 1 \)
Step 1: Add 3 to both sides. \[ \sqrt{5x + 7} = 4 \]
Step 2: Square both sides. \[ (\sqrt{5x + 7})^2 = 4^2 \ 5x + 7 = 16 \]
Step 3: Subtract 7 from both sides. \[ 5x = 9 \]
Step 4: Divide by 5. \[ x = \frac{9}{5} \]
Answer: \( x = \frac{9}{5} \)
5. Solve \( \sqrt{4x - 1} + 1 = 6 \)
Step 1: Subtract 1 from both sides. \[ \sqrt{4x - 1} = 5 \]
Step 2: Square both sides. \[ (\sqrt{4x - 1})^2 = 5^2 \ 4x - 1 = 25 \]
Step 3: Add 1 to both sides. \[ 4x = 26 \]
Step 4: Divide by 4. \[ x = \frac{26}{4} = \frac{13}{2} \]
Answer: \( x = \frac{13}{2} \)
6. Solve \( \sqrt{2x + 3} = x - 1 \)
Step 1: Square both sides. \[ (\sqrt{2x + 3})^2 = (x - 1)^2 \ 2x + 3 = x^2 - 2x + 1 \]
Step 2: Rearrange everything to one side. \[ 0 = x^2 - 4x - 2 \]
Step 3: Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1, b = -4, c = -2 \). \[ x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot -2}}{2 \cdot 1} \ x = \frac{4 \pm \sqrt{16 + 8}}{2} \ x = \frac{4 \pm \sqrt{24}}{2} \ x = \frac{4 \pm 2\sqrt{6}}{2} \ x = 2 \pm \sqrt{6} \]
Answer: \( x = 2 + \sqrt{6} \) or \( x = 2 - \sqrt{6} \)
7. Solve \( \sqrt{7x + 5} = 3x - 1 \)
Step 1: Square both sides. \[ 7x + 5 = (3x - 1)^2 \ 7x + 5 = 9x^2 - 6x + 1 \]
Step 2: Rearrange everything to one side. \[ 0 = 9x^2 - 13x - 4 \]
Step 3: Use the quadratic formula: \[ x = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 9 \cdot (-4)}}{2 \cdot 9} \ x = \frac{13 \pm \sqrt{169 + 144}}{18} \ x = \frac{13 \pm \sqrt{313}}{18} \]
Answer: \( x = \frac{13 \pm \sqrt{313}}{18} \)
8. Solve \( \sqrt{2x + 10} = x + 2 \)
Step 1: Square both sides. \[ 2x + 10 = (x + 2)^2 \ 2x + 10 = x^2 + 4x + 4 \]
Step 2: Rearrange everything to one side. \[ 0 = x^2 + 4x - 2x + 4 - 10 \ 0 = x^2 + 2x - 6 \]
Step 3: Use the quadratic formula: \[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot -6}}{2 \cdot 1} \ x = \frac{-2 \pm \sqrt{4 + 24}}{2} \ x = \frac{-2 \pm \sqrt{28}}{2} \ x = \frac{-2 \pm 2\sqrt{7}}{2} \ x = -1 \pm \sqrt{7} \]
Answer: \( x = -1 + \sqrt{7} \) or \( x = -1 - \sqrt{7} \)
9. Solve \( \sqrt{3x + 6} = x + 4 \)
Step 1: Square both sides. \[ 3x + 6 = (x + 4)^2 \ 3x + 6 = x^2 + 8x + 16 \]
Step 2: Rearrange everything to one side. \[ 0 = x^2 + 8x - 3x + 16 - 6 \ 0 = x^2 + 5x + 10 \]
Step 3: Use the quadratic formula: \[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} \ x = \frac{-5 \pm \sqrt{25 - 40}}{2} \ x = \frac{-5 \pm \sqrt{-15}}{2} \] This means the solutions are complex. \[ x = \frac{-5 \pm i\sqrt{15}}{2} \]
Answer: \( x = \frac{-5 + i\sqrt{15}}{2} \) or \( x = \frac{-5 - i\sqrt{15}}{2} \)
10. Solve \( \sqrt{5 - x} = x + 1 \)
Step 1: Square both sides. \[ 5 - x = (x + 1)^2 \ 5 - x = x^2 + 2x + 1 \]
Step 2: Rearrange everything to one side. \[ 0 = x^2 + 3x - 4 \ 0 = (x + 4)(x - 1) \]
Step 3: Set each factor to zero: \[ x + 4 = 0 \quad \text{or} \quad x - 1 = 0 \ x = -4 \quad \text{or} \quad x = 1 \]
Final Answer: \( x = -4 \) or \( x = 1 \)
Each solution involved a combination of isolating the radical expression, squaring both sides to eliminate the square root, and then solving an algebraic equation that followed. In cases where the quadratic formula was needed, the solutions were derived while ensuring simplification was applied where feasible.