If \( (4-x) ; 3 \) and \( \sqrt{2 x+5} \) are the first three ter if an anthmetic Serquence a). Determine the value of \( x \).

To determine the value of \( x \) such that \( (4 - x) \), \( 3 \), and \( \sqrt{2x + 5} \) are the first three terms of an arithmetic sequence, follow these steps: ### Step 1: Understand the Arithmetic Sequence Condition In an arithmetic sequence, the difference between consecutive terms is constant. Therefore: \[ \text{Second term} - \text{First term} = \text{Third term} - \text{Second term} \] Substituting the given terms: \[ 3 - (4 - x) = \sqrt{2x + 5} - 3 \] ### Step 2: Simplify the Equation \[ 3 - 4 + x = \sqrt{2x + 5} - 3 \] \[ -1 + x = \sqrt{2x + 5} - 3 \] Add 3 to both sides to isolate the square root: \[ x + 2 = \sqrt{2x + 5} \] ### Step 3: Solve for \( x \) Square both sides to eliminate the square root: \[ (x + 2)^2 = 2x + 5 \] \[ x^2 + 4x + 4 = 2x + 5 \] \[ x^2 + 2x - 1 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2} \] ### Step 4: Verify Potential Solutions 1. **For \( x = -1 + \sqrt{2} \):** - Compute \( \sqrt{2x + 5} = \sqrt{2(-1 + \sqrt{2}) + 5} = \sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2} \) - The sequence becomes \( 5 - \sqrt{2} \), \( 3 \), \( 1 + \sqrt{2} \) - The differences: \( 3 - (5 - \sqrt{2}) = -2 + \sqrt{2} \) and \( (1 + \sqrt{2}) - 3 = -2 + \sqrt{2} \) - The differences are equal, so \( x = -1 + \sqrt{2} \) is valid. 2. **For \( x = -1 - \sqrt{2} \):** - Compute \( \sqrt{2x + 5} = \sqrt{3 - 2\sqrt{2}} = \sqrt{2} - 1 \) - The sequence becomes \( 5 + \sqrt{2} \), \( 3 \), \( \sqrt{2} - 1 \) - The differences: \( 3 - (5 + \sqrt{2}) = -2 - \sqrt{2} \) and \( (\sqrt{2} - 1) - 3 = \sqrt{2} - 4 \) - The differences are not equal, so \( x = -1 - \sqrt{2} \) is extraneous. ### Conclusion The valid solution is: \[ x = \sqrt{2} - 1 \] **Answer:** After simplifying, the valid solution is x = √2 minus one. In symbols, x = √2 − 1