If \( (1-x) \); 3 and \( \sqrt{2 x+3} \) are the first there trei of an arithruetic sequmee a) Determine the value of \( x \).

To determine the value of \( x \) such that \( (1-x) \), 3, and \( \sqrt{2x+3} \) form an arithmetic sequence, we need to use the property that the difference between consecutive terms is constant. Specifically, we can set up the equation: \[ 3 - (1-x) = \sqrt{2x+3} - 3 \] Now, simplifying the left side, we have: \[ 3 - 1 + x = 2 + x \] So, our equation becomes: \[ 2 + x = \sqrt{2x + 3} - 3 \] Adding 3 to both sides gives us: \[ 5 + x = \sqrt{2x + 3} \] Next, we square both sides to eliminate the square root: \[ (5 + x)^2 = 2x + 3 \] Expanding the left side yields: \[ 25 + 10x + x^2 = 2x + 3 \] Rearranging everything to one side results in: \[ x^2 + 8x + 22 = 0 \] Now we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 1 \), \( b = 8 \), and \( c = 22 \): \[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 1 \cdot 22}}{2 \cdot 1} \] \[ x = \frac{-8 \pm \sqrt{64 - 88}}{2} \] \[ x = \frac{-8 \pm \sqrt{-24}}{2} \] Since we have a negative number under the square root, there are no real solutions for \( x \) in this scenario. Thus, the set of values does not exist in the realm of real numbers.