To solve the problem, we need to find the value of \( \alpha^2 + \beta^2 \) given the conditions \( \alpha + \beta = 2 \) and the distance between the points \( (1, \alpha) \) and \( (\beta, 1) \) is 3 units. ### Step 1: Understanding the Distance Formula The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) in a Cartesian plane is given by the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] In our case, the points are \( (1, \alpha) \) and \( (\beta, 1) \). Therefore, we can substitute these points into the distance formula: \[ d = \sqrt{(\beta - 1)^2 + (1 - \alpha)^2} \] ### Step 2: Setting Up the Equation We know that the distance \( d \) is 3 units, so we can set up the equation: \[ \sqrt{(\beta - 1)^2 + (1 - \alpha)^2} = 3 \] To eliminate the square root, we square both sides: \[ (\beta - 1)^2 + (1 - \alpha)^2 = 9 \] ### Step 3: Expanding the Equation Now, we expand both squares: \[ (\beta - 1)^2 = \beta^2 - 2\beta + 1 \] \[ (1 - \alpha)^2 = 1 - 2\alpha + \alpha^2 \] Substituting these into our equation gives: \[ \beta^2 - 2\beta + 1 + 1 - 2\alpha + \alpha^2 = 9 \] Combining like terms results in: \[ \beta^2 + \alpha^2 - 2\beta - 2\alpha + 2 = 9 \] Subtracting 2 from both sides: \[ \beta^2 + \alpha^2 - 2\beta - 2\alpha = 7 \] ### Step 4: Using the Sum of \( \alpha \) and \( \beta \) We know from the problem statement that \( \alpha + \beta = 2 \). We can express \( \beta \) in terms of \( \alpha \): \[ \beta = 2 - \alpha \] Now, we substitute \( \beta \) into our equation: \[ (2 - \alpha)^2 + \alpha^2 - 2(2 - \alpha) - 2\alpha = 7 \] Expanding \( (2 - \alpha)^2 \): \[ (2 - \alpha)^2 = 4 - 4\alpha + \alpha^2 \] Substituting this back gives: \[ 4 - 4\alpha + \alpha^2 + \alpha^2 - 4 + 2\alpha - 2\alpha = 7 \] This simplifies to: \[ 2\alpha^2 - 4\alpha = 7 \] Rearranging gives: \[ 2\alpha^2 - 4\alpha - 7 = 0 \] ### Step 5: Solving the Quadratic Equation Now we can use the quadratic formula to solve for \( \alpha \): \[ \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 2 \), \( b = -4 \), and \( c = -7 \): \[ \alpha = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot (-7)}}{2 \cdot 2} \] Calculating the discriminant: \[ \sqrt{16 + 56} = \sqrt{72} = 6\sqrt{2} \] Thus: \[ \alpha = \frac{4 \pm 6\sqrt{2}}{4} = 1 \pm \frac{3\sqrt{2}}{2} \] This gives us two possible values for \( \alpha \): 1. \( \alpha_1 = 1 + \frac{3\sqrt{2}}{2} \) 2. \( \alpha_2 = 1 - \frac{3\sqrt{2}}{2} \) ### Step 6: Finding \( \beta \) Using \( \beta = 2 - \alpha \): 1. If \( \alpha = 1 + \frac{3\sqrt{2}}{2} \), then \( \beta = 2 - (1 + \frac{3\sqrt{2}}{2}) = 1 - \frac{3\sqrt{2}}{2} \) 2. If \( \alpha = 1 - \frac{3\sqrt{2}}{2} \), then \( \beta = 2 - (1 - \frac{3\sqrt{2}}{2}) = 1 + \frac{3\sqrt{2}}{2} \) ### Step 7: Calculating \( \alpha^2 + \beta^2 \) Using the identity: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \] We already know \( \