{x^2+ y^2 =65;{xy=8

You have a system of equations with the following: 1. \( x^2 + y^2 = 65 \) 2. \( xy = 8 \) To solve this system, we can use the identities related to the sums and products of the roots. We can let \( s = x + y \) and \( p = xy \). From the equations, we know \( p = 8 \). We also have the identity relating the sum of squares to the sum and product of the roots: \[ x^2 + y^2 = (x + y)^2 - 2xy \] Substituting from what we know: \[ 65 = s^2 - 2 \cdot 8 \] \[ 65 = s^2 - 16 \] \[ s^2 = 65 + 16 \] \[ s^2 = 81 \] \[ s = 9 \quad \text{or} \quad s = -9 \] Now, we have two cases for \(s\): 1. \( s = 9 \) (which means \( x + y = 9 \)) 2. \( s = -9 \) (which means \( x + y = -9 \)) Next, we can use the quadratic formula to find \(x\) and \(y\). The corresponding quadratic equation for both cases is given by: \[ t^2 - st + p = 0 \] ### Case 1: \( s = 9 \) The equation becomes: \[ t^2 - 9t + 8 = 0 \] Calculating the discriminant: \[ D = (-9)^2 - 4 \cdot 1 \cdot 8 = 81 - 32 = 49 \] Finding the roots: \[ t = \frac{9 \pm \sqrt{49}}{2} = \frac{9 \pm 7}{2} \] This gives us: \[ t = \frac{16}{2} = 8 \quad \text{or} \quad t = \frac{2}{2} = 1 \] Thus, we have \( x = 8 \) and \( y = 1 \) (or vice versa). ### Case 2: \( s = -9 \) The equation becomes: \[ t^2 + 9t + 8 = 0 \] Calculating the discriminant: \[ D = 9^2 - 4 \cdot 1 \cdot 8 = 81 - 32 = 49 \] Finding the roots: \[ t = \frac{-9 \pm \sqrt{49}}{2} = \frac{-9 \pm 7}{2} \] This gives us: \[ t = \frac{-2}{2} = -1 \quad \text{or} \quad t = \frac{-16}{2} = -8 \] Thus, we have \( x = -1 \) and \( y = -8 \) (or vice versa). ### Solutions The solutions to your system of equations are: 1. \( (x, y) = (8, 1) \) or \( (1, 8) \) 2. \( (x, y) = (-1, -8) \) or \( (-8, -1) \) Feel free to ask if you have further questions!