{x^2+ y^2 =65;{xy=8
You have a system of equations with the following:
- \( x^2 + y^2 = 65 \)
- \( xy = 8 \)
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 \(s\):
- \( s = 9 \) (which means \( x + y = 9 \))
- \( s = -9 \) (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:
- \( (x, y) = (8, 1) \) or \( (1, 8) \)
- \( (x, y) = (-1, -8) \) or \( (-8, -1) \)
Feel free to ask if you have further questions!
