Two friend are \( x \) and \( y \) years old. The sum of their age is 36 . In five yeas time the Product of their age will be 528 . Fund \( x \) and

1. We are given that the sum of their ages is \[ x + y = 36. \] 2. In five years, their ages will be \( x + 5 \) and \( y + 5 \). The product of their ages in five years is given by \[ (x+5)(y+5) = 528. \] 3. Expanding the product, we have \[ (x+5)(y+5) = xy + 5x + 5y + 25. \] 4. From step 1, since \( x + y = 36 \), we substitute to obtain \[ xy + 5(36) + 25 = xy + 180 + 25 = xy + 205. \] Setting the expression equal to 528, we write \[ xy + 205 = 528. \] 5. Solving for \( xy \), we subtract 205 from both sides: \[ xy = 528 - 205 = 323. \] 6. We now have a system with the sum and product of \( x \) and \( y \): \[ x + y = 36 \quad \text{and} \quad xy = 323. \] The corresponding quadratic equation whose roots are \( x \) and \( y \) is \[ t^2 - (x+y)t + xy = 0, \] or \[ t^2 - 36t + 323 = 0. \] 7. We solve the quadratic equation using the quadratic formula \[ t = \frac{36 \pm \sqrt{36^2 - 4 \cdot 323}}{2}. \] 8. First, compute the discriminant: \[ 36^2 - 4 \cdot 323 = 1296 - 1292 = 4. \] 9. Since the discriminant is positive, the roots are \[ t = \frac{36 \pm \sqrt{4}}{2} = \frac{36 \pm 2}{2}. \] 10. Calculating both solutions: - When using the positive square root: \[ t = \frac{36 + 2}{2} = \frac{38}{2} = 19. \] - When using the negative square root: \[ t = \frac{36 - 2}{2} = \frac{34}{2} = 17. \] 11. Thus, the two friends are \( 17 \) and \( 19 \) years old. So, the solution for \( x \) is either \( 17 \) or \( 19 \), and accordingly, \( y \) is \( 19 \) or \( 17 \).