1. **Problem 1: Solve the system of equations:** $$x^2 + y^2 = 25$$ $$xy = 12$$ 2. Use substitution or elimination. Here, we use the identity for $(x+y)^2$: $$ (x+y)^2 = x^2 + 2xy + y^2 $$ 3. Substitute known values: $$ (x+y)^2 = 25 + 2(12) = 25 + 24 = 49 $$ 4. So, $$ x + y = \pm 7 $$ 5. Also, from the product: $$ xy = 12 $$ 6. The system can be seen as roots of the quadratic equation: $$ t^2 - (x+y)t + xy = 0 $$ 7. Substitute $x+y$ and $xy$: $$ t^2 - 7t + 12 = 0 $$ 8. Factor or use quadratic formula: $$ (t-3)(t-4) = 0 $$ 9. So, $$ t = 3 \text{ or } t = 4 $$ 10. If $x+y=7$, then $x$ and $y$ are 3 and 4 in some order. 11. If $x+y=-7$, then the quadratic is: $$ t^2 + 7t + 12 = 0 $$ 12. Factor: $$ (t+3)(t+4) = 0 $$ 13. So, $$ t = -3 \text{ or } t = -4 $$ 14. Thus, the solutions for $(x,y)$ are: $$(3,4), (4,3), (-3,-4), (-4,-3)$$ --- 15. **Problem 2: Find all real roots of the quartic equation:** $$ x^4 - 4x^3 + 6x^2 - 4x + 1 = 0 $$ 16. Recognize this as a binomial expansion: $$ (x-1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 $$ 17. So, $$ (x-1)^4 = 0 $$ 18. The only real root is: $$ x = 1 $$ **Final answers:** - System solutions: $(3,4), (4,3), (-3,-4), (-4,-3)$ - Quartic root: $x=1$