Find all pairs $(x,y)$ of real numbers $(x,y)$ such that $x + y = 10$ and $x^2 + y^2 = 56$.
The roots of the quadratic equation $z^2 + az + b = 0$ are $-7 + 2i$ and $-7 - 2i$. What is $a+b$?
1) x + y = 10 ---> y = 10 - x
x2 + y2 = 56 ---> x2 + (10 - x)2 = 56 ---> x2 + 100 - 20x + x2 = 56
2x2 - 20x - 44 = 0
x2 - 10x - 22 = 0
Using the quadratic formula: x = [ 10 +/- sqrt(100 - 88) ] / 2 ---> x = [ 10 +/- sqrt(12) ] / 2
---> x = [ 10 +/- 2sqrt(3) ] / 2 ---> x = 5 +/- sqrt(3)
When x = 5 + sqrt(3), y = 5 - sqrt(3)
and if x = 5 - sqrt(3), y = 5 + sqrt(3)
2) If -7 + 2i is an answer ---> z - (-7 + 2i) is a factor ---> z + 7 - 2i is a factor
If - 7 - 2i is an answer ---> z - (-7 - 2i) is a factor ---> z + 7 + 2i is a factor
Therefore, the equation is: ( z + 7 - 2i ) · (z + 7 + 2i ) = 0
Simplify: z2 + 14z + 53 = 0
---> a= 14 and b = 53