1. The roots of the quadratic equation $z^2 + az + b = 0$ are $-7 + 2i$ and $-7 - 2i$. What is a+b?
2. Find all pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 56$.
For example, to enter the solutions (2,4) and (-3,9), you would enter "(2,4),(-3,9)" (without the quotation marks).
+130511 1. ( z - ( -7+ 2i) ) ( z - ( -7 - 2i) ) =
z^2 - z ( -7 + 2i) - z(-7 -2i) + (-7+2i)(-7 - 2i) =
z^2 + 7z - 2iz + 7z + 2iz + 49 - 4i^2 =
z^2 + 14z + 49 + 4 =
z^2 + 14z + 53
a = 14 b = 53
a + b = 67
2. x + y =10 ⇒ y = 10 - x (1)
x^ 2 + y^2 = 56 (2)
Sub (1) into (2)
x^2 + (10 - x)^2 = 56
x^2 + x^2 - 20x + 100 = 56
2x^2 -20x + 44 = 0 divide through by 2
x^2 -10x + 22 = 0 complete the square on x
x^2 - 10x + 25 = -22 + 25
(x - 5)^2 = 3 take both roots
x - 5 = ± √3
x = ± √3 + 5
When x = √ 3 + 5 y = - √3 + 5
When x = - √3 + 5 y = √3+ 5
