Write a quadratic eq. with rel coefficients having the following solutions. a) solutions are 3 and -5?

Question

final answer should be in the form ax^2 + bx + c = 0
CHALLENGE
b) one solution is 6 + 4i

Answer 1

if x = 3, x - 3 = 0
if x = -5, x + 5 = 0

so (x-3)(x+5) = 0
x² + 2x - 15 = 0

if 6 + 4i is a solution, so is 6 - 4i, so
(x - 6 - 4i)(x - 6 + 4i) = 0
x² - 12x + 36 -16i² = 0
x² - 12x + 52 = 0

Answer 2

a) solutions 3 and -5
so
P(x) = (x-3)(x+5)
=x^2 - 3x + 5x -15
=x^2 +2x -15

in the form ax^2 + bx + c
a= 1, b=2, c=-15

b)
conjugate root theorem says
if P(z) has only real coefficients then the solutions (roots) are in conjugate pairs (ie x+yi and x-yi)

fundamental theorem of algebra,
a polynomial of degree n has n solutions over C (provided repeated solutions are counted the appropriate number of times
so
there are 2 solutions
if
6 + 4i is a solution
then
6 - 4i must also be a solution

P(x)=(x-(6+4i))(x-(6-4i))
=x^2 - x(6+4i) - x(6-4i) + (6+4i)(6-4i)
=x^2 - (6x+4xi) - (6x-4xi) + (36-16i^2)
=x^2 - 6x - 4xi - 6x + 4xi (36+16)
=x^2 - 12x + 52

in the form ax^2 + bx + c
a=1 b=-12 c=52
(NOTE: you would usually use z instead of x for complex numbers)

Answer 3

x^2-2x-15=0


x^2+Sx+P=0
where S= sum of 3 and -5
P= product of 3 and -5