how is 7-3i + 7+3i = 14/1

we're doing sum and product of roots, and i don't get how this is equal to 14

sorry i needed to add more info for this to make sense

the problem is asking to write the quadratic equation that has those roots.
so in this case i'd use those roots to find -b/a
but i don't get how that correlates to -b/a?

if you want a quadratic that has 7-3i and 7+3i as roots, build it the way you normally do

say the roots were 3 and -2.You would say
y = (x-3)(x+2)

so, here we have

y = (x-(7-3i))(x-(7+3i))

looks messy, but with a little regrouping, and recalling that (a-bi)(a+bi) = a^2+b^2,

y = ((x-7)-3i)((x-7)+3i)
= (x-7)^2 + 9
= x^2 - 14x + 58

To understand why 7 - 3i + 7 + 3i equals 14, we need to simplify the expression.

First, let's combine the real numbers (7 + 7) and the imaginary numbers (-3i + 3i):

7 - 3i + 7 + 3i

= (7 + 7) + (-3i + 3i)

= 14 + 0i

The term (-3i + 3i) evaluates to 0 because the positive and negative imaginary components cancel each other out. Thus, we are left with:

= 14

So, the result is 14, not 14/1 as you mentioned in the question. There was no need to divide by 1 because 14 is already a whole number.