Question

The roots of the quadratic equation $z^2 + az + b = 0$ are $2 - 3i$ and $2 + 3i$. What is $a+b$?

Answer 1

We know for our problem that the zeroes of our quadratic equation are
`(2-3i)` and
`(2+3i)`, which means that the solutions for our equation are
`x=2-3i` and
`x=2+3i`. We are going to use those solutions to express our quadratic equation in the form
`a x^(2) +bx+c`; to do that we will use the zero factor property in reverse:

`x=2-3i`

`x-2=-3i`

`x-2+3i=0`


`x=2+3i`

`x-2=3i`

`x-2-3i=0`

Now, we can multiply the left sides of our equations:

`(x-2+3i)(x-2-3i)= x^(2) -2x-3ix-2x+4+6i+3ix-6i-3^2i^2`
=
`x^(2)-4x+4-9i^(2)`
=
`x^(2) -4x+4+9`
=
`x^(2) -4x+13`
Now that we have our quadratic in the form
`a x^(2) +bx+c`, we can infer that
`a=1` and
`b=-4`; therefore, we can conclude that
`a+b=1+(-4)=1-4=-3`.