Question

How do you find the roots, real and imaginary, of `y=(x-5)(x-2)` using the quadratic formula?

Answer 1

`x_1=5+0i`

`x_2=2+0i`

Explanation:

`y=(x-5)(x-2)=x^2-7x+10`

Let `y=0`

`x^2-7x+10=0`

Using Descartes' rule for `f(x)`, we know that there are two positive roots. Using it for `f(-x)`, we know that there can be zero or two negative roots.

`Delta=(-7)^2-4(10)(1)=49-40=9`

`x=(7+-sqrt(9))/2=(7+-3)/2=5, 2`

There are no negative roots and a quadratic can only have, at most, two real roots.

So the real values of `x` are `5` and `2`. These are also the complex roots are we can write them as `5+0i` and `2+0i`.