Question

How do you find the roots, real and imaginary, of `y= (x + 1) (x - 4)` using the quadratic formula?

Answer 1

Roots are 4 and -1.

Explanation:

The roots are those numbers that assigned to the `x` give as result zero. Because in this equation you have the multiplication of two quantities `(x+1)` and `(x-4)` it is enough that one is zero to have the roots.

`(x+1)` is equal to zero when `x=-1` and `(x-4)` is equal to zero when `x=4`. So your roots are `-1, 4`.

But you ask about to apply the quadratic formula, To do that you have to multyply the two terms together:

`(x+1)(x-4)=x^2-4x+x-4=x^2-3x-4`.

Now we have the equation in the format

`ax^2+bx+c` where `a=1`, `b=-3` and `c=-4`.

The formula is: `x=(-b\pm sqrt(b^2-4ac))/(2a)` where you can substitute your numbers
`x=(3 \pm sqrt((-3)^2-4*1*(-4)))/(2*1)=(3\pm sqrt(9+16))/2=(3\pm 5)/2`.
The solutin with the `+` gives `4`, the one with the `-` gives `-1` as we expected from the initial consideration.