Question

How do you find the zeros, real and imaginary, of `y=- x^2-22x+6` using the quadratic formula?

Answer 1

Zeros are `-11+sqrt127` and `-11-sqrt127`

Explanation:

The roots of general form of equation `ax^2+bx+c=0` are zeros of the general form of equation `y=ax^2+bx+c`.

In the given equation `y=-x^2-22x+6`, we see that `a=-1`, `b=-22` and `c=6`.

As discriminant `b^2-4ac=(-22)^2-4(-1)(6)=484+24=508`

As discriminant is positive but not a complete square, we have real but irrational roots.

Hence, using quadratic formula `(-b+-sqrt(b^2-4ac))/(2a)`,

the zeros are the given equation are `x=(-(-22)+-sqrt508)/(2*(-1))`

or `x=-(22+-sqrt508)/2=-(22+-2sqrt127)/2` i.e.

zeros are `-11+sqrt127` and `-11-sqrt127`