Question

How do you find the zeros, real and imaginary, of `y=-x^2-2x+11` using the quadratic formula?

Answer 1

This equation has 2 real (irrational) solutions.

Explanation:

To find the zeros of a quadratic equation you use a formula:

`x_{1,2}=(-b+-sqrt(Delta))/(2a)`, where

`Delta=b^2-4ac`

This zeroes are:

• real and different `iff Delta>0`
• real and equal `iff Delta=0`
• conjugate complex numbers `iff Delta<0`

To calculate `x_{1,2}` in the last case you have to remember, that if `a<0` then `sqrt(a)=sqrt(abs(a))*i`, where `i` is an imaginary unit.

If we use this rule for our function we get:

`Delta=4-4*(-1)*11`

`Delta=4+44=48`

In this place we can say that the roots of the equation are real (and irrational) because `48>0` (and `sqrt(48)` is not a natural number).

Now we are looking for `x_1` and `x_2`

`x_{1}=(-b+sqrt(Delta))/(2a)=(2-4sqrt(3))/(-2)=-1+2sqrt(3)`

`x_{2}=(-b-sqrt(Delta))/(2a)=(2+4sqrt(3))/(-2)=-1-2sqrt(3)`

Note that we can say that this function has complex zeros (because all real numbers are also complex `RR subset CC`), but we cannot say that the zeros are imaginary (because an imaginary number is a complex number with real part equal to zero according to http://mathworld.wolfram.com/ImaginaryNumber.html)