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

1 Answer
Nov 20, 2015

Answer:

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)