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

1 Answer
Daniel L.
Nov 20, 2015

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)

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