How do you find the zeros, real and imaginary, of #y=x^2-8x-3# using the quadratic formula?

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Answer 1 · 2016-06-09T15:59:09

This function hs #2# real zeros:

#x_1=4-sqrt(19)#

#x_2=4+sqrt(19)#

To calculate the zeros of a quadratic function first you calculate the determinant:

#y=x^2-8x-3#

#Delta = b^2-4ac=8^2-4xx1xx(-3)=64+12=76#

#sqrt(Delta)=sqrt(76)=sqrt(4xx19)=2sqrt(19)#

#Delta >=0#, so the function has #2# real zeros. (If #Delta<0# then the zeros would be complex numbers)

#x_1=(-b-sqrt(Delta))/(2a)#

#x_1=(8-2sqrt(19))/(2)=4-sqrt(19)#

#x_2=(-b+sqrt(Delta))/(2a)#

#x_1=(8+2sqrt(19))/(2)=4+sqrt(19)#