How do you find the zeros, real and imaginary, of #y= -9x^2-2x-3# using the quadratic formula?

1 Answer
Feb 11, 2017

Answer:

Zeros are #1/9-isqrt26/9# and #1/9+isqrt26/9#, two complex conjugate numbers

Explanation:

Zeros of a quadratic function #y=f(x)=-9x^2-2x-3# are those values of #x# for which #y=0#.

Using quadratic formula for a function #f(x)=ax^2+bx+c#,

zeros are #(-b+-sqrt(b^2-4ac))/(2a)#

and hence for #y=f(x)=-9x^2-2x-3#,

zeros are #(-(-2)+-sqrt((-2)^2-4xx(-9)xx(-3)))/(2xx(-9))#

i.e. #(2+-sqrt(4-108))/(-18)=(2+-sqrt(-104))/18#

= #(2+-sqrt(2^2xx26xxi^2))/18# - as #i^2=-1#

= #(2+-i2sqrt26)/18=1/9+-isqrt26/9#

Hence zeros are #1/9-isqrt26/9# and #1/9+isqrt26/9#, two complex conjugate numbers