How do you find the zeros, real and imaginary, of #y=x^2 +18x +81# using the quadratic formula?

1 Answer
Feb 11, 2017

Answer:

We have only one zero #x=-9#.

Explanation:

Zeros of a quadratic function #y=f(x)=x^2+18x+81# 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)=x^2+18x+81#,

zeros are #(-18+-sqrt(18^2-4xx1xx81))/(2xx1)#

i.e. #(-18+-sqrt(324-324))/2=(-8+-0)/2=-9#

Check - We can write #y=f(x)=x^2+18x+81#

= #x^2+2xx9xx x+9^2# - and using #a^2+2ab+b^2=(a+b)^2#

= #(x+9)^2#

and hence, #y=0# only when #x=-9#

Hence we have only one zero #x=-9#.