Question

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

Answer 1

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`.