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

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Answer 1 · 2017-02-11T07:17:22

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

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 $x=-9$ .

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