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

Feb 11, 2017

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

Explanation:

Zeros of a quadratic function $y = f \left(x\right) = {x}^{2} + 18 x + 81$ are those values of $x$ for which $y = 0$.

Using quadratic formula for a function $f \left(x\right) = a {x}^{2} + b x + c$,

zeros are $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

and hence for $y = f \left(x\right) = {x}^{2} + 18 x + 81$,

zeros are $\frac{- 18 \pm \sqrt{{18}^{2} - 4 \times 1 \times 81}}{2 \times 1}$

i.e. $\frac{- 18 \pm \sqrt{324 - 324}}{2} = \frac{- 8 \pm 0}{2} = - 9$

Check - We can write $y = f \left(x\right) = {x}^{2} + 18 x + 81$

= ${x}^{2} + 2 \times 9 \times x + {9}^{2}$ - and using ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$

= ${\left(x + 9\right)}^{2}$

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

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