How do you find the zeros, real and imaginary, of $y=x^2 +36x +81$ using the quadratic formula?

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Answer 1 · 2016-04-20T06:15:35

Zeros are $-18-9sqrt3$ and $-18+9sqrt3$

Zeros of $y=x^2+36x+81$ are given by solution to the equation

$x^2+36x+81=0$

Hence using quadratic formula $x$ is given by $(-b+-sqrt(b^2-4ac))/(2a)$

Hence $x=(-36+-sqrt(36^2-4xx1xx81))/2=(-36+-sqrt(1296-324))/2$

= $(-36+-sqrt(972))/2=(-36+-sqrt(3xx3xx3xx3xx3xx2xx2))/2=(-36+-18sqrt3)/2$

= $-18+-9sqrt3$

Hence zeros are $-18-9sqrt3$ and $-18+9sqrt3$

How do you find the zeros, real and imaginary, of y=x^2 +36x +81y=x^2 +36x +81 using the quadratic formula?