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 · 2015-11-20T17:56:56

Only one solution $x=9$
As $("Re"+"Im")-> (9+0i)$

Given: $y=x^2-18x+81$

Using std form of $y=ax^2+bx+c$

Where $x=(-b+-sqrt(b^2-4ac))/(2a)$

$a=1$
$b=(-18)$
$c=81$

Thus

$x=(-(-18)+-sqrt((-18)^2-4(1)(81)))/(2(1))$

$x=(18+-sqrt(324-324) )/2$

$x=18/2 =9$

Tony B

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