If f(x)=x^2-72 What are the zeros?
x^2 -72 = 0
x^2 = 72
x = +- sqrt (72)
\(f(x) = x^2-72\). y = 0 when x^2 = 72. Square rooting both sides, we find that \(x = \pm\sqrt{72}\) or \(\boxed{x = \pm 6\sqrt{2}}\).
Hope this helps,
- PM
graph:
https://www.desmos.com/calculator/knf4bbbesb
+773 
