What are the zero(s) #1x^2-6x+20=0#?

1 Answer
Mar 20, 2018

There are no zeros for the function specified.


I first attempted to solve this using the quadratic formula:


However, the #4ac# term ends up being much larger than #b^2#, making the term under the radical negative and therefore imaginary.

My next thought was to plot and just check if the graph crosses the x-axis:

graph{x^2-6x+20 [-37.67, 42.33, -6.08, 33.92]}

As you can see, the plot doesn't cross the x-axis, and therefore has no 'zeros'.