How do you find all zeros of #f(x)=1/2x^2+5/2x-3/2#?

1 Answer
Jul 30, 2017

Answer:

#x=-5/2+-1/2sqrt37#

Explanation:

#"take out common factor of "1/2#

#rArrf(x)=1/2(x^2+5x-3)=0#

#"solve "x^2+5x-3" using the "color(blue)"quadratic formula"#

#"with "a=1,b=5,c=-3#

#rArrx=(-5+-sqrt(25+12))/2=(-5+-sqrt37)/2#

#rArrx=-5/2+-1/2sqrt37larrcolor(red)" exact solutions"#

#x~~ -5.54" or " x~~ 0.54" to 2 dec/ places"#
graph{1/2x^2+5/2x-3/2 [-10, 10, -5, 5]}