Question

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

Answer 1

`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]}