How do you solve #2u^2+5u>=12# using a sign chart?

1 Answer
Nov 28, 2017

Answer:

The solution is #u in (-oo,-4] uu [3/2, +oo)#

Explanation:

Lets 's factorise the function

#f(u)=2u^2+5u-12=(2u-3)(u+4)#

The equation we have to solve is

#f(u)=(2u-3)(u+4)>=0#

Let's build the sign chart

#color(white)(aaaa)##u##color(white)(aaaa)##-oo##color(white)(aaaa)##-4##color(white)(aaaaaaaaaa)##3/2##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##u+4##color(white)(aaaaaa)##-##color(white)(aa)##0##color(white)(aaaa)##+##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##2u-3##color(white)(aaaaa)##-##color(white)(aa)####color(white)(aaaaa)##-##color(white)(aaaa)##0##color(white)(aaa)##+#

#color(white)(aaaa)##f(u)##color(white)(aaaaaaa)##+##color(white)(aa)##0##color(white)(aaaa)##-##color(white)(aaaa)##0##color(white)(aaa)##+#

Therefore,

#f(u)>=0# when #u in (-oo,-4] uu [3/2, +oo)#

graph{2x^2+5x-12 [-29.27, 28.45, -13.66, 15.22]}