How do you solve #2v^2+7v<4# using a sign chart?

1 Answer
Dec 31, 2016

Answer:

The answer is #v in ] -4,1/2 [ #

Explanation:

Let's rewrite the inequality as

#2v^2+7v-4<0#

Let #f(v)=2v^2+7v-4#

The domain of #f(v)# is #D_f(v)=RR#

Let's factorise the expression

#2v^2+7v-4=(2v-1)(v+4)#

Now we establish the sign chart

#color(white)(aaaa)##v##color(white)(aaaaa)##-oo##color(white)(aaaaa)##-4##color(white)(aaaaa)##1/2##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##v+4##color(white)(aaaaaaa)##-##color(white)(aaaaa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##2v-1##color(white)(aaaaaa)##-##color(white)(aaaaa)##-##color(white)(aaaaa)##+#

#color(white)(aaaa)##f(v)##color(white)(aaaaaaaa)##+##color(white)(aaaaa)##-##color(white)(aaaaa)##+#

Therefore,

#f(v)<0# when #v in ] -4,1/2 [ #