How do you solve the inequality #2x^2+25x+63<=0#?

1 Answer
Feb 1, 2017

Answer:

The answer is #x in [-9, -7/2]#

Explanation:

Let's factorise the LHS

#2x^2+25x+63<=0#

#(2x+7)(x+9)<=0#

Let #f(x)=(2x+7)(x+9)#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-9##color(white)(aaaa)##-7/2##color(white)(aaaa)##-oo#

#color(white)(aaaa)##x+9##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##2x+7##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)<=0#, when #x in [-9, -7/2]#