How do you solve #9x^2+3x-2>=0# using a sign chart?

Question

1383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-13T16:19:20

The answer is #x in ] -oo,-2/3 ] uu [1/3, +oo[ #

Let's factorise the LHS

#9x^2+3x-2=(3x-1)(3x+2)#

Let #f(x)=(3x-1)(3x+2)#

Now we can make the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-2/3##color(white)(aaaaaa)##1/3##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##3x+2##color(white)(aaaaaa)##-##color(white)(aaaaaa)##+##color(white)(aaaaaa)##+#

#color(white)(aaaa)##3x-1##color(white)(aaaaaa)##-##color(white)(aaaaaa)##-##color(white)(aaaaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##+##color(white)(aaaaaa)##-##color(white)(aaaaaa)##+#

Therefore,

#f(x)>=0#, when #x in ] -oo,-2/3 ] uu [1/3, +oo[ #