How do you solve #x^3<=4x^2+3x# using a sign chart?

1 Answer
Feb 9, 2017

Answer:

The answer is #x in ]-oo,2-sqrt7]uu [0,2+sqrt7]#

Explanation:

Let's rewrite and factorise the inequality

#x^3-4x^2-3x<=0#

#x(x^2-4x-3))<=0#

We calculate the roots of the equation #x^2-4x-3=0#

#Delta=(-4)^2-4*1*(-3)=16+12=28#

#x_1=(4+sqrt28)/2=(4+2sqrt7)/2=2+sqrt7#

#x_2=(4-sqrt28)/2=(4-2sqrt7)/2=2-sqrt7#

Let #f(x)=x(x-(2+sqrt7))(x-(2-sqrt7))#

Now, we build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaaa)##-oo##color(white)(aaaa)##2-sqrt7##color(white)(aaaa)##0##color(white)(aaaa)##2+sqrt7##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-x_2##color(white)(aaaaaaa)##-##color(white)(aaaaaaa)##+##color(white)(aaaa)##+##color(white)(aaaaaaa)##+#

#color(white)(aaaa)##x-x_1##color(white)(aaaaaaa)##-##color(white)(aaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaaaaa)##+#

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

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

Therefore,

#f(x)<=0# when #x in ]-oo,2-sqrt7]uu [0,2+sqrt7]#