How do you solve and graph #x^3<=4x^2+3x#?

1 Answer
Jun 25, 2017

Answer:

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

Explanation:

We solve this inequality with a sign chart.

#x^3<=4x^2+3x#

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

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

We need the roots of the quadratic equation

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

The discrimenant is

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

As, #Delta>0#, there are 2 real roots

#x_2=(-b+sqrtDelta)/2=1/2(4+sqrt28)=2+sqrt7=4.646#

#x_1=(-b-sqrtDelta)/2=1/2(4+sqrt28)=2-sqrt7=-0.646#

Let #f(x)=x(x-x_1)(x-x_2)#

We can build the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##x_1##color(white)(aaaa)##0##color(white)(aaaaaa)##x_2##color(white)(aaaa)##+oo#

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

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

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

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

Therefore,

#f(x)<=0# when #x in (-oo, 2-sqrt7] uu [0, 2+sqrt7]#
graph{x^3-4x^2-3x [-12.34, 12.97, -9.01, 3.65]}