Find the solutions that satisfy both x^3+x^2-20x

x^3+x^2-20x<or equal to 0 and x^3-4x^2-11x+30<0

Solution:

Consider, (x^3+x^2-20x) and (x^3-4x^2-11x+30)

Therefore both function values needs to be negative but the 1st includes "an equal to zero condition". We consider below these two functions for their signs by factorising them:

x^3+x^2-20x = x(x^2+x-20)= x(x+5)(x-5).

x value      :           sign of (x+5)x(x-5)

above5      :  positve

At x=5       : 0

0<=x<=5     :   0 or negative...............................(1)

x=0           : 0

-5<x<=0      : postive

x= -5          : 0

x<=-5           :  0 or negative..............................(2)

The second function, x^3-4x^2-11x+30, has roots x=2 as it satisfies  the function and similarly, x=5 ans x=-3 also satisfies it. Therefore, x^3-4x^2-11x+30 = (x+3)(x-2)(x-5)

Now let us examine for the sign of the funtion:

Value        : Sign of x^2-4x^2-11x+30=(x+3)(x-2)(x-5)

x>5           :All factors +ve . So, function is +ve

x=5           :  one faxtor is zero. So, function is zero

2<x<5       : (+ve)(+ve)-ve). Function is -ve............(3)

x=2           : one factor zero. So, function is zero.

-3<x<2      : (+ve)(-ve)(-ve). So, functiona value is +ve.

x=-3         : one factor is zero. So, function  is zero.

x<-3          : All 3 factors -ve.So, the product is -ve.........(4)

From  siuations at (1) or (2) and (3) or (4), we could have:

{ 0<=x<=5 or x<-5 } AND {2<x<5 or x <-3} from which we t the valid interval for to satisfy the given conditions.

2<x<5  OR x < 5 .

Note function value equals zero and second function value strictly less than zero are exluseve situation and can't happens simultaneously. or

In the conditions, x^3+x^2-20x < or equal to 0 and x^3+x^2-20x < 0 , we have to remove  the condition x^3+x^2-20x = 0 and retain only, x^3+x^2-20x < 0  and x^3+x^2-20x < 0