Find the solutions that satisfy both x^3+x^2-20x<or equal to 0 and x^3-4x^2-11x+30<0

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

 

 

 

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

It is very easy to solve the first inequation, finding out the intervals where the function has negative sign(meaning that it's < or equal to 0).

After that, we have to consider that the solutions for the first inequality, has to be verified in the second one. The roots of the first inequality which, substituted into the second one, do not change the sign of the second one, are the common solutions for both of them.

For the first inequality, we'll write it as a product of 2 factors, so that the negative sign will be obtained from a product of 2 factors with opposite signs.

(x^3)+(x^2)-20x<0 will become

x*(x^2 + x - 20)<0

Pay attention at the calculation of the roots!

It is obvious that we'll obtain 3 roots, because of the fact that the inequality has the third grade:

x*(x^2 + x - 20)=0

x1=0

x^2 + x - 20=0

x2=[-1+sqrt(1-4*1*(-20)]/2=(-1+9)/2=4

x3=[-1+sqrt(1-4*1*(-20)]/2=(-1-9)/2=-5

We know that between the roots, a second grade function, has the sign opposite to the sign of the coefficient of the term with the highest grade (in this case, the coefficient of x^2, which has the value=1), so the sign will be negative, between the values -5 and 4!

With the values -5, 4 and values from this interval, we'll verify if the sign of the second inequality is changing:

For x=4

4^3 - 4*(4^2)-11*4+30=64-64-44+30=-14<0

For x=-5

(-5)^3 - 4*(-5)^2 - 11*(-5)+ 30=-125-100+55+30=-140<0

For values between 0 and 2, the second inequality is positive, or equal to 0 for x=2,which is not true!

So the common sollution are:

All the set of values < or equal to -5

For x=4, the inequality is true, also!

 

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