Solve Inequality Problem: x^3 + 3x^2 - 16x - 48

[Sheneron]

Homework Statement

$$x^3 + 3x^2 - 16x - 48 <= 0$$

factored it down to get this.

$$(x+3)(x^2 -16) <= 0$$

$$x + 3 <= 0$$
$$x <= -3$$

$$x^2 - 16 <= 0$$
$$x <= 4$$
$$x <= -4$$

But, that is not right, looking at a graph x needs to be greater than or equal to -3, not less than or equal to. But the other one is right. How come?

 

[Mentallic]

$$x^2 - 16 <= 0$$ can you see this is a difference of 2 squares?
Mainly: $$a^{2}-b^{2}=(a+b)(a-b)$$

 

[Ray 4]

$$x^2 - 16 <= 0$$ can you see this is a difference of 2 squares?
Mainly: $$a^{2}-b^{2}=(a+b)(a-b)$$

thanks but why does the graph show it as x >= -3 when the equation yields x<= - 3

 

[Mark44]

thanks but why does the graph show it as x >= -3 when the equation yields x<= - 3

What you're dealing with isn't an equation; it's an inequality, and they're trickier to work with than equations, since you have in take into account the the two factors (or three, if you factor the x² - 16 expression) can be zero or can be negative.

It's easy to determine where (x + 3)(x² - 16) is equal to zero, but takes a little more work to find where it is less than zero. For the above to be negative, it must be that either
a) x + 3 < 0 and x² - 16 > 0, or that
b) x + 3 > 0 and x² - 16 < 0

What do you get if you follow this reasoning for the two pairs of inequalities?

You can also approach it using the three linear (i.e., first degree in x) factors. To get a negative product, one factor has to be negative and the other two have to be positive. Taking that approach, you'll have three sets of three inequalities, but it might be that one of the sets has no solution.

Hope that helps.
Mark

 

[Mentallic]

for the $$x^{2}\leq 16$$ part of the inequality, this is obviously a quadratic that has roots of $$\pm 4$$. This quadratic is concave up, so $$f(\frac{-b}{2a})<0$$ (the y-value of the vertex). Just picturing the graph of the quadratic, the partial solution will be for the x-values where their y-values < 0. This means that this partial solution is $$-4\leq x \leq 4$$.
Can you now see how you got this part wrong firstly?

So now we have, $$(x+3)(x^{2}-16)\leq 0$$
through further factorisation $$(x+3)(x+4)(x-4)\leq 0$$

So the partial solutions we have here are:
$$x+3\leq 0, x\leq -3$$ (1)
$$x+4\leq 0, x\leq -4$$ (2)
$$x-4\leq 0, x\leq 4$$ (3)

But there is a problem here, there are intersections of all 3 solutions, which ones are correct and which are trashed?
Well it all has to do with how these terms are multiplied. For all values $$-3< x< 4$$ only one of the partial solutions is satisfied, mainly (3). So the inequality is multiplied by the 3 factored terms, one of which is negative [the factor (x-4)] so the inequality is satisfied. Once the x-values become $$-4< x<-3$$, 2 of the factors become negative [the factors (x+3)(x-4)] and when 2 negatives are multiplied together, it becomes positive, so the inequality is not satisfied in this range. Lastly, once $$x<-4$$ all 3 factors are satisfied, and 3 negative terms multiplied together is still negative, so that partial solution is satisfied.
Be sure to include the roots of the function since this is where the function = 0 and this satisfies the inequality.

It is always easier just to visual the graph and understand for which x-values, the y-values are negative.

 

[Sheneron]

What you're dealing with isn't an equation; it's an inequality, and they're trickier to work with than equations, since you have in take into account the the two factors (or three, if you factor the x² - 16 expression) can be zero or can be negative.

It's easy to determine where (x + 3)(x² - 16) is equal to zero, but takes a little more work to find where it is less than zero. For the above to be negative, it must be that either
a) x + 3 < 0 and x² - 16 > 0, or that
b) x + 3 > 0 and x² - 16 < 0

What do you get if you follow this reasoning for the two pairs of inequalities?

You can also approach it using the three linear (i.e., first degree in x) factors. To get a negative product, one factor has to be negative and the other two have to be positive. Taking that approach, you'll have three sets of three inequalities, but it might be that one of the sets has no solution.

Hope that helps.
Mark

So, assuming one is negative and the other is positive, which it must be for that inequality to be less than 0, you end up with 2 inequalities (a and b from your post). So one of those is right, and the other is wrong, and you just have to check to see which is which?

 

[Mark44]

So, assuming one is negative and the other is positive, which it must be for that inequality to be less than 0, you end up with 2 inequalities (a and b from your post). So one of those is right, and the other is wrong, and you just have to check to see which is which?

It's not necessarily so that one will be right and one wrong, although this is possible. You have to check both and see what the implications are for values of x.

 

[HallsofIvy]

One way to do this is just as you say: if (x- 4)(x+ 4)< 0, then x-4 and x+4 must have different signs. Either x-4< 0 and x+ 4> 0 or x-4>0 and x+ 4< 0.

The first pair gives x< 4 and x> -4 so -4< x< 4.

The second pair give x> 4 and x< -4 which is impossible. The only answer is -4< x< 4.

Another way to do it is to solve the related equation (x- 4)(x+ 4)= 0 which gives x= -4 and x= 4. Since (x-4)(x+4) is a continuous function, it can change from "< 0" to "> 0", and vice-versa, only where it is "= 0", i.e. at x= -4 and 4. That means that in each of the three intervals it cannot change sign. If x= -5, -5< -4, (-5-4)(-5+4)= (-9)(-1)= 9> 0 so (x- 4)(x+ 4)> 0 for all x< -4. For x= 0, -4< 0< 4, (0-4)(0+4)= (-4)(4)= -15< 0 so (x-4)(x+4)< 0 for all -4< x< 4. Finally, if x= 5, 5> 4, (5-4)(5+4)= (1)(9)= 9> 0 so (x-4)(x+4)> 0 for all x> 4.