# Solving Inequalities / Check my work?

• Oct 7th 2013, 09:23 AM
tdotodot
Solving Inequalities / Check my work?
Can someone please check my work / make sure I am doing these correctly? Really appreciate the help!

1) x^2 + x -2 > 0

....do I just factor then that what x can not be?
x(x+2)-1(x+2)
(x+2)(X-1)

x=-2 or x=1

2) x(x-2)(x-1)(x+1) ≤ 0

...it's already in factored form so am I just listing what can not be x?
x(x-2)=0 so x=0
(x-2)=0 so x=2
(x-1)=0 so x=1
(x+1)=0 so x=-1

3) x^3 ≥ 9x

x(x+3)(x-3)
≥ 0

x(x+3)=0 so x=0
(x+3)=x so x=-3
(x-3) = 0 so x=3

4) -x^3 + 2x^2 + 4x -8 ≥ 0

-x^2(x-2)+4(x-2)
(x-2)(-x^2+4)
(x-2)-(x-2)(x+2)

(x-2)=o so x=2
-(x-2)=0 so x=-2
(x+2)=0 so x=-2
• Oct 7th 2013, 10:52 AM
mathssolutions
Re: Solving Inequalities / Check my work?
This kind of work is easy to self check. Remember they are inequalities and not equations.

1) x^2 + x -2 > 0

....do I just factor then that what x can not be?
x(x+2)-1(x+2)
(x+2)(X-1)

x=-2 or x=1

In this example x equals neither -2 nor 1 but if you plug in a value of x to the left or right of those values say x = -3 you will be able to tell where the less than and greater than signs go.

(-3)^2 +(-3) -2 = 4> 0 is true so it is safe to say that the region to the left of x=-2 is true for all values and by default (but you can check) region to the right of x=1 will also be true for this equality.
Hence x<-2 and x>1 satisfies this inequality.
You are right to treat them all as equations to get the critical values of x however.
• Oct 7th 2013, 11:11 AM
tdotodot
Re: Solving Inequalities / Check my work?
Quote:

Originally Posted by mathssolutions
This kind of work is easy to self check. Remember they are inequalities and not equations.

1) x^2 + x -2 > 0

....do I just factor then that what x can not be?
x(x+2)-1(x+2)
(x+2)(X-1)

x=-2 or x=1

In this example x equals neither -2 nor 1 but if you plug in a value of x to the left or right of those values say x = -3 you will be able to tell where the less than and greater than signs go.

(-3)^2 +(-3) -2 = 4> 0 is true so it is safe to say that the region to the left of x=-2 is true for all values and by default (but you can check) region to the right of x=1 will also be true for this equality.
Hence x<-2 and x>1 satisfies this inequality.
You are right to treat them all as equations to get the critical values of x however.

So when it says "solve the equalities" what would my solution be?
x^2 + x -2 > 0 ....solution is: x < -2 and x>1? or (x+2)(X-1)?

So then what we are looking for is to see what region of the graph lies between the zero's? We do this by plugging in a value of x, so that we see where the ">/<" signs go? That would be our answer?

For questions 2through4, I want to simplify as much as possible (which I did, right?), then substitute a value into x, then put those into the equality (as in _x_ > __ or _x_ < __)
• Oct 7th 2013, 12:03 PM
SlipEternal
Re: Solving Inequalities / Check my work?
For all of these, solving the associated equality will give you critical numbers. These are the values where the polynomial actually equals 0. For any other values of x, it will not be zero. Since these functions can be drawn without lifting a pencil (in mathematical terms, they are called continuous functions), the graph of the function cannot jump from above the x-axis to below the x-axis without crossing it. Each time it crosses the x-axis, you have a solution to the equation. Since you are finding ALL possible solutions, there is no other value that the polynomial can be zero. So, we put the roots in order. For problem #2, that would be -1,0,1,2. Now, we look at values of x to the left of -1. Then to the right of -1, but to the left of 0. Then to the right of 0, but left of 1. Then to the right of 1, but left of 2. Finally, we look at values of x to the right of 2. Let's consider one of these intervals: $x<-1$. I can choose any value (such as -2), and if the polynomial is positive for that one value, then it is positive for any value where $x<-1$ (similarly for negative). Next, I pick a value between -1 and 0 (such as -0.5). If the polynomial is positive on that one value, it is positive for every value between -1 and 0 (similarly for negative). Then, keep going. Find all intervals that satisfy the given inequality. The solution would be writing out all such intervals.

To see the pattern more clearly, you can put the terms in the same order as the roots associated to them. So, $-1,0,1,2$ are the roots for $(x+1),x,(x-1),(x-2)$. So, for values $x<-1$, we have $(x+1)<0, x<0, x-1<0, x-2<0$ respectively. The product of two negatives is a positive, the product of two positives is a positive, and the product of a positive and negative is a negative. So, we have -*-*-*- = (-*-)*(-*-) = (+)*(+) = +. So, for $x<-1$, $(x+1)x(x-1)(x-2)>0$. Move $x$ to the right on the number line. When it hits -1, $x+1$ becomes zero, but the rest are all still negative. As soon as it goes to the right of -1, but to the left of 0, $x+1$ becomes positive, but the rest are all still negative. So, for $-1, you have +*-*-*- = (+*-)*(-*-) = (-)*(+) = -, and $(x+1)x(x-1)(x-2)<0$. Keep sliding x to the right, and like light switches, the terms go from negative to positive in order.
• Oct 7th 2013, 06:53 PM
Prove It
Re: Solving Inequalities / Check my work?
Quote:

Originally Posted by tdotodot
Can someone please check my work / make sure I am doing these correctly? Really appreciate the help!

1) x^2 + x -2 > 0

....do I just factor then that what x can not be?
x(x+2)-1(x+2)
(x+2)(X-1)

x=-2 or x=1

2) x(x-2)(x-1)(x+1) ≤ 0

...it's already in factored form so am I just listing what can not be x?
x(x-2)=0 so x=0
(x-2)=0 so x=2
(x-1)=0 so x=1
(x+1)=0 so x=-1

3) x^3 ≥ 9x

x(x+3)(x-3)
≥ 0

x(x+3)=0 so x=0
(x+3)=x so x=-3
(x-3) = 0 so x=3

4) -x^3 + 2x^2 + 4x -8 ≥ 0

-x^2(x-2)+4(x-2)
(x-2)(-x^2+4)
(x-2)-(x-2)(x+2)

(x-2)=o so x=2
-(x-2)=0 so x=-2
(x+2)=0 so x=-2

Here's a word of advice, when solving QUADRATIC inequalities like the first, the most direct method of solution is to complete the square...

\displaystyle \begin{align*} x^2 + x - 2 &> 0 \\ x^2 + x &> 2 \\ x^2 + x + \left( \frac{1}{2} \right) ^2 &> 2 + \left( \frac{1}{2} \right) ^2 \\ \left( x + \frac{1}{2}\right) ^2 &> \frac{8}{4} + \frac{1}{4} \\ \left( x + \frac{1}{2} \right) ^2 &> \frac{9}{4} \\ \left| x + \frac{1}{2} \right| &> \frac{3}{2} \\x+ \frac{1}{2} &< -\frac{3}{2} \textrm{ or } x+ \frac{1}{2} > \frac{3}{2} \\x &< -2\textrm{ or } x > 1 \end{align*}

However, this is only possible with quadratic equations, for higher polynomials that aren't quadratics in disguise, you will have to use the more indirect method of finding the x-intercepts and testing each region bounded by them...
• Oct 8th 2013, 07:07 AM
tdotodot
Re: Solving Inequalities / Check my work?
Appreciate the help guys! I think I'm getting it down now!

Can we take a look at example2:

2) x(x-2)(x-1)(x+1) ≤ 0

x(x-2)=0 so x=0
(x-2)=0 so x=2
(x-1)=0 so x=1
(x+1)=0 so x=-1

using a number line, I put x+1 (x=-1) and x-1 (x=1) on the number lines.

-----------(-1)--------------------(+1)-------------

Tested with -1 in equation and it did result in a number less then or equal to zero.
Tested with (+2) in equation and it did NOT result in a number less then or equal to zero. SO I tried with +1.
Tested with +1 in equation and it did result in a number less then or equal to zero.

Then I chose a number to represent each section of the number line:
before -1 to -1............value did not equal less than or equal to 0 (did not work)
from -1 to 1............... value equaled 0..... which is less than or equal to zero. (DID work).
from 1 after 1............value did not equal less than or equal to 0 (did not work)

Now that I have done this. How would I write out my solution?

would it be?: (-1, infiniti)u(infiniti,1)

Also for 3:
3) x^3 ≥ 9x

x(x+3)(x-3)
≥ 0

x(x+3)=0 so x=0
(x+3)=x so x=-3
(x-3) = 0 so x=3
---------(-3)------------(3)---------
Tested -3 in equation, did equal zero. (works)
Tested 3 in equation, did equal zero. (works)

Picked a # less than -3. Between -3 and 3. After 3. Tested each.
-10 ..... -910 > 0 (DID NOT WORK)
0 ..... 0 > 0 (Works..it's equal)
10 ..... 910 > 0 (Works, larger than zero)

Now what? How do I write out my solution, I am seeming to have trouble figuring this out.

Need to have these answered within the next hour and a half!

Finally #4.
4) -x^3 + 2x^2 + 4x -8 ≥ 0

-x^2(x-2)+4(x-2)
(x-2)(-x^2+4)
(x-2)-(x-2)(x+2)

(x-2)=o so x=2
-(x-2)=0 so x=-2
(x+2)=0 so x=-2

---------(-2)----------(2)----------

-2 into equation, equals 0. 0=0 (works)
2 into equation, equals 0. 0=0 (works)

Choose a #, before -2, between -2 and 2, after 2.

-10, =108. 108>0. Works.
0, =2 . 2>0. Works.
10, = -88. -88NOTGREATERThan0. DoesNOTwork.

From here, I don't know what to do.
• Oct 8th 2013, 07:53 AM
tdotodot
Re: Solving Inequalities / Check my work?
• Oct 8th 2013, 07:58 AM
tdotodot
Re: Solving Inequalities / Check my work?

a) (∞,-2)u(1,∞)

b) (-1,∞)u(-∞,0)

c) (-3,∞)u(-∞,0)

d) (-∞,2)
• Oct 8th 2013, 09:58 AM
SlipEternal
Re: Solving Inequalities / Check my work?
For question 1, yes.
For question 2, you found four roots. Why do you only have two numbers on your number line?
$-\infty \rule[0.5ex]{2em}{1pt} -\! 1 \rule[0.5ex]{2em}{1pt} 0 \rule[0.5ex]{2em}{1pt} 1 \rule[0.5ex]{2em}{1pt} 2 \rule[0.5ex]{2em}{1pt} \infty$
Now plug in -2 (-*-*-*- = +)
Plug in -1 (0*-*-*- = 0)
Plug in -0.5 (+*-*-*- = -)
Plug in 0 (+*0*-*- = 0)
Plug in 0.5 (+*+*-*- = +)
Plug in 1 (+*+*0*- = 0)
Plug in 1.5 (+*+*+*- = -)
Plug in 2 (+*+*+*0 = 0)
Plug in 3 (+*+*+*+ = +)

Perhaps making a table would help.
 img.top {vertical-align:15%;} $x+1$ img.top {vertical-align:15%;} $x$ img.top {vertical-align:15%;} $x-1$ img.top {vertical-align:15%;} $x-2$ img.top {vertical-align:15%;} $(x+1)x(x-1)(x-2)$ img.top {vertical-align:15%;} $(x+1)x(x-1)(x-2)\le 0$ img.top {vertical-align:15%;} $x=-2$ -1 -2 -3 -4 24 FALSE img.top {vertical-align:15%;} $x=-1$ 0 -1 -2 -3 0 TRUE img.top {vertical-align:15%;} $x=-\tfrac{1}{2}$ img.top {vertical-align:15%;} $\tfrac{1}{2}$ img.top {vertical-align:15%;} $-\tfrac{1}{2}$ img.top {vertical-align:15%;} $-\tfrac{3}{2}$ img.top {vertical-align:15%;} $-\tfrac{5}{2}$ img.top {vertical-align:15%;} $-\tfrac{15}{8}$ TRUE img.top {vertical-align:15%;} $x=0$ 1 0 -1 -2 0 TRUE img.top {vertical-align:15%;} $x=\tfrac{1}{2}$ img.top {vertical-align:15%;} $\tfrac{3}{2}$ img.top {vertical-align:15%;} $\tfrac{1}{2}$ img.top {vertical-align:15%;} $-\tfrac{1}{2}$ img.top {vertical-align:15%;} $-\tfrac{3}{2}$ img.top {vertical-align:15%;} $\tfrac{9}{8}$ FALSE img.top {vertical-align:15%;} $x=1$ 2 1 0 -1 0 TRUE img.top {vertical-align:15%;} $x=\tfrac{3}{2}$ img.top {vertical-align:15%;} $\tfrac{5}{2}$ img.top {vertical-align:15%;} $\tfrac{3}{2}$ img.top {vertical-align:15%;} $\tfrac{1}{2}$ img.top {vertical-align:15%;} $-\tfrac{1}{2}$ img.top {vertical-align:15%;} $-\tfrac{15}{8}$ TRUE img.top {vertical-align:15%;} $x=2$ 3 2 1 0 0 TRUE img.top {vertical-align:15%;} $x=3$ 4 3 2 1 24 FALSE

Every value for $-1\le x \le 0$ satisfies the inequality. Every value $1\le x \le 2$ also satisfies it. So the intervals should be:
$[-1,0] \cup [1,2]$ (these are closed intervals, so they include their endpoints)

For problem 3, $x^3 \ge 9x$. Let's make another table:
 img.top {vertical-align:15%;} $x^3$ img.top {vertical-align:15%;} $9x$ img.top {vertical-align:15%;} $x^3\ge 9x$ img.top {vertical-align:15%;} $x=-10$ -1000 -90 FALSE img.top {vertical-align:15%;} $x=-3$ -27 -27 TRUE img.top {vertical-align:15%;} $x=-1$ -1 -9 TRUE img.top {vertical-align:15%;} $x=0$ 0 0 TRUE img.top {vertical-align:15%;} $x=1$ 1 9 FALSE img.top {vertical-align:15%;} $x=3$ 27 27 TRUE img.top {vertical-align:15%;} $x=10$ 1000 90 TRUE

So, the intervals are $x \in \left([-3,0] \cup[3,\infty) \right)$ or $(-3 \le x \le 0 \mbox{ or } 3 \le x)$

For problem 4, you are close enough, so I will help you finish it: $(x-2)(-1)(x^2-4) = -(x-2)^2(x+2) \ge 0$. Add $(x-2)^2(x+2)$ to both sides to get $0\ge (x-2)^2(x+2)$. Since $(x-2)^2 \ge 0$ for all values of $x$, this inequality is satisfied when $(x-2)^2=0$ and when $x+2\le 0$. So, the solution would be $x \in (-\infty,-2]\cup \{2\}$. Or you could also write: $(x\le -2\mbox{ or } x = 2)$. These say the same thing, just different notation.