[SPLIT, MOVED] solving x(2x + 7) >= 0

[abel muroi]

I'm starting to understand now. So what if I get another problem in the form x(2x + 7) >= 0

do i have to multiply the x with the 2x (which equals 2x²) and the 7 (which equals 7x) and then factor from there?

 

[Ishuda]

I'm starting to understand now. So what if I get another problem in the form x(2x + 7) >= 0

do i have to multiply the x with the 2x (which equals 2x²) and then factor from there?

No. You already have the two terms, factor x and factor 2x+7. Both must be non-positive or both must be non-negative.

 

[Steven G]

I'm starting to understand now. So what if I get another problem in the form x(2x + 7) >= 0

do i have to multiply the x with the 2x (which equals 2x²) and then factor from there?

It's already is factored. You could factor out the 2 if you like but that is not necessary.

Since this product is >=0 then both factors are both positive or both factors are both negative. We have two cases to consider.

case 1: x>=0 AND 2x+7>=0

OR

Case 2: x<= 0 AND 2x + 7 < 0

You need to solve both case 1 and case 2 and then combine them via OR.

So please tell us 1)the sol to case 1, 2)the sol to case 2 and 3) the final answer.

 

[abel muroi]

It already is factored. You could factor out the 2 if you like but that is not necessary.

Since this product is >=0 then both factors are both positive or both factors are both negative. We have two cases to consider.

case 1: x>=0 AND 2x+7>=0

OR

Case 2: x<= 0 AND 2x + 7 < 0

You need to solve both case 1 and case 2 and then combine them via OR.

So please tell us 1)the sol to case 1, 2)the sol to case 2 and 3) the final answer.

ok so first i have to solve 2x + 7 = 0 and i got x = -7/2 which is -3 1/2 on the number line and the other value for x is 0

am i right SO FAR?

 

[Steven G]

ok so first i have to solve 2x + 7 = 0 and i got x = -7/2 which is -3 1/2 on the number line and the other value for x is 0

am i right SO FAR?

Yes, what you wrote is correct but why are you doing this?

1st solve x>=0 AND 2x+7>=0.

 

[abel muroi]

Yes, what you wrote is correct but why are you doing this?

1st solve x>=0 AND 2x+7>=0.

I thought i had to find two values for X so that I can put them on a number line and then i can test out the numbers. That's why i tried to solve 2x+ 7 >= 0 so that i can get a value

this is what i did, i put the values on a number line and i tested only one number for each side. so that i can see which numbers to include in my solution

I thought that was how you were supposed to do it.

anyway to see the method that i used just watch this video https://www.youtube.com/watch?v=uc_SciNQ2lo

 

[stapel]

Note: For some reason, the forum script is choking on my reply, so I'm splitting it into pieces.

ok so first i have to solve 2x + 7 = 0 and i got x = -7/2 which is -3 1/2 on the number line and the other value for x is 0

am i right SO FAR?

If this were only an "equals" statement, then yes (other than in converting to mixed-number format, which is not appropriate for algebraic computations). But, since this is an inequality, no. Please review (and study until you understand) the replies, explanations, and worked solutions you received in your other thread on this topic.

First, it appears that you are not familiar with what factoring is, since you did not know that "x(2x + 7)" was factored form. It would be a good idea to review that topic (start here and then go here).

Then, as was explained earlier, when you have two factors which result in a signed product, you have to consider the signs of the factors. In this case, you have the two factors "x" and "2x + 7", and you are given that their product is non-negative. For two numbers to multiply to a positive value, what must be true of the signs of the two numbers? Are they opposite signs? Or the same sign? (You can review multiplying with negative numbers here.)

 

[stapel]

As was illustrated in your other thread, two numbers multiply to a positive value when the two numbers have the same sign. You are given that these two factors multiply to (zero or) a positive number. This means that the two numbers have the same sign. This means that they are both positive or else they are both negative. In other words, the two factors (which are "x" and "2x + 7") are either both greater than zero or else both less than zero (or else one or the other is equal to zero, but that's for later). This means that you have two cases, like was shown earlier:

1) x > 0 and 2x + 7 > 0

2) x < 0 and 2x + 7 < 0

To find the solution, if any, you need to solve these sets of inequalities. (To learn how to solve linear inequalities, try here.)

Please reply with all of your steps to this point. Then we can see where to go from there. Thank you!

 

[HallsofIvy]

Yes, what you wrote is correct but why are you doing this?

1st solve x>=0 AND 2x+7>=0.

A continuous function can only change signs where it is equal to 0,
Solving the associated equation tells you where it is 0 so gives intervals where it is "> 0" or "< 0".

Solving for x= 0 and x= -7/2 gives three such intervals- x< -7/2, -7/2< x< 0, and 0< x.
You could then argue that x= x- 0 is negative for x< 0 and positive for x> 0 while 2x+ 7 is negative for x< -7/2 and
positive for x> -7/2. The product is then positive for x< -7/2, where both factors are negative, negative for -7/2< x< 0, where one factor is negative and the other is positive, and positive for x> 0 where both factors are positive.

Alternatively, check one value of x in each interval: -4< -7/2, and x(2x+7)= -4(-8+ 7)= 4> 0. Since x(7x+ 2) is not 0 for any x< 7/2, it is positive for all x< -7/2. -7/2< -1< 0, and x(2x+ 7)= (-1)(-2+ 7)= -5< 0. Since x(7x+ 2) is not negative between -7/2 and 0, so x(7x+ 2)< 0 for all x between -7/2 and 0. Finally, 1> 0 and x(7x+ 2)= 9> 0 so x(7x+2)> 0 for all x> 0.

 

[abel muroi]

Sorry I am still having some difficulty understanding any of this.

Maybe I'll understand more if you can walk me through the steps of the problem

x(2x + 7) >= 0

I thought the first step was to find two values from this inequality so that i can put them on the number line. Then the second step is to test one number from each interval.

what's next?

 

[Steven G]

I thought i had to find two values for X so that I can put them on a number line and then i can test out the numbers. That's why i tried to solve 2x+ 7 >= 0 so that i can get a value

this is what i did, i put the values on a number line and i tested only one number for each side. so that i can see which numbers to include in my solution

I thought that was how you were supposed to do it. Then why did I not give you those instructions?

anyway to see the method that i used just watch this video https://www.youtube.com/watch?v=uc_SciNQ2lo

The way you are now trying works perfectly. What I do not get is why we are showing you one method and you are trying to learn another method? I prefer using methods that I will remember for the rest of my life. Since I will always remember that (+*+)=+, (-*-)=+ and (+*-)=- I prefer the method which I showed you. Having said that I will also remember the method you are trying now. My concern is that you will not remember it.

Here is my method worked out: x>=0 (already solved). 2x+7>= 0 then x>-7/2 or -3 1/2. We want all numbers, if any, that are greater than or equal to 0 AND greater than or equal to -3 1/2. x>=0 is that solution.

Now x< =0 (already solved for x). 2x+7<=0. Then x<=-7/2 or -3 1/2. We want all numbers, if any, that are less than or equal to 0 AND less than or equal to -3 1/2. x>=0 is that solution. x<= -3 1/2 is the solution.

Now to get the final answer we need to consider x>=0 OR x<= -3 1/2(these are the solutions from the paragraphs above) . Sometimes you can write this as one interval and sometimes you can't. This time you can't and the final solution is x>=0 OR x<= -3 1/2

Here are some other examples.
x>5 or x>7. This combines to x>5
x<-2 or x<-5 combines to x<-2

 

[Ishuda]

Sorry I am still having some difficulty understanding any of this.

Maybe I'll understand more if you can walk me through the steps of the problem

x(2x + 7) >= 0

I thought the first step was to find two values from this inequality so that i can put them on the number line. Then the second step is to test one number from each interval.

what's next?

Hi abel

Yes, you can do it that way to solve the problem and, if that is all you want to do, that is fine. However, most of us here think that understanding the problem more thoroughly is better and will help you much more as you progress to higher levels in mathematics.

That is what most of the advice you have been getting is based on: If the product of two numbers is positive, then both numbers must be positive OR both numbers must be negative. That is an important concept and should be learned. If you do learn this and really understand it, you can do away with the the last three steps and use what has been presented here.

 

[abel muroi]

The way you are now trying works perfectly. What I do not get is why we are showing you one method and you are trying to learn another method? I prefer using methods that I will remember for the rest of my life. Since I will always remember that (+*+)=+, (-*-)=+ and (+*-)=- I prefer the method which I showed you. Having said that I will also remember the method you are trying now. My concern is that you will not remember it.

Here is my method worked out: x>=0 (already solved). 2x+7>= 0 then x>-7/2 or -3 1/2 (these are the solutions from the paragraph above). We want all numbers, if any, that are greater than or equal to 0 AND greater than or equal to -3 1/2. x>=0 is that solution.

Now x< =0 (already solved for x). 2x+7<=0. Then x<=-7/2 or -3 1/2. We want all numbers, if any, that are less than or equal to 0 AND less than or equal to -3 1/2. x>=0 is that solution. x<= -3 1/2 is the solution.

Now to get the final answer we need to consider x>=0 OR x<= -3 1/2. Sometimes you can write this as one interval and sometimes you can't. This time you can't and the final solution is x>=0 OR x<= -3 1/2

Here are some other examples.
x>5 or x>7. This combines to x>5
x<-2 or x<-5 combines to x<-2

I will definitely consider using my method since it's really the only method i can understand properly

but what if a non linear inequality has more than 2 variables?

for example:

3x² - 3x < 2x² + 4
-2x² + 3x² - 3x - 4 < 0

what can i do to get the two values from this inequality so that i can put them on the number line?

 

[Steven G]

I will definitely consider using my method since it's really the only method i can understand properly

but what if a non linear inequality has more than 2 variables?

for example:

3x² - 3x < 2x² + 4
-2x² + 3x² - 3x - 4 < 0

what can i do to get the two values from this inequality so that i can put them on the number line?

Now combine like terms.
I'd bring the terms to the side that has the largest coefficient of x^3 (left side)
Then x^2 -3x - 4 <0
Factor and then use your method

 

[abel muroi]

Now combine like terms.
I'd bring the terms to the side that has the largest coefficient of x^3 (left side)
Then x^2 -3x - 4 <0
Factor and then use your method

Ah i understand. So what about a inequality that is in the form (x - 2)² (x - 3) (x + 1) <= 0

since this is already factored, all i have to do is get the 3 values right?

and the 3 values are -2, 3 and -1?

 

[Steven G]

Ah i understand. So what about a inequality that is in the form (x - 2)² (x - 3) (x + 1) <= 0

since this is already factored, all i have to do is get the 3 values right?

and the 3 values are -2, 3 and -1?

I guess that you listed 3 because you know that 3 - 3 = 0. But you don't know that 2 -2 = 0?? Some mistakes are ok, after all we are human, but knowing that 2-2 = 0 should be as natural as knowing your own name.

Since (x-2)^2 is always positive ( it's 0 when x=2, so x=2 will be in the solution set) we can ignore partitioning the number line at 2 since the sign of (x - 2)² (x - 3) (x + 1) will not change around x=2.

 

[Steven G]

I will definitely consider using my method since it's really the only method i can understand properly

but what if a non linear inequality has more than 2 variables?

for example:

3x² - 3x < 2x² + 4
-2x² + 3x² - 3x - 4 < 0

what can i do to get the two values from this inequality so that i can put them on the number line?

By the way in the future can you start a new thread when you pose a new question.

On another topic, x and x^2 have the same variable! 3x +5y =7 has two variables

 

[HallsofIvy]

Ah i understand. So what about a inequality that is in the form (x - 2)² (x - 3) (x + 1) <= 0

since this is already factored, all i have to do is get the 3 values right?

and the 3 values are -2, 3 and -1?

As has already been pointed out, the "3 values" are x= 2 (not -2), x= 3, and x= -1. But do you know what to do with them after you have those values? Those are the values that make the left side, (x- 2)²(x- 3)(x+ 1) equal to 0. You still need to determine where that product is negative.

Here's one way to continue. Put those in increasing order, -1, 2, 3. If x< -1 then all three of x+ 1= x- (-1), x- 2 and x- 3 are negative because we are subtracting each number from a smaller number. Actually, "x- 2" is not important because it is squared and a square is never negative. So for x< -1, (x- 2)²(x- 3)(x+ 1) is the product of four negative numbers and so positive. For -1< x< 2. x+ 1 is positive but the other three factors, (x- 2)² and x- 3 are all still negative. The product of three negatives is negative. If 2< x< 3, the x+ 1 and x- 2 are positive but x- 3 is still negative. Since we now have three positive factors and one negative, the product is still negative. Finally, if x> 3, all four factors are positive and the product is positive. That is, (x- 2)²(x+ 1)(x- 3) is less than or equal to 0 for \(\displaystyle -1\le x\le 3\).

Another way to do this is to check one point in each interval bounded by those three numbers, x< -1, -1< x< 2, 2< x< 3, and 3< x. -2 is less than -1. (-2- 2)²(-2- 3)(-2+ 1)= (-4)²(-5)(-1)= 80 which is larger than 0. 0 lies between -1 and 2 and (0- 2)²(0- 3)(0+ 1)= -12 which is less than 0. 2.5 lies between 2 and 3 and (2.5- 2)²(2.5- 3)(2.5- 1)= -1.875 which is less than 0. 4 is larger than 3 and (4- 2)²(4- 3)(4- 1)= 12 which is positive. Again (x- 2)²(x+ 1)(x- 3) is less than or equal to 0 for \(\displaystyle -1\le x\le 3\).

 

[Deleted member 4993]

Sorry I am still having some difficulty understanding any of this.

Maybe I'll understand more if you can walk me through the steps of the problem

x(2x + 7) >= 0

I thought the first step was to find two values from this inequality so that i can put them on the number line. Then the second step is to test one number from each interval.

what's next?

Before that - since this is inequality - you need to establish conditions of solution

x(2x + 7) ≥ 0 implies x≥ 0 and (2x+7) ≥ 0 [both the factors are positive] ..................................(1)

OR

x(2x + 7) ≥ 0 implies x ≤ 0 and (2x+7) ≤ 0 [both the factors are negative] ................................(2)

Now the next step is "to find two values from this inequality so that i can put them on the number line. - in each solution conditions [(1) and (2)]

from condition (1)

x ≥ 0 ....................................................(1a)

AND

(2x+7) ≥ 0 → x ≥ (-7/2)............................(1b)

both of the conditions will be satisfied with x ≥ 0 [ since a point say x = -1 satisfies (1b) but doesn't satisfy (1a) - since it is "AND" we need to satisfy both]

Then

from condition (2)

x ≤ 0 ....................................................(2a)

AND

(2x+7) ≤ 0 → x ≤ (-7/2)............................(2b)

both of the conditions will be satisfied with x ≤ (-7/2) [ since a point say x = -1 satisfies (2a) but doesn't satisfy (2b) - since it is "AND" we need to satisfy both]

Conditions (1) and (2) are connected by OR

so the solution is

x ≤ (-7/2) OR x ≥ 0

The numbers -7/2 < x < 0 are not included in the solution set.

 

[Deleted member 4993]

Got some Baby Blue ink at xmas?

No... got it for upcoming Holi....

http://en.wikipedia.org/wiki/Holi