With |a| + |b| = c, there are always max. 3 cases you have to check because each of the absolute value has one zero point, then using them you divide the real number set into intervals and solve for each
Absolutely right. Case 1 2 and 4 are the ones to be assumed and after getting each according solution a checkup is necessary, to whether the solution agrees to the prerequisite. The solution of case 2 fails prerequisite and was abandoned, while solution of case 1 and 4 are accepted.
@@phoenixarian8513 Due to the behaviour of the absolute value function, the function f(x)=|ax+b|+|cx+d| can only have a maximum of 2 solutions when f(x)=e where a,b,c,d,e are elements of the real numbers provided that a≠c. I'm sure there are other ways but this is how I interpret it: Consider the 3 cases; Case 1 : x>=-b/a Case 2 : -d/c
By the definition of absolute value, |x| = x if x >= 0 and -x if x < 0; similarly, |2x+3| = 2x+3 if x >= -3/2 and -2x-3 if x < -3/2 (you can work that out by setting the inner expression to 0 and solving for x. Since it's a linear function with a positive slope, all x values to the left of the root will produce a negative value). That gives 3 cases to consider: x being bigger than 0, x being smaller than -3/2 and x being between -3/2 and 0. For each case, use the appropriate rendition of each absolute value term and solve for x. You'll find that the third case gives you a nonsensical answer of x = 1, so it's really just two cases (x = 1 is a nonsensical answer because in the third case, x is constrained by -3/2 =< x < 0)
@@suryanshrastogi1340 he's not wrong saying 4 cases. By the definition of absolute value on real line, we DO get 2 cases for each absolute sign, and it then turns into a combinatorial question: how many possible pairs do we get without any restrictions of analysis? It's like asking if we're going buy 2 different types of food and each type has 2 flavors (options).
The way I did it was to find the critical points first--where the sign of x changes when taking the absolute value. Those are x=0 and 2x+3=0→x=-3/2. Then I checked the cases where x < -3/2, -3/2≤x
Look at the intended audience of this channel. While most of the alternative methods proposed below are more efficient, will they work for the intended audience? I am a retired high school mathematics teacher. Getting started is the most difficult part for the inexperienced math student. The method used here: Set up all the cases and then check for extraneous answers works not only for this problem but for many other types. As the student builds mathematical knowledge and confidence, then some of the methods proposed below would be appropriate.
This is a fair point. But I would have preferred in each case he write down the conditions on x and check that the value found satisfies these conditions rather than plugging in the value of x and finding whether the equation is satisfied. The method shown in the video gives no clue as to why the extraneous solutions arose.
To further elaborate on your explanation we have |2x+3| + |x| = 4 and we have the x< - 3/2 and we have till x=0 and we have x>0 . For each of these 3 "areas" we have an equation without the | | ....
Case 3 is impossible. x < -3/2 and x > 0 at the same time is impossible. So, case 1, 2, 4 are the only cases, and in case 2 where x € [-3/2, 0] the x solution is outside the interval so we can reject that solution
My thought for this was only solve for case 1 and 4 because x is in both parts of the absolute value, and one part of x doesn't make sense to be negative while the other part stays positive. This is what I mean. |-1(x+2x+3)| and |+1(x+2x+3)|. Just solve for that. I know I didn't provide it, but the original equation can be rewritten as |x+2x+3|. Then apply the same idea to the other side. |-1|4 and |+1|4. The other 2 cases don't make sense to spend time solving because they don't follow the rule (I forget the name). This is only tricky because the absolute value isn't on both sides of the equation. However, do not combine the like terms in the rewritten equation before distributing the +1 or -1. Doing so will come up with an incorrect answer, and original equation must be followed to verify and come up with a correct answer.
What would be tricky in my opinion would be adding a non absolute value term to the absolute value side. Something like this. x^2-|x|+|2x+3|=4. I would rewrite it like this. |x+2x+3|=4-x^2. Then solve for 2 possible answers. -x-2x-3=4-x^2 and x+2x+3=4-x^2. x^2-3x-7=0 and x^2+3x-1=0. Then use the quadratic formula. a=1, b=3,-3 and c=-1,-7. X=(-3/2)+-(13^1/2÷2) and X=(3/2)+-(37^1/2÷2). Then verify the 4 solutions for validity.
Is this just a coincidence, or is it something you can actually do? If you combine the terms inside of the seperate absolute values to just one, giving abs(3x+3) = 4, and then solve by exchanging the absolute value for a plus/minus infront of the 4, and do simple algebra to solve for x, you get the two correct solutions. Is this just a coincidence, or is it something you can actually do?
I'll start with the critical values where the absolute values are 0, which is x = 0, and 2x + 3 = 0, or x = -3/2 1. |x| - |2x + 3| = 4, critical values x = 0, -3/2. If x = 0, LHS = -3 != 4, if x = -3/2, LHS = 3/2 != 4, so the critical values are not solutions x < -3/2 -x - (-2x - 3) = 4 -x + 2x + 3 = 4 x = 1, no as it is not less than -3/2 -3/2 < x < 0 -x - 2x - 3 = 4 -3x = 7 x = -7/3, no as it is not on interval above x > 0 x - 2x - 3 = 4 -x = 7 x = -7, no as it is not greater than 0 No solutions at all. 2. |2x + 3| - |x| = 4, critical values x = 0, -3/2, If x = 0, LHS = 3 != 4, If x = -3/2, LHS = -3/2 != 4, so the critical values are not solutions x < -3/2 -2x - 3 + x = 4 -x = 7, x = -7, ok -3/2 < x < 0 2x + 3 + x = 4 3x = 1 x = 1/3, no as it does not lie in above interval x > 0 2x + 3 - x = 4 x = 1, ok x = -7, 1 are the solutions
Wow what a strange comment section. No, doing more work to have a more sophisticated case analysis for the domain of x to come up only 3 lines rather than 4 easy cases without regard to domain isn't more efficient. If you're looking for the quickest argument, observe that the function is convex so there are going to be at most 2 solutions, then simply guess the two solutions based on graphical evidence, plug in and verify that you have all 2 solutions. Without already knowing, you absolutely! must check at least 3 cases, and check whether the extrapolated solution contradicts the case assumption (or plug the x value back in to try to get 4)
Case 2 doesn’t make sense bc it’s not the same equation when you put only one abs negative. You need to put the whole of both side negative which gives x - |2x+3|=-4 Right? Idk if this is right or wrong, it’s my logic Edit: Also case 3 is also just wrong because of the same reason
Right conclusion (it doesn't make sense) but your logic is wrong. It's not about the form the equation takes after removing the absolute values. It doesn't make sense because the answer you get for x in that case lies outside the interval constraining it.
thats wrong approach, you find when the absolute switch sign, and the find your cases which will be only 3 and also you would already have a limit for x so if you get an x above that there is no solution
to solve any of these: move the absx to the other side square both sides you get something like this: (2x + 3)^2 = 16 + abs8x + x^2 or, almost the same thing: (2x + 3)^2 = 16 - abs8x + x^2 rewrite it as: 4x^2 + 12x + 9 = 16 + x^2 +- abs(8x) move everything to LHS 3x^2 + 12x +-8absx - 7 = 0 split this into two cases: 3x^2 + 12x + 8x - 7 = 0 3x^2 + 12x - 8x - 7 = 0 which is to say: 3x^2 + 20x - 7 = 0 3x^2 + 4x - 7 = 0 now it's as simple as applying the quadratic formula. You get the same 4 possible answers for all 3 versions of this problem. So all you have to do is plug in those 4 possible solutions, see which ones work, and you're done. You don't have to consider any other values of x.
This method is wrong and inefficient. You have to analyse the sign of the single functions inside the absolute value (using a graph of sign) and then you are left with the three acceptable cases
Sometimes math is very tricky. This time is's because of absolute values are artificial 'creatures'. They don't exist in nature and our world is running just on 'real' math, not such created by people if you know what I mean. It is something similiar to negative numbers, they exist on bank accounts statuses only. You cant have negative number of shoes in your home for example. Some of you can say... wait a moment... temperatures can be under zero and that is true but we can use new range other than °C or F or K, where all the measured temp-s in our location are always 0+.
This smacks a bit too much of mathematical realism for my taste 🤣 ALL maths is just the product of the human mind, so the concept of magnitude (which is what the absolute value is) is just as natural as any other
My guess is because, since x is in both absolute values, you can't only make only one of them negative at a time like case 2 and 3. Since they both have x in them, x is either going to be positive or negative, and you can't have a situation where x is positive in one and negative in the other.
If I may comment on this great video, you can think of solutions as x-intercepts of the graph y=|x|+|2x+3|-4. The graph consists of 3 distinct linear graphs, divided into 3 regions: x
This is a very simple problem. American literature is simply bad making it unnecessarily difficult. Make an easy calculation of the absolute value to get the breaking point then study the intervals with the correct terms and you will get the correct answer by simply looking if the answer is in agreement to the interval you are working in. It is easiest visualized if you draw a real line and place every breaking point on the line then calculate the equation in its correct interval with the help of the definition of absolute values; using the breaking points. Why this is better is because you shouldn't just learn how to just calculate, you need to be able to visualize and understand what is happening on the real line. Absolute values are distances and simply a magnitude and you are looking at what solutions fulfills the equation and/or inequality problems.
Why can’t you just get rid of the absolute value signs and solve with the equation set to 4 and then solve again when set to -4? I got the two correct answers and no extraneous answers.
No, you can just once solve for 4 and another for -4, what about when x lies between -3/2 and 0? Then |2x+3|=2x+3 but |x|=-x. Then you get -x+2x+3=4-->x=1, since x lies between -3/2 and 0, so the answer can't be 1. You need to follow the method.
So the fundamental problem is that you can't have the two terms to have different signs. Since they are both absolute values, the terms (x) and (2x+3) are always additive, which means you have to either add x+2x or -x-2x. With the absolute values, you can never have x be positive and 2x be negative, or vice versa. I think that's my interpretation.
Case 1: x>=0
Case 2: -3/2
That was the first thing I did. Through this consideration, I directly ignored case 3. In addition, you don't have to do the test.
With |a| + |b| = c, there are always max. 3 cases you have to check because each of the absolute value has one zero point, then using them you divide the real number set into intervals and solve for each
Absolutely right. Case 1 2 and 4 are the ones to be assumed and after getting each according solution a checkup is necessary, to whether the solution agrees to the prerequisite. The solution of case 2 fails prerequisite and was abandoned, while solution of case 1 and 4 are accepted.
@@phoenixarian8513 Due to the behaviour of the absolute value function, the function f(x)=|ax+b|+|cx+d| can only have a maximum of 2 solutions when f(x)=e where a,b,c,d,e are elements of the real numbers provided that a≠c.
I'm sure there are other ways but this is how I interpret it:
Consider the 3 cases;
Case 1 : x>=-b/a
Case 2 : -d/c
By the definition of absolute value, |x| = x if x >= 0 and -x if x < 0; similarly, |2x+3| = 2x+3 if x >= -3/2 and -2x-3 if x < -3/2 (you can work that out by setting the inner expression to 0 and solving for x. Since it's a linear function with a positive slope, all x values to the left of the root will produce a negative value).
That gives 3 cases to consider: x being bigger than 0, x being smaller than -3/2 and x being between -3/2 and 0. For each case, use the appropriate rendition of each absolute value term and solve for x. You'll find that the third case gives you a nonsensical answer of x = 1, so it's really just two cases (x = 1 is a nonsensical answer because in the third case, x is constrained by -3/2 =< x < 0)
Suppose x>=0, then 2x + 3>=0 + 3= 3> 0, so in fact we just really need 3 cases.
You re right
Yes I got confused when he said 4 cases would be formed
@@suryanshrastogi1340 he's not wrong saying 4 cases. By the definition of absolute value on real line, we DO get 2 cases for each absolute sign, and it then turns into a combinatorial question: how many possible pairs do we get without any restrictions of analysis? It's like asking if we're going buy 2 different types of food and each type has 2 flavors (options).
@@kobethebeefinmathworld953 Yeah I know he is not wrong, this is the most fundamental way of solving this, I just expressed my thoughts
With that same logic, we could also eliminate the case where we have 2x+3 = 0 since we know if 2x+3
With absolue values , you need to study the sign of the functions involved :)
The way I did it was to find the critical points first--where the sign of x changes when taking the absolute value. Those are x=0 and 2x+3=0→x=-3/2. Then I checked the cases where x < -3/2, -3/2≤x
My method involves trying to figure out what f(x) = |x| + |2x+3| looks like. Fortunately, everything is linear. f(x) = 3x+3 for x >= 0, x+3 if -3/2
This is so much easier just finding out where the 0's of each moduli and studying the case for intervals between these values
Look at the intended audience of this channel. While most of the alternative methods proposed below are more efficient, will they work for the intended audience? I am a retired high school mathematics teacher. Getting started is the most difficult part for the inexperienced math student. The method used here: Set up all the cases and then check for extraneous answers works not only for this problem but for many other types. As the student builds mathematical knowledge and confidence, then some of the methods proposed below would be appropriate.
This is a fair point. But I would have preferred in each case he write down the conditions on x and check that the value found satisfies these conditions rather than plugging in the value of x and finding whether the equation is satisfied. The method shown in the video gives no clue as to why the extraneous solutions arose.
To further elaborate on your explanation we have |2x+3| + |x| = 4 and we have the x< - 3/2 and we have till x=0 and we have x>0 . For each of these 3 "areas" we have an equation without the | | ....
Case 3 is impossible. x < -3/2 and x > 0 at the same time is impossible. So, case 1, 2, 4 are the only cases, and in case 2 where x € [-3/2, 0] the x solution is outside the interval so we can reject that solution
Technically it makes sense since there are no solutions
If x ≥0 than: 2x+3 >0 and therefore Case 3 is logically impossible and should not be analyzed at first place. Besides, nice explanation.
U can also analyse it through graph
This video helped me the most, thank you!!!!
shortly x + 2x+3 = ±4
Would work in every scenario, right?
thank you you are the only one that explains it well ❤
My thought for this was only solve for case 1 and 4 because x is in both parts of the absolute value, and one part of x doesn't make sense to be negative while the other part stays positive. This is what I mean. |-1(x+2x+3)| and |+1(x+2x+3)|. Just solve for that. I know I didn't provide it, but the original equation can be rewritten as |x+2x+3|. Then apply the same idea to the other side. |-1|4 and |+1|4. The other 2 cases don't make sense to spend time solving because they don't follow the rule (I forget the name). This is only tricky because the absolute value isn't on both sides of the equation. However, do not combine the like terms in the rewritten equation before distributing the +1 or -1. Doing so will come up with an incorrect answer, and original equation must be followed to verify and come up with a correct answer.
What would be tricky in my opinion would be adding a non absolute value term to the absolute value side. Something like this. x^2-|x|+|2x+3|=4. I would rewrite it like this. |x+2x+3|=4-x^2. Then solve for 2 possible answers. -x-2x-3=4-x^2 and x+2x+3=4-x^2. x^2-3x-7=0 and x^2+3x-1=0. Then use the quadratic formula. a=1, b=3,-3 and c=-1,-7. X=(-3/2)+-(13^1/2÷2) and X=(3/2)+-(37^1/2÷2). Then verify the 4 solutions for validity.
Use graph and critical values . Much more easier . Idk why no body is talking about graphical method
graphing abs(srt.line) pretty easy, but might be more helpful to visualize more difficult functions
It seems to me like the x value will always both have to be positive or both have to be negative since they represent the same X
Is this just a coincidence, or is it something you can actually do? If you combine the terms inside of the seperate absolute values to just one, giving abs(3x+3) = 4, and then solve by exchanging the absolute value for a plus/minus infront of the 4, and do simple algebra to solve for x, you get the two correct solutions. Is this just a coincidence, or is it something you can actually do?
you are really good thank you so much l love you
Dang the way u made it look so simple 😅☺️
Good evening. We can make the calculation of the signs of the conditions, then confront the solutions with them. It is more intuitive.
Why not directly do x(2x+3)≥0?
4:00
how did 3 turn into 6/3
I'll start with the critical values where the absolute values are 0, which is x = 0, and 2x + 3 = 0, or x = -3/2
1. |x| - |2x + 3| = 4, critical values x = 0, -3/2. If x = 0, LHS = -3 != 4, if x = -3/2, LHS = 3/2 != 4, so the critical values are not solutions
x < -3/2
-x - (-2x - 3) = 4
-x + 2x + 3 = 4
x = 1, no as it is not less than -3/2
-3/2 < x < 0
-x - 2x - 3 = 4
-3x = 7
x = -7/3, no as it is not on interval above
x > 0
x - 2x - 3 = 4
-x = 7
x = -7, no as it is not greater than 0
No solutions at all.
2. |2x + 3| - |x| = 4, critical values x = 0, -3/2, If x = 0, LHS = 3 != 4, If x = -3/2, LHS = -3/2 != 4, so the critical values are not solutions
x < -3/2
-2x - 3 + x = 4
-x = 7,
x = -7, ok
-3/2 < x < 0
2x + 3 + x = 4
3x = 1
x = 1/3, no as it does not lie in above interval
x > 0
2x + 3 - x = 4
x = 1, ok
x = -7, 1 are the solutions
I never knew that x=1/3 can be lying
Wow what a strange comment section. No, doing more work to have a more sophisticated case analysis for the domain of x to come up only 3 lines rather than 4 easy cases without regard to domain isn't more efficient. If you're looking for the quickest argument, observe that the function is convex so there are going to be at most 2 solutions, then simply guess the two solutions based on graphical evidence, plug in and verify that you have all 2 solutions.
Without already knowing, you absolutely! must check at least 3 cases, and check whether the extrapolated solution contradicts the case assumption (or plug the x value back in to try to get 4)
Case 2 doesn’t make sense bc it’s not the same equation when you put only one abs negative. You need to put the whole of both side negative which gives x - |2x+3|=-4
Right? Idk if this is right or wrong, it’s my logic
Edit: Also case 3 is also just wrong because of the same reason
Right conclusion (it doesn't make sense) but your logic is wrong. It's not about the form the equation takes after removing the absolute values. It doesn't make sense because the answer you get for x in that case lies outside the interval constraining it.
And if we'll had it, |x|+|2x+3|>4, or
graphically i am getting -7/3 and 1/3, am I wrong?
thats wrong approach, you find when the absolute switch sign, and the find your cases which will be only 3
and also you would already have a limit for x so if you get an x above that there is no solution
to solve any of these: move the absx to the other side
square both sides
you get something like this:
(2x + 3)^2 = 16 + abs8x + x^2
or, almost the same thing:
(2x + 3)^2 = 16 - abs8x + x^2
rewrite it as:
4x^2 + 12x + 9 = 16 + x^2 +- abs(8x)
move everything to LHS
3x^2 + 12x +-8absx - 7 = 0
split this into two cases:
3x^2 + 12x + 8x - 7 = 0
3x^2 + 12x - 8x - 7 = 0
which is to say:
3x^2 + 20x - 7 = 0
3x^2 + 4x - 7 = 0
now it's as simple as applying the quadratic formula. You get the same 4 possible answers for all 3 versions of this problem.
So all you have to do is plug in those 4 possible solutions, see which ones work, and you're done. You don't have to consider any other values of x.
Sketch the graph its easy u get the correct values
1. No solutions.
2. x=1 and x=-7
A graph would help
A quick sketch indicates two solutions
....2.333333 -
Is a solution
Edit: The last case solution looked like a -1/3
to me instead of -7/3
我全式平方，之後解出來是一樣答案，這樣是對的嗎？
I thing
1is 1/3
2 is 1
E'm i right broo?
Realistically you only consider 3 cases because in case 3, x is positive while 2x+3 is negative, which is of course impossible
A) x=1
This method is wrong and inefficient. You have to analyse the sign of the single functions inside the absolute value (using a graph of sign) and then you are left with the three acceptable cases
Can u plot graph and solve ???
Take y= LHS and y=RHS and plot them both, and find x corresponding to intersections.
I got all solutions too. Nice.
3x+3=4
3x=1
x=1/3
i watch this:helpful legit 💀
Your video is much ❤❤❤
He did that mistake on purpose didnt he? He wants us shouting nein nein nein nein nein!
The fastest way to take down this problem is to look at how big and how small 2x+3 can be. We have | 2x+3|
Sir you did wrong last step verification
Thnku i did with same method
Number line number line number line number line only number line just make a number line number line
my first instict is to square em all 😂
Sometimes math is very tricky. This time is's because of absolute values are artificial 'creatures'. They don't exist in nature and our world is running just on 'real' math, not such created by people if you know what I mean. It is something similiar to negative numbers, they exist on bank accounts statuses only. You cant have negative number of shoes in your home for example. Some of you can say... wait a moment... temperatures can be under zero and that is true but we can use new range other than °C or F or K, where all the measured temp-s in our location are always 0+.
Negative magnitudes are real as vectors, so they exist in nature.
This smacks a bit too much of mathematical realism for my taste 🤣 ALL maths is just the product of the human mind, so the concept of magnitude (which is what the absolute value is) is just as natural as any other
You need to check the absolute value of X and no need to check the full value of the equality
I didn't know that you can get extraneous solutions in absolute value equations.
My guess is because, since x is in both absolute values, you can't only make only one of them negative at a time like case 2 and 3. Since they both have x in them, x is either going to be positive or negative, and you can't have a situation where x is positive in one and negative in the other.
If I may comment on this great video, you can think of solutions as x-intercepts of the graph y=|x|+|2x+3|-4. The graph consists of 3 distinct linear graphs, divided into 3 regions: x
@@stellacollector i did the same thing
This is a very simple problem. American literature is simply bad making it unnecessarily difficult. Make an easy calculation of the absolute value to get the breaking point then study the intervals with the correct terms and you will get the correct answer by simply looking if the answer is in agreement to the interval you are working in.
It is easiest visualized if you draw a real line and place every breaking point on the line then calculate the equation in its correct interval with the help of the definition of absolute values; using the breaking points.
Why this is better is because you shouldn't just learn how to just calculate, you need to be able to visualize and understand what is happening on the real line. Absolute values are distances and simply a magnitude and you are looking at what solutions fulfills the equation and/or inequality problems.
Isn't it a rule - either both absolutes are positive or both are negative?
A method for storage.
Why can’t you just get rid of the absolute value signs and solve with the equation set to 4 and then solve again when set to -4? I got the two correct answers and no extraneous answers.
No, you can just once solve for 4 and another for -4, what about when x lies between -3/2 and 0? Then |2x+3|=2x+3 but |x|=-x.
Then you get -x+2x+3=4-->x=1, since x lies between -3/2 and 0, so the answer can't be 1.
You need to follow the method.
In these type of problems you need to make 3 cases, there can be atmost 3 solutions, but your method will only give the extreme two.
@@Shreyas_Jaiswal 4 cases.
@@azzteke no, only 3 cases.
First, x≤-3/2
Second, -3/20
I don't see any other case possible.
x=1/3;
👍👍👍👍
Square both sides, it is easier to solve in that way, I think.
Easy
So the fundamental problem is that you can't have the two terms to have different signs. Since they are both absolute values, the terms (x) and (2x+3) are always additive, which means you have to either add x+2x or -x-2x. With the absolute values, you can never have x be positive and 2x be negative, or vice versa.
I think that's my interpretation.
X+2x+3=4
X+2x+3=-4
Using number line system get the right answer faster
This method is not always correct. From the triangle inequality, we know that | a + b |
You look like my classmate
Copy paste
Real big brains realize the second absolute value is around the plus sign. 🧠
First absolute value.
1x |+| 2x+31 = 4
;)