Why solving a rational inequality is tricky!
HTML-код
- Опубликовано: 2 окт 2024
- Be careful when we solve a rational inequality! This tutorial will help you with your algebra 2 or precalculus classes. Subscribe for more math tutorials. @bprpmathbasics
💪 Full playlist to my precalculus class review#3: • Precalculus Review #3
Shop my math t-shirts & hoodies on Amazon: 👉 amzn.to/3qBeuw6
My blue jacket: 👉 amzn.to/3qBeuw6
-----------------------------
I help students master the basics of math. You can show your support and help me create even better content by becoming a patron on Patreon 👉 / blackpenredpen . Every bit of support means the world to me and motivates me to keep bringing you the best math lessons! Thank you!
-----------------------------
#math #algebra #mathbasics
bro said "thats just life" what has he gone through 😭😭🙏
shush
Great knowledge on inequality problems and being very careful as needed.
Case I : 3 x + 4 > 0
x^2 - 3 x - 4 > 0
( x - 4) ( x + 1) > 0
Either x < - 1 or x > 4
with a precondition x > - 4/3
Hereby x > 4 is only set of feasible solution
Case II : 3 x + 4 < 0
x^2 - 3 x - 4 < 0
( x - 4) ( x + 1) < 0
- 1 < x < 4
with a precondition x < - 4/3
This designates a NULL set
Another way would be to multiply both sides of (a) by (3x+4)^2 instead so that we can be sure the inequality direction is preserved so we get
x^2(3x+4)>(3x+4)^2
x^2(3x+4)-(3x+4)^2>0
(3x+4)[x^2-(3x+4)]>0
(3x+4)(x^2-3x-4)>0
(3x+4)(x+1)(x-4)>0
And then do the sign chart
I don’t really know which way is better, I like my way because we at least dont have to deal with fractions.
I kinda like this one too.
@@dinonugget2479 yes
You did it wrong 😅
@@Hgggghhhfria1593 No? This is still a proper method that works all the time.
@@Ninja20704 Not always , imagine a question where we deal with odd exponential power , this method is only valid with always positive variable (x)²'⁴'⁶''' etc.
The other way to think about this is to consider a graph, as if it were an equation.
For B, we have the equatiob x² -3x -4 = 0. This will be a parabola with the ends sticking up. We know the intersection of the parabola with the X axis from his solution, so it is easy to see that the ends sticking upward are the regions that are above zero.
For a, we see that we can transform the expression into the form of (the parabola from equation b) over (the linear expression 3x+4).
When you graph he linear expression, you get a line with a Y intercept at (0, 4) and an X intercept of (-4/3, 0)
Now draw both lines on the same chart. The equation of interest is the parabola divided by the line.
From the sign analysis you can see the same results as g8ven:
Less than -4/3 the parabola is positive and the line is negative, so the result is negative.
Between there and -1 both the parabola and the line are positive, so the result is too.
Between -1 and 4 the parabola is negative and th line is positive, so the result is negative, and greater than 4 both are positive.
A few other things to note is are:
1. where the line and parabola intersect, the rational function will have the value 1.
2. Where the parabola has the value zero the rational function will as well
3. Where the line goes through zero, the rational function will have a vertical asymtote. As you divide by smaller numbers the rational function's value will approach positive or negative infinity.
These last don't really help solve the problem at hand, but are interesting to yhink about.
When you say "we care about these values," the reasoning is weak or unexplained. It's better to evaluate the inequality when 3x+4 is positive, negative, and zero separately.
The bigger explanation is because sign changes can occur when a graph approaches 0 or an asymptote, we check those points on the number line to find where the signs are positive and negative and then answer from there.
1:24 Maybe tell people why we care about those numbers. These are the zeroes of the polynomial, i.e. where it can change sign, and we're looking for the positive solutions.
Also worth mentioning that depending on zero's (or root's) multiplicity the sign can stay the same. Imagining it as a parabola graph really helps, it made me understand it.
Usually when it comes to this, I consider the two possibilities (when the denominator is negative, and when it is positive) and make an interval for both (sort of like solving a modulus equation). So, for example, the denominator here would be negative for x < -4/3, and positive for x > -4/3. Then, I'd multiply the denominator and solve as normal in the positive case, and flip the sign in the negative case, and then find the intersection between the interval I defined earlier and the solution interval for both cases
yea i do the same
im a french student and its pretty funny bc you guys do the "sign table" in a completly different method than in france! and btw in what grade do you learn this ?`
I was so stuck! perfect explaining Thanks
3:45 all else being equal (and it isn't, that's why you made this video), the solution for (a) cannot be x = -4/3. But on problem (b), -4/3 is a valid solution. Yowch.
You can also notice thqt since x^2 >=0, 3x+4 has to be postive too for the LHS > 0 (and its >1 so it has to be greater than zero). Then you can just multpily ny denominator and solve it like the other inequality (just keeping in mind that 3x+4>0)
what are you on bro why does 3x+4 have to be positive?I don't think you understand the question.It asks us for which values of x that equation is true it's not stating that 3x+4>1 .Maybe x can be -2 (for example) and see that x² is positive but 3x+4 aint
@@ts9dream are we both talking about the inequality a?
as I wrote "3x+4 has to be positive too for the LHS > 0"
we have x^2/(3x+4)>1
we now that x^2>=0 for any value of x in R
let's assume 3x+4
Thank you so much. I'm in College Algebra. I got all my homework right after watching this.
I actually teach algebra and calculus at a community college and watch these videos to hone my teaching methods!
sorry for asking but in what country is this kind of exercises college level?Because in Greece this is like 10th grade stuff. I also happen to study mathematics in university and none of our classes(43 completed classes needed in order to graduate) are even remotely as easy as this stuff
thank you so much for this.
Excellent. Thank you for clarifying that!!! _"Just do it the safe way! That's it!"_
x²/(3x+4) > 1
If 3x+4 is pos, then the inequality is
x² > 3x+4
x²-3x-4 > 0
(x-4)(x+1) > 0
pos neg pos
-------(-1)-------(4)--------
x < -1 ∪ x > 4
If 3x+4 is neg, then the inequality is
x² < 3x+4
x² -3x -4 < 0
(x-4)(x+1) < 0
pos neg pos
-------(-1)-------(4)--------
-1 < x < 4
So in sum, we have
If 3x+4 > 0
3x > -4
x > -4/3
Then x < -1 ∪ x > 4. But we already said x > -4/3, so the first part becomes -4/3 < x < -1 and we have -4/3 < x < -1 ∪ x > 4.
If 3x+4 < 0
3x < -4
x < -4/3
Then -1 < x < 4. But we already said x < -4/3, so this whole thing is gone
So what we have in the end is -4/3 < x < -1 ∪ x > 4.
Wavy curve for the clutch
It’s the only real way to solve for domain and range, but what do we call the z and t or nth dimension where the function exists?
Or are we claiming it’s axiomatic that range is the dependent variable when only two variables are considered?
u can multiple by (3x+4)² on both sides
can you just use a sign chart?
yes but sign chart is different for them
Uhh wasn't he using that...?
Someone didn't watch the vid
@@simplebutpowerfulwhich is okay.
A sign table/chart would me waaaaay better
for the left ineq>1, 3X+4 must >0 because x^2>0 , so just indictical to the right ineq with additional condition: 3X+4>0, its much easier to understand
I never commented on a math video before but I never felt more triggered just multiply both sides to get the second equation😂 3:22
You can’t because you don’t know what x is. Since x could be a negative number, and multiplying by a negative number flips the sign, it doesn’t work
@@DogoBoy He even addressed that fact.
Earlier he pointed out why that doesn't work (and he said *exactly* the same thing as the other person in this thread). He's also gone over something similar regarding why one function or inequality doesn't equal another when on the surface you could just multiply out the denominator to get the same result. I don't remember what the functions were, but I think it had something to do with square roots.
case 1
3x+4 > 0, x > -4/3
x^2 > 3x + 4
x^2 -3x -4 > 0
x < -1 or x > 4
-4/3 < x < -1 or x > 4
case 2
3x+4 < 0, x < -4/3
x^2 < 3x + 4
x^2 -3x -4 < 0
-1 < x < 4
no solution
combining,
-4/3 < x < -1 or x > 4
you sometimes forget to say 'care about' when saying 'care', probably because you can just say 在乎 when speaking chinese
Is this correct: -3.5/3?
I never saw an ordinary fraction written like this.
Wouldn't it be -7/6?
It's still the same thing 🙂
Much easier is this method:
(x^2) / (3x + 4) > 1 ------> D: x must be different than -4/3
(x^2) / (3x + 4) - (3x + 4) / (3x + 4) > 0
(x^2 - 3x - 4) / (3x + 4) > 0
now we can multiply both sides by(3x + 4) ^2 (which is a positive number)
(x^2 - 3x - 4) (3x+4) > 0
(x - 4) (x+1) (3x+4) > 0
We don't need to test any numbers, because it is just wasting of time. Just draw the number line and a "rough-draft" of the parabola and draw a line through the point -4/3 (an increasing line because m > 0)
a > 0 -----> the arms of parabola directed above the x-axis
/
| / |
I I
/ I I
----------------------o-------o------o--------------> x
- ∞ / I _ I +∞
/ -4/3 -1 4
we instantly can see x-intercepts and the part of parabola and the line which are
above the x-axis (y > 0)
so the solution is:
x = ( -4/3, -1) v (4, ∞ )
u drew the whole graph 💀💀 📈
@@Hgggghhhfria1593 Bro just plugged this into ChatGPT and then tried to use it to flex. That's how ChatGPT draws graphs lol. While this method is fine, I think inequalities are much easier to solve without trying to graph them. Yeah, you have to plug in numbers, but it's quick.
@@Hgggghhhfria1593
Not exactly. We don't know where is the vertex of parabola. That's why it is only a "rough-draft" of it.
Wow. Very good!!
when you use this neg pos chart, just check one and when you come a care point it changes and goes like neg pos neg pos but when you have 2 of this number like you got number 4 2 times that means you wont change and it goes like neg pos neg pos pos neg pos neg...
you multiply both sides by 3x+4 while adding the restriction (x> -3/4) then instead of (-inf,-1)U... you have (-3/4,-1)U.... etc
edit: and for the sign chart, you could just see the value between, -1 and 4 (0 for example), and since it's a second degree polynomial you know it's a parabola so you inverse the signs outside the (-1,4) interval.
You mean -4/3 but yeah
@@simplebutpowerful yeah haha
I would have divided it into two based on the divisor and then multiplied it away. The first oen woud give x-4/3 and the results of the right side problem.
could u explain the linear inequalities that are having the modulous function
Why union? Why not that upside down "u" sign?
Sorry for the late reply! The upside down U is the intersection, or the values shared between the two intervals. The Union is all of the values in both of the intervals, which is what we want.
Awesome.
x²/(3x+4)>1
(x²-3x-4)/(3x+4)>0
(x²-3x-4>0 iff 3x+4>0) and neither are 0
(iff means both are true or both are false, AKA XNOR)
((x-4)(x+1)>0 iff x>-4/3) and x≠4 and x≠-1 and x≠4/3
((x4) iff x>-4/3) and x≠4 and x≠-1 and x≠4/3
x is either greater than -4/3 and over the range (-∞,-1)U(4,∞) or it's less than -4/3 and over the range (-1,4). The latter is impossible. Therefore x is in the range (-4/3,-1)U(4,∞).
If you look at the graph, you'll see y=x²/(3x+4) is negative if x is less than -4/3, undefined at x=-4/4, and positive otherwise. In the positive range, you see y goes down from ∞, dips below 1 at x=-1, starts increasing at x=0, and then dips above 1 at x=4 and increases indefinitely. So...I'm right.
This is because
Multiply both side (3x+4)
We did not care about pos or neg
If we care it we can get a solution
for 3x+4>0, we have x^2 -3x - 4 > 0
we get x > -4/3 and x >4 and x < -1 => x = (-4/3, -1) and (4, inf)
for 3x+4 -1 => x = nothing
you only need to know what will happend when divide neg number
6:26 Okay, I was kind of expecting a simple fraction between -1 and -4/3 like -7/6 for example (which was technically used, but not the simple way).
My approach was just to plug -1.2 into the original equation. (-1.2)^2/(3*(-1.2)-4) = 1.44/.4 > 1. Since the inequality direction never flipped, I think it was safe for me to assume that the sign chart would be the same.
Plain answer will be:
"Because it is not an equation.
Thus, you cannot perform anything you want on both sides"
I love you so much
Vwry interesting... In school days, I always confused why those are not same.
To get from x^2-3x-4 to (x-4)(x+1) you *test in your head* which numbers a and b give a*b=-4 and a+b = +1?
3:30 important
solve for
5-3x
-------------- < 2
x²-4x-10
A/B>0 if and only if A*B>0 ...
4 and -1 are excluded because its not *or* equal
thank you so much😏😏
Why not include ±Infinity?
because we don't like infinity
You cannot include infinity
@@arcane-2947 i'll ask again: WHY?
which grade is this
11 if ur from india
We use a simple method, which we call the "snake method."
We transform the inequality into a form where one side is 0 and the other side is a rational function by moving all terms to one side, as in The movie.
We find the zeros of the numerator and the denominator and determine their multiplicities.
On the number line, we mark the zeros of the numerator with solid small circles and the zeros of the denominator with empty small circles.
We check the sign of the limit at positive infinity. Depending on the result, we start from the right side either from the top if it is positive or from the bottom if it is negative, drawing a "snake" through all the circles. We change the sign for roots with odd multiplicities and "bounce off the axis" for even multiplicities. The solution becomes visible.
This method is permissible in exams.
To clarify, while this specific method might not be widely recognized by name in English-speaking countries, students there often learn similar techniques for determining the sign of rational functions over intervals. The key steps of identifying zeros, plotting them on a number line, and analyzing the intervals are common practices.
This method helps students easily remember the rules for determining the sign of the function and is acceptable in exams in Poland
The ‘snake’ method is taught some in US. But the ‘test points’ method that BPRP does is more generalizable to more complex functions (non-polynomials/rationals) where it may be difficult to recognize whether they have an even or odd multiplicity pattern at the roots.
@@stephenbeck7222 You are right.
My markers dissapeared, I wonder who took them 💀
Multiple both sides by 3x+4
Hey! It would be really kind of you to listen to this and actually clear my doubt which could’ve been of many people but they just accepted it at school and forgot the why?
My question is when we represent numbers like rt(3) on a number line we can easily do that by using pythagoras theorem.Its also very intuitive
But,
(Now comes the actual part , for the sake of conv. I’ve stated it here)pls make a video for this
The things is when we encounter numbers like rt(9.3) oof that number takes a bit of construction like making a line then adding 1 unit to it then making circle marking the point on that circle then taking that point making another curve which finally marks the point rt(9.3)
Hoping that you are familiar with it , kindly explain this strange magical seeming concept about circles.
🙏🏼
Let's say we are representing (x)^½ on number line
We mark x {name this point X} and when we add one to it the distance becomes (x+1) {name this point A} now we bisect (x+1) , the distance of this point from 0 is (x+1)/2 {Name this point M}. MA=(x+1)/2 since M is mid point of OA. MA is also the radius of the drawn circle. now we draw a perpendicular at X,let it cut the circle at Y. MY is also radius of the circle hence its length is also (x+1)/2. MX length is (x-1)/2.triangle MXY is a right angle triangle with angle X=90° ,by Pythagoras theorem XY=(x)^½
Hope this helps
Aiureli!
The trial and error way takes so much time thats not how you do it
How would you do it then? Imo this is a really safe way to do it, and in ur head you can definitely do it faster than how he does it here
@@swordofjustice7444 the numbers you care about are the zeros, so you put for each of them pluses on the right between each number and minuses on the left, once youve constructed a table you multiple or divide the signs and obtain where the function is negative and where it is positive
@@swordofjustice7444 Make the table of the roots from left to right as they increase. Draw single or double circles at the borders depending on the root's multiplicity. Then look at the main coefficient's sign, if it's positive put (+) at the most-right and then change it at every root except when it has even multiplicity.
@@swordofjustice7444 After making the graph and denoting the multiplicity of the roots with single or double rings, start the rightmost with the sign of the main coefficient, if it's all in factors multiply their x'es and put that sign and then change the sign at every ring (odd multiplicity) or keep it the same at double rings (even multiplicity)
a) is wrong.... for so many cases. So fix that first and thrn we xan ask whether we have to think about multiplying by 3x + 4.
... And for those cases for which a) is true, you can simply multiply by 3x + 4 without worrying about the relation sign.
When you test the equation with -2 you must get positive and not negative. Your entire Notation is INCORRECT.
Apologies, you are correct. -2 does not satisfy the equation.👍👍
Lol,me solving these easily (I'm Indian)
very nice
its in indian dna
This is a real stumper for me because they should be the same equation and therefore had the same answers but I started to think what you are doing when you divide both sides by 3x - 4. If all 3x - 4 were non-zero answers then there shouldn't be an issue but when you divide by 3x - 4 you are introducing a zero with x = -3/4 and dividing by zero is not possible therefore it creates problems in the equation.
And that would be fine if the were equations. But they are inequalities.
Solving inequalities is not tricky