Why solving a rational inequality is tricky!

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Жыл бұрын

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
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#math #algebra #mathbasics

Пікірлер: 80

@Ninja20704 Жыл бұрын

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.

@Vikdeb25502 Жыл бұрын

I kinda like this one too.

@Ninja20704 Жыл бұрын

@@dinonugget2479 yes

@Hgggghhhfria1593 2 ай бұрын

You did it wrong 😅

@Ninja20704 2 ай бұрын

@@Hgggghhhfria1593 No? This is still a proper method that works all the time.

@Hgggghhhfria1593 2 ай бұрын

@@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.

@akarooyyy 4 күн бұрын

bro said "thats just life" what has he gone through 😭😭🙏

@infswrld 4 күн бұрын

shush

@user-ue9td4hr4v 9 ай бұрын

Thank you so much. I'm in College Algebra. I got all my homework right after watching this.

@ManjulaMathew-wb3zn 3 ай бұрын

Great knowledge on inequality problems and being very careful as needed.

@cyansorcerer6491 Ай бұрын

thank you so much for this.

@miacoolfacex 9 ай бұрын

I was so stuck! perfect explaining Thanks

@matdryz 3 күн бұрын

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)

@invexical Жыл бұрын

can you just use a sign chart?

@tmd4378 Жыл бұрын

yes but sign chart is different for them

@jofx4051 Жыл бұрын

Uhh wasn't he using that...?

@bandamkaromi Жыл бұрын

Wow. Very good!!

@fatgrandpa9376 Ай бұрын

could u explain the linear inequalities that are having the modulous function

@Lordmewtwo151 11 ай бұрын

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).

@Hanible 5 ай бұрын

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.

@ValidatingUsername 8 күн бұрын

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?

@okaro6595 Жыл бұрын

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.

@gheffz 11 ай бұрын

Excellent. Thank you for clarifying that!!! _"Just do it the safe way! That's it!"_

@sadeqirfan5582 3 күн бұрын

Awesome.

@elionsakshith3508 Жыл бұрын

Wavy curve for the clutch

@Robert-er5wq 3 күн бұрын

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.

@tonyennis1787 3 ай бұрын

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.

@Pengochan 11 күн бұрын

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.

@dangerous_woman 3 күн бұрын

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.

@James-dv9wz 6 ай бұрын

thank you so much😏😏

@dragonmaster909 6 ай бұрын

Why union? Why not that upside down "u" sign?

@swordofjustice7444 Ай бұрын

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.

@xXJ4FARGAMERXx Жыл бұрын

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.

@DRAAi_wHity Жыл бұрын

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

@DogoBoy 11 ай бұрын

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

@Lordmewtwo151 11 ай бұрын

@@DogoBoy He even addressed that fact.

@Lordmewtwo151 11 ай бұрын

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.

@indiasingh6891 6 күн бұрын

which grade is this

@chaoticfn2644 6 күн бұрын

11 if ur from india

@Infernalith Жыл бұрын

4 and -1 are excluded because its not *or* equal

@OrenLikes 3 ай бұрын

Why not include ±Infinity?

@triconax3956 2 ай бұрын

because we don't like infinity

@shauryakaushik8879 Жыл бұрын

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. 🙏🏼

@abhimanyusingh9489 Жыл бұрын

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

@dumitrudraghia5289 5 күн бұрын

Aiureli!

@black_earth1996 Жыл бұрын

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

@lechaiku 14 күн бұрын

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, ∞ )

@Hgggghhhfria1593 8 күн бұрын

u drew the whole graph 💀💀 📈

@philj9594 8 күн бұрын

@@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.

@lechaiku 7 күн бұрын

@@Hgggghhhfria1593 Not exactly. We don't know where is the vertex of parabola. That's why it is only a "rough-draft" of it.

@kevinseptember2917 5 күн бұрын

When you test the equation with -2 you must get positive and not negative. Your entire Notation is INCORRECT.

@kevinseptember2917 5 күн бұрын

Apologies, you are correct. -2 does not satisfy the equation.👍👍

@chucksucks8640 10 ай бұрын

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.

@michaels333 9 ай бұрын

And that would be fine if the were equations. But they are inequalities.

@boguslawszostak1784 13 күн бұрын

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

@stephenbeck7222 9 күн бұрын

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.

@boguslawszostak1784 8 күн бұрын

@@stephenbeck7222 You are right.

@SmashingCapital Ай бұрын

The trial and error way takes so much time thats not how you do it

@swordofjustice7444 Ай бұрын

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

@SmashingCapital Ай бұрын

@@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

@dangerous_woman 3 күн бұрын

@@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.

@tombraidering 3 күн бұрын

@@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)