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

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.

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.

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

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

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

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

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)

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 ?`

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

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.

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

can you just use a sign chart?

thank you so much for this.

I was so stuck! perfect explaining Thanks

Is this correct: -3.5/3? I never saw an ordinary fraction written like this. Wouldn't it be -7/6?

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.

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

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.

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?

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, ∞ )

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.

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?

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.

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

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

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...

u can multiple by (3x+4)² on both sides

Wavy curve for the clutch

Vwry interesting... In school days, I always confused why those are not same.

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.

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

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).

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.

Wow. Very good!!

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

you sometimes forget to say 'care about' when saying 'care', probably because you can just say 在乎 when speaking chinese

I love you so much

3:30 important

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

solve for 5-3x -------------- < 2 x²-4x-10

A/B>0 if and only if A*B>0 ...

Awesome.

Plain answer will be: "Because it is not an equation. Thus, you cannot perform anything you want on both sides"

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.

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

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. 🙏🏼

thank you so much😏😏

which grade is this

Why not include ±Infinity?

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

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.

Multiple both sides by 3x+4

My markers dissapeared, I wonder who took them 💀

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

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

Lol,me solving these easily (I'm Indian)

Solving inequalities is not tricky