Why solving a rational inequality is tricky!

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

[ManjulaMathew-wbzn]

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

[akarooyyy]

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

[edwardblair]

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.

[miacoolfacex]

I was so stuck! perfect explaining Thanks

[OmnipresentPotato]

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

[Ninja]

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.

[uetdhrv]

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

[leealex]

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

[matdryz]

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)

[honestadministrator]

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

[butterspread]

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 ?`

[mryip]

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

[張大刀-gy]

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<0, we havex^2 -3x - 4 < 0
we get x < -4/3 and x < 4 and x > -1 => x = nothing
you only need to know what will happend when divide neg number

[teddybearcn]

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.

[Hanible]

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.

[wassollderscheiss]

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?

[gheffz]

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

[tonyennis]

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.

[ValidatingUsername]

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?

[Pengochan]

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.