I'd love to buy this merch but with Amazon the international shipping is almost $20. I really wish you'd open more suppliers near Asia but I think that's asking for too much lol
Blackpenredpen, Can you do videos teaching irrational numbers?, Please? I would like to know how number e to the power i*pi works solving irrational equations
Ever since I have discovered this channel I have been learning so many incredibly valuable things !! Cannot put it in words how grateful I am for this and finding my fascination for mathematics to be growing and growing ♡
i just started watching your videos and i love them , whenever my brother used to watch you i used to think who is this weird dude with a pokeball lol but honestly tysm , i have gotten to learn many new things
Another problem that would have been good to have as a bonus would be x^4+3x^2=1. It isn't necessarily a quadratic equation but it resembles one and it uses the numbers 1, 2, 3, and 4.
Ah yes, the hidden quadratic equations, inside higher order polynomials. Simply let W = x^2, and it becomes a quadratic equation in terms of W. Then, our solutions for x, becomes both of the square roots of both solutions for W
@@seanbastian4614 I came across a similar problem like this, setting up an example for Lagrange multipliers. The problem stumped Wolfram Alpha's computation time, but with a change of variables, letting V replace x^2 and W replace y^2, it was much easier to solve.
@@seanbastian4614 Think of it this way. You're given a hill's shape, with its z-position, as a function of x and y horizontal coordinates. You are also given a trail on this hill, also as a function of x and y. You'd like to find the highest point on the trail, or the lowest point. What you do is look for places where the trail is locally parallel to a contour line, which is where you'll have a local extreme point in the elevation on the path. The Lagrange multiplier is dummy variable you don't ultimately care about It helps you set up a system of equations, to look for alignment with the hill's gradient, and the path's normal vector.
Could you please form a playlist in which u will teach us calculus from basic to standard The questions of quadratic are very easy for us but we get problems in calculus only so teach us calculus if possible For that i shall be very thankful to you 😊
I wish teachers would explain that the sqrt(x^2) is |x| (absolute value of x) and that is why to solve the absolute value equation, we need to put down the plus minus. Right now, it’s two mistakes that confuse students. By doing the sqrt cancel and put down the plus minus, students might make 2 wrong connections in their brain. That sqrt(x^2) is x and sqrt(4) for example is plus minus 2.
@@dananajj Koen isn't saying otherwise. They actually point out that many people make the mistakes mentioned due to a lack of clarity in classes regarding how rooting works
@@skc4188 Here's an interesting experiment for you to try with rooting. Try calculating (-8)^(1/3) and (-8)^0.3333 in Google Calculator, and observe that you will get significantly different results, even though you are ultimately solving the same approximate problem. This is why less-sophisticated calculators can give you an error when you attempt the same problem, even though the solution is a real number. So, can you explain why you get significantly different results, and why each result is equally valid (approximately) to the original expression of cube root of -8? I know the answer. Let me know if you can figure it out.
@@Apollorion True, but what are those other solutions, and why does the exponent of 0.3333 cause the calculator to default on a solution than is nowhere near x=-2? I'll give you a clue. If you try this with 5th roots, such as (-32)^(1/5), you will get the real solution of -2, whether the exponent is 1/5, or whether the exponent is 0.2. But there is a reason why 3rd roots get stuck on a complex solution, if the calculator rounds the fractional exponent of 1/3.
For Q3, you can actually take the square roots of both sides and it would be quicker too. This would give you (3x-1) = 2x^(1/2) , --> 3x -2x^(1/2) -1 = 0 , which you can treat as a quadratic by substituting y=x^(1/2) to give y^2-2x-1=0 (I didnt take into account the other root as i would get the same anwser anyway). This can be solve to give (3y-1)(y+1) , (if you had considered the other root, ie -2x^(1\2), you would have the same expression with the signs swapped in the factored form, but since you are solving for y^2, ie x, the signs swap doesnt matter as the the result would be positive regardless of sign). y^2=(1/3)^2 = 1/9 and y^2=(-1)^2 = 1
1:06:03 Meanwhile, I start by taking the square root of both sides, and somehow arrive at the correct answer in a way that is arguably easier than your method.
For Q2, arent you supposed to check if the solutions are valid by inserting the x-values in the original equation and see if left side = right side since there can be false solutions when squaring both sides?
When you're not certain you'll always do well to verify the solutions in the original equation. Those false solutions you're thinking of come from (-A)(-A)=(A)(A) and that hence (A)(A)=(B)(B) can be solved not only by A=B but also by A=-B. If you begin an equation equivalent to A=B and wish to find the solutions by squaring both sides then you'd indeed added the false solution A=-B. The Q2 however already in the beginning is of the form (A)(A)=(B)(B), and what BPRP does is not square both sides but taking the square root of both sides to find both A=B and A=-B as solutions. (Think of A and B as some arbitrary expressions.)
@@VenkataB123 Hashtags are used in C programming, to include libraries. Like #include math, will include the mathematics package, that allows you to access your transcendental functions, beyond the functions that come standard with the programming language. You determine which library to include, and which ones not to include, in order to make your program more memory efficient.
@@VenkataB123 Another significance. Suppose you want to specify that e refers to a generic variable, that has nothing to do with Euler's number. Or that gamma has nothing to do with the Euler-Mascheroni constant. The hashtag could override a default use of a variable.
@@carultch They asked specifically in Mathematics, so your programming argument is irrelevant. As for the other comment, no one would ever use "e" as a variable. e is always used for Euler's number. Using letters cautiously is enough. We don't need to unecessarily complicate things, especially not in Math, where we aim to simplify stuff.
I think you mean quadratic inequalities. Equations such as -x^2 + 5*x > 6. It's just like solving the quadratic equation, to find the boundaries of the valid domain of x-values that satisfy this inequality, which in this case are x=2 and x=3. In order to determine how to establish whether it is between these bounds, or outside these bounds, you need to consider the nature of this eqaution. The negative on x^2 tells you that it is concave-down, and therefore peaks out above the line y=6, in the interval of x E (2, 3). So this means the solution is x E (2, 3) or 2 < x < 3. It is an exclusive interval, excluding the 2 and 3, because the original inequality uses an exclusive greater than sign.
Hey plz read my comment... I have a great qn Suppose u have 1 yellow colour highlighter. U can produce only yellow colour. Means 1 colour is produced with 1 highlighter. If u have yellow and blue highlighter then u can produce blue , yellow and (yellow + blue=) green colour. Means if if u have 2 highlighters then u can produce 3 colours. Similarly if u have 3 colour highlighter then u can produce 7 colours So, how many colour can u produce with x number of highlighter(of diff colours)?????
Shop my math t-shirts and hoodies on Amazon now: www.amazon.com/shop/blackpenredpen/list/290XVHH0MDUO9?ref_=aip_sf_list_spv_ofs_mixed_d
I'd love to buy this merch but with Amazon the international shipping is almost $20.
I really wish you'd open more suppliers near Asia but I think that's asking for too much lol
Blackpenredpen,
Can you do videos teaching irrational numbers?, Please?
I would like to know how number e to the power i*pi works solving irrational equations
Hi My Inspiring Teacher @blackpenredpen please Check your Gmail Thank you for your time
It's always a wonderful day when blackpenredpen uploads
Yeah
Tbh I've only noticed today that his channel name is blackpenredpen and not blackpendragon
It is always a blackpenredpen uploads when wonderful day
#MathHelpers#MathHelpers
@PandaGMD #MathHelpers#MathHelpers
Ever since I have discovered this channel I have been learning so many incredibly valuable things !! Cannot put it in words how grateful I am for this and finding my fascination for mathematics to be growing and growing ♡
#MathHelpers#MathHelpers
i just started watching your videos and i love them , whenever my brother used to watch you i used to think who is this weird dude with a pokeball lol but honestly tysm , i have gotten to learn many new things
PROF what i learned from THIS video is factring and compliting square thanks for that i knew only the quadratic formula before
#MathHelpers#MathHelpers
Wow, that was an incredible lesson! Almost 100 minutes of math 🇧🇷✌️
Watch his 100 series
100 integral
100 calculus
100 derevative
The titles are something like that and they are of minimum 5 hrs
@@im.rishabh_ofc 🤣
#MathHelpers#MathHelpers
@@elevolucionestoica #MathHelpers#MathHelpers
Another problem that would have been good to have as a bonus would be x^4+3x^2=1. It isn't necessarily a quadratic equation but it resembles one and it uses the numbers 1, 2, 3, and 4.
Ah yes, the hidden quadratic equations, inside higher order polynomials. Simply let W = x^2, and it becomes a quadratic equation in terms of W. Then, our solutions for x, becomes both of the square roots of both solutions for W
@@carultch Exactly my thought. I always feel students need to see those hidden quadratics.
@@seanbastian4614 I came across a similar problem like this, setting up an example for Lagrange multipliers.
The problem stumped Wolfram Alpha's computation time, but with a change of variables, letting V replace x^2 and W replace y^2, it was much easier to solve.
@@carultch Ok. I don’t understand Lagrange Multipliers that much but still nice application.
@@seanbastian4614 Think of it this way. You're given a hill's shape, with its z-position, as a function of x and y horizontal coordinates. You are also given a trail on this hill, also as a function of x and y. You'd like to find the highest point on the trail, or the lowest point.
What you do is look for places where the trail is locally parallel to a contour line, which is where you'll have a local extreme point in the elevation on the path. The Lagrange multiplier is dummy variable you don't ultimately care about It helps you set up a system of equations, to look for alignment with the hill's gradient, and the path's normal vector.
Could you please form a playlist in which u will teach us calculus from basic to standard
The questions of quadratic are very easy for us but we get problems in calculus only so teach us calculus if possible
For that i shall be very thankful to you 😊
#MathHelpers#MathHelpers
Next : 12 cubic equations questions
IT IS A PLEASURE TO SEE YOUR VIDEOS SIR.
#MathHelpers#MathHelpers
Great service to science
good job, 76 more quadratic equations to go!
#MathHelpers#MathHelpers
I wish teachers would explain that the sqrt(x^2) is |x| (absolute value of x) and that is why to solve the absolute value equation, we need to put down the plus minus. Right now, it’s two mistakes that confuse students. By doing the sqrt cancel and put down the plus minus, students might make 2 wrong connections in their brain. That sqrt(x^2) is x and sqrt(4) for example is plus minus 2.
Sqrt(4) is just 2, positive 2. It's not plus minus 2.
@@dananajj Koen isn't saying otherwise. They actually point out that many people make the mistakes mentioned due to a lack of clarity in classes regarding how rooting works
@@skc4188 Here's an interesting experiment for you to try with rooting. Try calculating (-8)^(1/3) and (-8)^0.3333 in Google Calculator, and observe that you will get significantly different results, even though you are ultimately solving the same approximate problem. This is why less-sophisticated calculators can give you an error when you attempt the same problem, even though the solution is a real number.
So, can you explain why you get significantly different results, and why each result is equally valid (approximately) to the original expression of cube root of -8? I know the answer. Let me know if you can figure it out.
@@carultch Well, 1/3 isn't 0.3333 and xxx = -8 has more solutions than just x = -2.
@@Apollorion True, but what are those other solutions, and why does the exponent of 0.3333 cause the calculator to default on a solution than is nowhere near x=-2?
I'll give you a clue. If you try this with 5th roots, such as (-32)^(1/5), you will get the real solution of -2, whether the exponent is 1/5, or whether the exponent is 0.2. But there is a reason why 3rd roots get stuck on a complex solution, if the calculator rounds the fractional exponent of 1/3.
I WISH WISH WISH there were videos just like this but for Physics questions....
Wonderful work. Thank you.
#MathHelpers#MathHelpers
make a video on the different completing the square method! super nice!
#MathHelpers#MathHelpers
Please make complex numbers study guide.
You are amazing keep uploading I love your math videos
#MathHelpers#MathHelpers
I might be watching pausa with but full video no minute wasted PROF 👍🤗
For Q3, you can actually take the square roots of both sides and it would be quicker too. This would give you (3x-1) = 2x^(1/2) , --> 3x -2x^(1/2) -1 = 0 , which you can treat as a quadratic by substituting y=x^(1/2) to give y^2-2x-1=0 (I didnt take into account the other root as i would get the same anwser anyway). This can be solve to give (3y-1)(y+1) , (if you had considered the other root, ie -2x^(1\2), you would have the same expression with the signs swapped in the factored form, but since you are solving for y^2, ie x, the signs swap doesnt matter as the the result would be positive regardless of sign). y^2=(1/3)^2 = 1/9 and y^2=(-1)^2 = 1
#MathHelpers#MathHelpers
You teach very good. I want to get better at math and the more basic things are lacking a bit. 33:23 I want to do all the questions.
#MathHelpers#MathHelpers
Thank you, Sir. I like this video
33 23 im here the night before my exam!
Respect to this man...he is trying hard to speak English...respect sir 👍
I want AI to progress the point where I can type in a maths problem and a blackpenredpen avatar will run through the explanation.
Please tell me sir about board size, ❤❤the lecture
Thanks PROF more videos like this 10 hour better👏
#MathHelpers#MathHelpers
i was watching at 33:24 :)
I am still here, sir 33:23
😇😇😇
Thank you!
Welcome Sir.
1:36:43 i multiplied by 3 and i came up with the same answer its realy cool
#MathHelpers#MathHelpers
Plz make a video on logarithmic functions
24 questions, 1 x 2 x 3 x 4, this was not a coincidence, was it?
you're my hero
mine as well. very well done, master!
33:23 here :)
Awesome! Thanks!
1:06:03
Meanwhile, I start by taking the square root of both sides, and somehow arrive at the correct answer in a way that is arguably easier than your method.
In sweden we do
PQ Formula
x²+px-q=0
For Q2, arent you supposed to check if the solutions are valid by inserting the x-values in the original equation and see if left side = right side since there can be false solutions when squaring both sides?
When you're not certain you'll always do well to verify the solutions in the original equation. Those false solutions you're thinking of come from (-A)(-A)=(A)(A) and that hence (A)(A)=(B)(B) can be solved not only by A=B but also by A=-B. If you begin an equation equivalent to A=B and wish to find the solutions by squaring both sides then you'd indeed added the false solution A=-B. The Q2 however already in the beginning is of the form (A)(A)=(B)(B), and what BPRP does is not square both sides but taking the square root of both sides to find both A=B and A=-B as solutions. (Think of A and B as some arbitrary expressions.)
@@Apollorion Okay thank you for clarifying! I got the two mixed up!
33:25
5:53 is # ever going to mean "hastag" in mathematics?
What significance would it have by meaning that?
@@VenkataB123 Hashtags are used in C programming, to include libraries. Like #include math, will include the mathematics package, that allows you to access your transcendental functions, beyond the functions that come standard with the programming language. You determine which library to include, and which ones not to include, in order to make your program more memory efficient.
@@VenkataB123 Another significance. Suppose you want to specify that e refers to a generic variable, that has nothing to do with Euler's number. Or that gamma has nothing to do with the Euler-Mascheroni constant. The hashtag could override a default use of a variable.
@@carultch They asked specifically in Mathematics, so your programming argument is irrelevant.
As for the other comment, no one would ever use "e" as a variable. e is always used for Euler's number. Using letters cautiously is enough. We don't need to unecessarily complicate things, especially not in Math, where we aim to simplify stuff.
Can you do a video on quadratic in-equations, i’ve got a uni entrance exam in 3 days so I’d appreciate it
I think you mean quadratic inequalities. Equations such as -x^2 + 5*x > 6.
It's just like solving the quadratic equation, to find the boundaries of the valid domain of x-values that satisfy this inequality, which in this case are x=2 and x=3. In order to determine how to establish whether it is between these bounds, or outside these bounds, you need to consider the nature of this eqaution. The negative on x^2 tells you that it is concave-down, and therefore peaks out above the line y=6, in the interval of x E (2, 3). So this means the solution is x E (2, 3) or 2 < x < 3. It is an exclusive interval, excluding the 2 and 3, because the original inequality uses an exclusive greater than sign.
@@carultch yes yes i meant that
Now black-red-blue pens 😂👏👏👏
Why couldn’t this come before the Algebra I and 2 Regents :(
23:40 That would have been great, though 😂🎤
33:22
33:37 yep in the class
I really liked your video on quadratic equations, what suggestion do you give me to learn the English language?
#MathHelpers#MathHelpers
4XX + 25X + 36 = 0
*. XX + 6,25X + 9 = 0
9 = 2x4,5= 2,25x4 = 3x3
2,25 + 4 = 6,25 = (b')
** 2,25: 4(2,25x2,25)+25(2,25)+36
20,25 + 56,25 + 36 # 0
-2,25: 4(-2,25x2,25) + 25(-2,25) + 36
20,5 - 56,25 + 36 = 0
** 4: 4(4x4) + 25x4 + 36
64 + 100 + 36 # 0
-4 : 4(-4x-4)+25(-4)+36
64 - 100 + 36 = 0
*** X' = -2,25, X" = -4./.
You forgot a semi colon in the time stamp for the last question (Description)
😁😁We miss you, Mr. Shaw. Why haven't you posted videos for a long time
Hey plz read my comment... I have a great qn
Suppose u have 1 yellow colour highlighter. U can produce only yellow colour. Means 1 colour is produced with 1 highlighter.
If u have yellow and blue highlighter then u can produce blue , yellow and (yellow + blue=) green colour. Means if if u have 2 highlighters then u can produce 3 colours.
Similarly if u have 3 colour highlighter then u can produce 7 colours
So, how many colour can u produce with x number of highlighter(of diff colours)?????
This ultimately is a question of the nature of human vision, rather than any kind of mathematical combination of the number of highlighters you have.
33:20 still watching
Thank you!
The number 1 is always the “ghost” number. Woooo - the horrors of math. The ghost 1! Lmao kidding.
PROF if U buy smart board U don't have to iraze anymore😊
Apply the Halloween theme to math. Horrors! 😁😈
Can someone please explain to me how x^2+4x+4 gets turned into (x+2)^2
it is simple , just we can expand the (x+2)^2 to (x+2) (x+2) and multiply it and we get x^2+4x+4 and we can do the reverse.
(X + 1)(X + 3) = 24
*. XX + 3X + X + 3 = 24
XX + 4X - 21 = 0
21 = 3 x 7
7 - 3 = 4 = (b)
**. 7 : 7x7 + 4x7 - 21
49 + 28 - 21 # 0
-7: (-7x-7) + 4(-7) - 21
49 - 28 - 21 = 0
**. 3: 3x3 + 4x3 - 21
9 + 12 - 21 = 0
***. X' = 3 , X" = -7./.
Bro has a whole stock of markers 🤣
Me: *Pulls out calculator*
for a moment i thought it was the quartic formula💀
Yo
Yo
Yo
Yo
You look younger
∫
Hey can you help me with this. Solve the Inequality (log₃x)²
33:23