Very well done. It is so helpful to avoid using any quadratic formula for getting students to actually understand what they are doing. However when the coefficient "a" is not 1 it is common to just teach "divide both side by a", but I really like this idea of rather factoring out the "a". Although the calculation will be similar as we are completing the square in the same manner, the division by "a" becomes more natural the way it is shown here compared to "just begin by dividing both side by a".
Congratulations that you’re smart enough that you consider this material “elementary”. I guess if you’re worried that his videos aren’t sophisticated enough, you should find someone else’s video channel; no one should feel obligated to teach all things at all levels.
Actually it's easier to solve by multiplying a and c and factoring b into that. i.e. in this case :: a = 4, b = 8, c = 5 .............. so a*c = 20 = 10 * 2 & b = 8 = 10 - 2 Thus equation becomes :: 4x^2 + 10x - 2x - 5 = 0 2x (2x + 5) - 1(2x + 5) = 0 (2x - 1) (2x + 5) = 0 i.e. x = 1/2 or -5/2
Please upload ur making learning math actually engaging sepescialy with someone with ADHD,Aspergers,and fatigue, along with a line of shitty math teachers, i would love to see more math videos involving chemistry,physics,or any math problems regarding fractions,imaginary number problems and solutions, and more
Hello sir,😊 I am your new subscriber sir Please answer the question Log((1-x)÷(1+x)) and limits -1 to +1 Integrate respect to x. Please answer the question its my maths exam creative question
Given: integral log((1 - x)/(1 + x)) dx from -1 to +1 Assuming log means natural log. If it doesn't, the answer is just different by a constant coefficient of 1/ln(10). Use integration by parts, with the log function differentiated, and 1 integrated: S _ _ _ D _ _ _ _ _ _ _ _ _ _ I + _ _ ln((1-x)/(1+x)) _ _ _ 1 - _ _ 2/(x^2 - 1) _ _ _ _ _ x Construct result: x*ln((1-x)/(1+x)) - integral 2*x/(x^2 - 1) dx Use u-substitution to unpack the integral: u = x^2 - 1 du = 2*x Thus it becomes: integral du/u, which integrates as: ln(|u|) = ln(|x^2 - 1|) Thus, our result is: x*ln((1-x)/(1+x)) - ln(|x^2 - 1|) + C Rewrite with log properties: x*ln(1-x) - x*ln(1+x) - ln(|x^2 - 1|) + C Evaluate from -a to +a, and take the limit as a approaches 1 from the left: [a*ln(1-a) - a*ln(1+a) - ln(|a^2 - 1|)] - [-a*ln(1 - -a) - a*ln(1 - a) - ln(|a^2 - 1|)] Plug in a=1, where possible to do so directly: ln(1-a) - ln(2) - ln(|a^2 - 1|) + ln(2) - ln(1- a) + ln(|a^2 - 1|) Observe that every term, has an equal and opposite negative term. This means, the result reduces to zero.
To visualize what we are doing in this example: Given 4*x^2 + 8*x - 5 = 0. Imagine constructing 4 copies of an actual square. The 4*x^2 term is represented by 4 squares that are x by x units. We now need to add a combination of shapes adding up to an area of 8*x to these 4 squares. And we need to do so in a way, that all our final shapes are identical squares, with the exception of the individual left over units (more on that later). Easiest way to do this, is to first divide 8*x by 4, to distribute equally to the squares. Thus, at any given square, we have to come up with shapes that add up in area to 2*x. This could be two rectangles, each 1 unit by x units. Add this smaller rectangle onto the right side and onto the bottom, of each of the 4 squares of x by x. Now, we ALMOST have four squares, of side length (x+1), but we are still missing something. Each of these squares is missing a 1x1 square in the corner. To do this, we add zero in a fancy way, by adding 4 qty 1x1 squares, and a debt of 4 area units. Have the debt of 4 area units join the existing debt of 5 area units, to get a debt of 9 area units remaining. This means we have: 4 squares of side length (x + 1), and a debt of 9 area units, corresponding to the form 4*(x + 1)^2 - 9 = 0
@@vansin001 Distance to the stars is found with stellar parallax. It works on the same principle as human depth perception, but at a much larger scale. We map the star's angular position in the sky on one day, and then again, 6 month's later. This gives us an isosceles triangle to work with, that has a known baseline of 300 million km, our planet's orbit diameter, and a known angle opposite the baseline. Solving for the unknown altitude of this triangle, tells us the distance to the star.
Can any one please explain how you got that (x^2 +2x + 1) ? when you are adding a value to a polynomial , shouldn't we add that to all terms ? Or direct me to some reference.
It's the perfect square identity. Given (a + b)^2, it expands as a*a + a*b + b*a + b*b, which simplifies to a^2 + 2*a*b + b^2. Let a = x and let b = 1, and you'll determine that (x + 1)^2 = x^2 + 2*x + 1. The idea in this example, is that you're given 4*x^2 + 8*x - 5 = 0. You'd like to rephrase this equation as 4 perfect squares, of side length (x + 1), and either add or subtract individual unit squares as necessary. So use that perfect square identity to find: 4*(x + 1)^2 = 4*x^2 + 8*x + 4, and line terms up with the original equation. You find that the essential difference is the final constant term, which is -5 instead of +4. This means we need to subtract 9, from our 4 copies of the perfect square identity, to get our original expression. 4*(x + 1)^2 - 9 = 4*x^2 + 8*x - 5.
A little complicated explanation. Based on my communication with students, those, who don't get this math, would loose the idea somewhere in the middle...
ax^2+bx+c=0 4x^2+8x-5=0 (2x-1)•(2x+5)=0x1,2= -b +-[(b^2-4ac)^1/2] ÷2a = -8 (+/-) (8^2-(4•4•-5)^1/2÷8= x1=0,5; x2=-2,5 The using of this two x-numbers in the first equal confims their correctness.
Because you are lining it up with the perfect square identity, (a + b)^2 = a^2 + 2*a*b + b^2. Given 4*x^2 + 8*x - 5 = 0, we'd like to line this up with 4 copies of the perfect square identity, and figure out how many unit squares, we either need to add or subtract to get it to balance. Let a = x, and expand out (x + b)^2, to get x^2 + 2*b*x + b^2. Then construct 4 copies of this: 4*(x + b)^2 = 4*x^2 + 8*b*x + 4*b^2 And you can see, that letting b = 1, this means 2 out of 3 of our coefficients are the same: 4*x^2 + 8*x + 4 Compare with: 4*x^2 + 8*x - 5 The essential difference is that our original is 9 units less than the four perfect squares. Thus, our square completes as: 4*x^2 + 8*x - 5 = 4*(x + 1)^2 - 9
Tomorrow is my final exam this came at the right time
Good luck on your final!
Thanks
@@mrhtutoringsir please
Do you have any video on differentiation and/or integration? 😢
The exam had 2 questions like about complete the square and i did this way
bro which type of exam you do ?
@@user-ts4zi4ud9k
I wish I was as good at math as Mr H. Your videos are a great help, thank you!
Very creative way of solving. Intuition.
Very well done. It is so helpful to avoid using any quadratic formula for getting students to actually understand what they are doing. However when the coefficient "a" is not 1 it is common to just teach "divide both side by a", but I really like this idea of rather factoring out the "a". Although the calculation will be similar as we are completing the square in the same manner, the division by "a" becomes more natural the way it is shown here compared to "just begin by dividing both side by a".
Great teacher.... energy. No fuzz
Congrutulations
Your videos are so effective to me. A new equation is set on my mind
Same here
Never heard of CTS in HS or college. Thanks for the revelation
Thank you Professor! I really need the basics so appreciate all your videos.
Glad to hear that!
Hey Mr H Im loving the videos! Can i ask for your name?
Mr h why do you only teach elementary stuff? I know basics are important but most are going to give undergrad exams in the next 2-3 years...
Because people don't know the "elementary" stuff!
@@GarethDaviesUKexactly! Like me 😭
Congratulations that you’re smart enough that you consider this material “elementary”. I guess if you’re worried that his videos aren’t sophisticated enough, you should find someone else’s video channel; no one should feel obligated to teach all things at all levels.
@@jimbobago yes, ur right 👍 I think blackpenredpen is a fun channel to relax on as it has some pretty crazy problems
Could u make a video of completing the cube
Great channel !
Glad you enjoy it!
Can u do a video for asvab math
Good work
When do you complete the square vs factoring? I was able to factor this just as easy using your method from another video and got same answer.
Actually it's easier to solve by multiplying a and c and factoring b into that. i.e. in this case ::
a = 4, b = 8, c = 5 .............. so a*c = 20 = 10 * 2 & b = 8 = 10 - 2
Thus equation becomes :: 4x^2 + 10x - 2x - 5 = 0
2x (2x + 5) - 1(2x + 5) = 0
(2x - 1) (2x + 5) = 0
i.e. x = 1/2 or -5/2
Your method is much easier !! Thanks
nice
nice
fun lesson. should have more likes.
I watch these videos and am at times confounded - so much so that I forget to like. I plan to do better. Thank you so... much Bro!
You are so welcome!
Pls make videos on trigonometry
Will do ~
Sir can you teach INTEGRATION and DIFFERENTIATION it's a challenge for us students🙏
exponents/logs quadratics... much practice needed. thanks
I thought the square root radical yields only a positive number. Why is this different?
Professor, do you a Twitter (X) account?!
Another way you can solve this is multiplying the 4 by -5
X^2+8x-20
(x+10)(x-2)
Add the 4 back in
(4x+10)(4x-2)
X=5/2 or x=1/2
Hello sir, may I know what kind of projector do use?
Please upload ur making learning math actually engaging sepescialy with someone with ADHD,Aspergers,and fatigue, along with a line of shitty math teachers, i would love to see more math videos involving chemistry,physics,or any math problems regarding fractions,imaginary number problems and solutions, and more
Why did you mention "its always a two" - are we always supposed to decide the middle term by 2? because of the " a^2+2ab+b^2) ??
Precisely. You always divide the middle term by 2, in order to match it to a perfect square identity.
ohh sir thank you🙂
4x^2+8x-5=0
4x^2+8x=5
4x^2+8x+4=4+5
(2x+2)^2=9
|2x+2|=3
2x+2=3
2x=1
x=0.5 ❤
2x+2=-3
2x=-5
x=-2.5 ❤
Hello sir,😊
I am your new subscriber sir
Please answer the question
Log((1-x)÷(1+x)) and limits -1 to +1
Integrate respect to x.
Please answer the question its my maths exam creative question
Given:
integral log((1 - x)/(1 + x)) dx from -1 to +1
Assuming log means natural log. If it doesn't, the answer is just different by a constant coefficient of 1/ln(10).
Use integration by parts, with the log function differentiated, and 1 integrated:
S _ _ _ D _ _ _ _ _ _ _ _ _ _ I
+ _ _ ln((1-x)/(1+x)) _ _ _ 1
- _ _ 2/(x^2 - 1) _ _ _ _ _ x
Construct result:
x*ln((1-x)/(1+x)) - integral 2*x/(x^2 - 1) dx
Use u-substitution to unpack the integral:
u = x^2 - 1
du = 2*x
Thus it becomes:
integral du/u, which integrates as:
ln(|u|) = ln(|x^2 - 1|)
Thus, our result is:
x*ln((1-x)/(1+x)) - ln(|x^2 - 1|) + C
Rewrite with log properties:
x*ln(1-x) - x*ln(1+x) - ln(|x^2 - 1|) + C
Evaluate from -a to +a, and take the limit as a approaches 1 from the left:
[a*ln(1-a) - a*ln(1+a) - ln(|a^2 - 1|)] - [-a*ln(1 - -a) - a*ln(1 - a) - ln(|a^2 - 1|)]
Plug in a=1, where possible to do so directly:
ln(1-a) - ln(2) - ln(|a^2 - 1|) + ln(2) - ln(1- a) + ln(|a^2 - 1|)
Observe that every term, has an equal and opposite negative term. This means, the result reduces to zero.
Add (b/2a)2 on both sides
0:50 why would he divide by 2? How can you just divide a number by 2 for no rhyme or reason without applying it to any other part of the equation?
👍🙏
Can you explain the science of how that works. Something about the stars I think? If you don't know Can you tell us?
To visualize what we are doing in this example:
Given 4*x^2 + 8*x - 5 = 0.
Imagine constructing 4 copies of an actual square.
The 4*x^2 term is represented by 4 squares that are x by x units. We now need to add a combination of shapes adding up to an area of 8*x to these 4 squares. And we need to do so in a way, that all our final shapes are identical squares, with the exception of the individual left over units (more on that later).
Easiest way to do this, is to first divide 8*x by 4, to distribute equally to the squares. Thus, at any given square, we have to come up with shapes that add up in area to 2*x. This could be two rectangles, each 1 unit by x units.
Add this smaller rectangle onto the right side and onto the bottom, of each of the 4 squares of x by x. Now, we ALMOST have four squares, of side length (x+1), but we are still missing something. Each of these squares is missing a 1x1 square in the corner.
To do this, we add zero in a fancy way, by adding 4 qty 1x1 squares, and a debt of 4 area units. Have the debt of 4 area units join the existing debt of 5 area units, to get a debt of 9 area units remaining. This means we have: 4 squares of side length (x + 1), and a debt of 9 area units, corresponding to the form 4*(x + 1)^2 - 9 = 0
@@carultch I know how to solve the equation, but I don't know how the knowledge used to calculate things like how far away is a star.
@@vansin001 Distance to the stars is found with stellar parallax. It works on the same principle as human depth perception, but at a much larger scale. We map the star's angular position in the sky on one day, and then again, 6 month's later.
This gives us an isosceles triangle to work with, that has a known baseline of 300 million km, our planet's orbit diameter, and a known angle opposite the baseline. Solving for the unknown altitude of this triangle, tells us the distance to the star.
Just divide everything by 4. And treat it like case of a=1
Sir, couldn't we have just split that middle term +8x and factorised the equation to arrive at the same answers...
ขอบคุณครับ
Can any one please explain how you got that (x^2 +2x + 1) ? when you are adding a value to a polynomial , shouldn't we add that to all terms ? Or direct me to some reference.
It's the perfect square identity. Given (a + b)^2, it expands as a*a + a*b + b*a + b*b, which simplifies to a^2 + 2*a*b + b^2. Let a = x and let b = 1, and you'll determine that (x + 1)^2 = x^2 + 2*x + 1.
The idea in this example, is that you're given 4*x^2 + 8*x - 5 = 0. You'd like to rephrase this equation as 4 perfect squares, of side length (x + 1), and either add or subtract individual unit squares as necessary. So use that perfect square identity to find:
4*(x + 1)^2 = 4*x^2 + 8*x + 4, and line terms up with the original equation. You find that the essential difference is the final constant term, which is -5 instead of +4. This means we need to subtract 9, from our 4 copies of the perfect square identity, to get our original expression.
4*(x + 1)^2 - 9 = 4*x^2 + 8*x - 5.
@@carultch Thanks a lot for your help, and the time you have taken for such a detailed explanation.
Nice video for everyone
Challenge you to solve iit jee question 😂😂😂😂😂
can you just quadratic equation this?
Of course.
Thanks for your support and continue
4x² + 8x - 5 = 0
x² + 2x - 5/4 = 0
x² + 2x = 5/4
x² + 2x + 1 = 5/4 + 1
(x+1)² = 9/4
x + 1 = ±√(9/4)
x = ±√(9/4) - 1
x = ± 3/2 - 1
In the fourth step, do you always add one to both sides? Otherwise much easier to follow. Thanks
@@RepentOrPerishL133 nope. the formula is: b² ÷ 4a² (b/2a)² , a = 1, b = 2
b² ÷ 4a² = 4 ÷ 4
= 1
So i'm adding 1 to have the left side factorable.
Other for example:
x² + 8x + 7 = 0
x² + 8x = -7
x² + 8x + ? = -7 + ?
? = b² ÷ 4a²
? = 16
x² + 8x + 16 = 9
(x+4)² = 9
x + 4 = ±3
x = ±3 - 4
Add (b/2a)2 on both sides
@@michucz Thank you
5/4+1= 9/4 how this?
A little complicated explanation. Based on my communication with students, those, who don't get this math, would loose the idea somewhere in the middle...
At 1:50 sir you forgot the +- on the right hand side
Next time I'll skip that that step.
ax^2+bx+c=0 4x^2+8x-5=0 (2x-1)•(2x+5)=0x1,2= -b +-[(b^2-4ac)^1/2] ÷2a = -8 (+/-) (8^2-(4•4•-5)^1/2÷8= x1=0,5; x2=-2,5 The using of this two x-numbers in the first equal confims their correctness.
0:47 but why can you do that
Because you are lining it up with the perfect square identity, (a + b)^2 = a^2 + 2*a*b + b^2.
Given 4*x^2 + 8*x - 5 = 0, we'd like to line this up with 4 copies of the perfect square identity, and figure out how many unit squares, we either need to add or subtract to get it to balance.
Let a = x, and expand out (x + b)^2, to get x^2 + 2*b*x + b^2. Then construct 4 copies of this:
4*(x + b)^2 = 4*x^2 + 8*b*x + 4*b^2
And you can see, that letting b = 1, this means 2 out of 3 of our coefficients are the same:
4*x^2 + 8*x + 4
Compare with:
4*x^2 + 8*x - 5
The essential difference is that our original is 9 units less than the four perfect squares. Thus, our square completes as:
4*x^2 + 8*x - 5 = 4*(x + 1)^2 - 9
4x^2 + 8x + 4 - 9 = 0
(2x + 2)^2 - 3^2 = 0
(2× + 5)(2x - 1) = 0
x = -5/2 or x = 1/2
( 2 x -2) ^2 = 2^2 + 5
2 x = 2 + 3, 2 -3
x = 5/2 , - 1/2
(2x+5)(2x-1)=0
... 4X^2 + 8X - 5 = 0 ... 4X^2 + 8X = 5 ... X^2 + 2X = 5/4 ... ( X + 1 )^2 - 1 = 5/4 ... ( X + 1 )^2 = 4/4 + 5/4 ... ( X + 1 )^2 = 9/4 ... X + 1 = - 3/2 v X + 1 = 3/2 ... X1 = - 5/2 v X2 = 1/2 .... S = { - 5/2 , 1/2 } ...
Nice
@@soundtracksfortheblind ... I'm glad it helped you ... thank you and good luck with maths, Jan-W
@@jan-willemreens9010 Thank you jan, same to you
U went too far 😂