Dear Sir, I found a simpler way to solve this : Add minus 4 to both sides of the given. use difference of two squares and simplify. Here (2-x) is on both sides which gives x=2 simplify more and the resulting cubic equation gives another x=2 . Finally solve a simple quadratic with the other two answers . I like your method too. Thanks for your GREAT videos 🙂
I've tried your method and it worked for me, it maybe not such elegant, than this substitution which came literally from nowhere (really impressed, I mean, I know such type of problems, still I was always wondering, how people could guess to move in that direction, obviously after trying another ways). Anyhow the version with factoring is also cool and subtracting 4 is also a brilliant idea, which doesn't really lies on the surface
There's been a little mistake on 5:02 and although I've replied to someone catching on it in the comments, I'm repeating it here because I see more people are also noticing. It was something I was going to write about, too. My reply was: "I've seen that too, and it's actually a "fortunate mistake". What I mean with this, the mistake carries from the previous step. He had xˆ4-2xˆ3+xˆ2+xˆ2 on the numerator. This equals to xˆ4-2xˆ3+2xˆ2. When he splits this in two parts, he writes xˆ4 on one side, and then the other side, he writes it as -(2xˆ3+2xˆ2) This is where the first mistake is done. The numerator grouping should equal to -(2xˆ3-2xˆ2), which is then equal to -2xˆ2(x-1) He writes, incorrectly, that grouping as -(2xˆ3+2xˆ2), but then he actually factors it correctly, fixing inadvertently the previous mistake. I was about to write about this too. I hope I helped with this. 😅 "
The part of my personality that resonates as autistic wants to pixelate the "+" you write at 4:22. I paused a long time when I first saw it. The next step corrected the mistake, relieving most of my anxiety, but the mistake itself continued to make me nervous. Worth it for the magical solution, though. That solution is what I clicked the video for!
Thank you for this video! As x is not 1 and 8-x² must be non-negative (as equal to a square), x must be between -2sqrt(2) and 2sqrt(2), 1 excluded. So if there is at least one solution in Z, it should be -2, -1, 0 or 2. Luckily 2 is solution, even twice actually so we are left with a second-degree polynom in the end. It doesn't work each time but it was worth the try here. 🙂
Now x² + (x/(x - 1))² = 8. So, x² and hence (x/(x - 1))² must both be small, and in fact -3 < x < 3, such that x ≠ 1. So if we try the possible integer solutions, we get lucky and find that x = 2 is a solution. Now as x² + (x/(x - 1))² = 8, then x²(x - 1)² + x² = 8(x - 1)² So, x⁴ - 2x³ - 6x² + 16x - 8 = 0 = f(x) say. Applying synthetic division to f(x) where x = 2, we obtain f(x) = (x - 2)(x³ - 6x + 4) = 0. Let's try and factorise x³ - 6x + 4 = 0 using the rational roots theorem. So possible rational roots are x = ±1, ±2 or ±4. Trying x = 2, we see it is a root. Using synthetic division on x³ - 6x + 4 = 0 with x = 2, we can factorise x³ - 6x + 4 = 0 as (x - 2)(x² + 2x +2)= 0 So f(x) = (x - 2)²(x² + 2x + 2) = 0 and it's easy to find the other two roots (solution) using this equation.
By inspection x=2 is one root. There must be a second real root. x^4-2x^3+x^2-8x^2+16x-8+x^2=x^4-2x^3-6x^2+16x-8=0 Divide by (x-2): x^3-6x+4=0; By inspection x=2 is a second real root. Divide by (x-2): x^2+2x-2=0; x=(-2±2√3)/2=-1±√3
There is a mistake in terms of the positive sign concerning the second term of the numerator of the second fraction at the second line exactly beforre time point 4:30.We should put -2X unstead of +2X as the two fration members of the numerator share the same negative outside sign similar they are between brakets.
The equation to solve is 8 − x² = (x/(x − 1))² These types of equations seem to be tremendously popular on TH-cam lately, but the proposed solution methods are often awkward at best. This equation can be solved elegantly (1) without having to mess with fractions and (2) without substitutions. The key here is to use an ancient technique which is often useful in solving algebraic equations, i.e. converting a product of quantities into a difference of squares. If we first bring x² over to the right hand side and then multiply both sides by (x − 1)² to eliminate the fraction (which is allowed since x ≠ 1) and then swap sides we can write the equation as x²(x − 1)² + x² = 8(x − 1)² I ultimately want to create a difference of two squares, and to do so I first incorporate the term x² at the left hand side into the product x²(x − 1)² which can be done since they both have a common factor x². Expanding (x − 1)² at the left hand side we have x²(x² − 2x + 1) + x² = 8(x − 1)² and taking out the common factor x² at the left hand side this gives x²((x² − 2x + 1) + 1) = 8(x − 1)² which is x²(x² − 2x + 2) = 8(x − 1)² We now have a product of two quantities x² and x² − 2x + 2 at the left hand side which we can turn into a difference of squares. The average (arithmetic mean) of x² and x² − 2x + 2 is half their sum which is x² − x + 1 and the difference between x² and x² − 2x + 2 is x² − (x² − 2x + 2) = 2x − 2 = 2(x − 1) so half their difference is (x − 1). So, we have x² = (x² − x + 1) + (x − 1) and x² − 2x + 2 = (x² − x + 1) − (x − 1) and we can therefore rewrite the equation as ((x² − x + 1) + (x − 1))((x² − x + 1) − (x − 1)) = 8(x − 1)² and applying the difference of two squares identity (a + b)(a − b) = a² − b² to the left hand side this can be written as (x² − x + 1)² − (x − 1)² = 8(x − 1)² and bringing the term 8(x − 1)² from the right hand side over to the left hand side this gives (x² − x + 1)² − 9(x − 1)² = 0 We now again have a difference of two squares on the left hand side and again applying the difference of two squares identity a² − b² = (a + b)(a − b) to the left hand side we have (x² − x + 1 + 3(x − 1))(x² − x + 1 − 3(x − 1)) = 0 which is (x² + 2x − 2)(x² − 4x + 4) = 0 and applying the zero product property this means x² + 2x − 2 = 0 ⋁ x² − 4x + 4 = 0 and solving these quadratic equations by completing the square we have (x + 1)² = 3 ⋁ (x − 2)² = 0 x = −1 + √3 ⋁ x = −1 − √3 ⋁ x = 2 and the equation is solved. Note that x = 2 is a double root which is why we have only three different solutions.
this is due to the previous line he made a mistake grouping the numbers with a minus, -(2x^3+2x^2) here should be -(2x^3-2x^2) so that we could take (x-1) out.
really that is just as hard if not harder to find than the straight forward working out. move 8-x^2 to other side as x^2-8. Next common denominator of (x-1)^2. simplify the numerator and you should get something nice.
4:22 it's 2 (times) cube of x minus(-) 2 (times) square of x (2x^3-2x^2). Acctually this reminds me of a classical math problem: th-cam.com/video/CZKD7rffF0M/w-d-xo.html which can be solved both in geometry way and in algebra way. @PrimeNewtons I have left a comment in that linked video showing how to solve this problem with algebras,just do some transformation to the equation and set something to a new "a" (in that case it was (x²+1)/x=a),and such that we get a new equations concerning a. but most of the times,I don't think people could easily find that "a" for a substitution,and then we usually get the standard power of 4th equation,in this case it's: x^4-2x^3-6x^2+16x-8=0 well,that doesn't mean your algebra way of solving this problem is going to end here,what I recommend is to do "the double cross" trick: a1 b1 c1 a2 b2 c2 a1a2=1(x^4);c1c2=-8(x^0);a1b2+a2b1=-2(x^3);b1c2+b2c1=16(x);a1c2+a2c1+b1b2=-6(x^2),with all the above condition satified,we get the double cross looking like the following: 1 -4 4 1 2 -2 then it's literally the exact factorization of the polynomium on the left (x^4-4x+4)(x^2+2x-2)=0 okay,it is a much easier case now,(x-2)^2(x^2+2x-2)=0 x={2,√3-1,-√3-1}
in that linked video of @MindYourDecisions,he also made a little mistake that the other shorter length should not be considered a solution,well it's not true,I think.if you imagine the ground was the wall,and take the wall as the ground,you will find that shorter length as one of the solutions.
To be honest even if we multiply and distribute everything out, we'll get an biquadratic equation ( a degree 4 but you can treat x² as a new variable) you'll just have to use the quadratic formula 2 times, nothing much
two errors at about 5:00 - it should be 2x^3-2x^2 but you wrote 2x^3+2x^2 ... then you wrote 2x^2(x-1) which is actually correct, but does not follow from the previous step
Sir, Would you please let us know how to solve the Korean National Entance Exam to Universitily problems, since they are notorious for its difficulties?
En la 2da. columna, en la fila 2, el signo en el numerador de la fracción es negativo no positivo !!!, en la 3ra. fila cuando factorizas, corriges el signo, esto me da a entender que el ejercicio lo haces de " memoria " NO ???
I have to say the more I look closely at you, the more I like your smile. It is, nevertheless, a symbol of pleasure and satisfaction. Once again, thanks.
So I somehow made a silly mistake by trying to distribute the denominator and still got the correct answer somehow. Math is magic and I shouldn’t be doing mental algebra at midnight
The (first) mistake was already in 4:24. The numerator of the second fraction should have already been 2x³-2x². But with the transcription error in 5:06 (second mistake) he compensated for his first mistake. As nice as the solution is, it unfortunately only works for very specific number constellations. Therefore, the method presented is unfortunately not generally usable, but only possible for specially tailored equations.
TED-Ed has a great video on logarithms. If you've already watched videos on logarithms and natural logarithms I would recommend you to play around with them in a graphing calculators to develop intuition.
I've seen that too, and it's actually a "fortunate mistake". What I mean with this, the mistake carries from the previous step. He had xˆ4-2xˆ3+xˆ2+xˆ2 on the numerator. This equals to xˆ4-2xˆ3+2xˆ2. When he splits this in two parts, he writes xˆ4 on one side, and then the other side, he writes it as -(2xˆ3+2xˆ2) This is where the first mistake is done. The numerator grouping should equal to -(2xˆ3-2xˆ2), which is then equal to -2xˆ2(x-1) He writes, incorrectly, that grouping as -(2xˆ3+2xˆ2), but then he actually factors it correctly, fixing inadvertently the previous mistake. I was about to write about this too. I hope I helped with this. 😅
Here you are some considerations, my dear friend: isn't it obvious at a simple glance that x=2 is a solution, right?...Okey dokey! Then, you have a natural solution to the equation, haven't you? ... Well, what if you apply Ruffini to the polinomial of 4 degree, yes? You'd get rapidly x=2 is a doble solution of the equation, thus you get a residual polinomial of degree 2, which is easily solvable. Best regards, man!
It is interesting and amazing. The solution is not difficult more than expected. but you made a mistake twice. so as a result your solution is right. please consider the comment of @AurynBeorn. thanks~~
Dear Sir, I found a simpler way to solve this : Add minus 4 to both sides of the given. use difference of two squares and simplify. Here (2-x) is on both sides which gives x=2 simplify more and the resulting cubic equation gives another x=2 . Finally solve a simple quadratic with the other two answers . I like your method too. Thanks for your GREAT videos 🙂
I've tried your method and it worked for me, it maybe not such elegant, than this substitution which came literally from nowhere (really impressed, I mean, I know such type of problems, still I was always wondering, how people could guess to move in that direction, obviously after trying another ways). Anyhow the version with factoring is also cool and subtracting 4 is also a brilliant idea, which doesn't really lies on the surface
Tried your method but ((x/(x-1))^2)-4, did you use substitution here to factor?
There's been a little mistake on 5:02 and although I've replied to someone catching on it in the comments, I'm repeating it here because I see more people are also noticing. It was something I was going to write about, too.
My reply was:
"I've seen that too, and it's actually a "fortunate mistake". What I mean with this, the mistake carries from the previous step. He had xˆ4-2xˆ3+xˆ2+xˆ2 on the numerator. This equals to xˆ4-2xˆ3+2xˆ2.
When he splits this in two parts, he writes xˆ4 on one side, and then the other side, he writes it as -(2xˆ3+2xˆ2)
This is where the first mistake is done. The numerator grouping should equal to -(2xˆ3-2xˆ2), which is then equal to -2xˆ2(x-1)
He writes, incorrectly, that grouping as -(2xˆ3+2xˆ2), but then he actually factors it correctly, fixing inadvertently the previous mistake.
I was about to write about this too. I hope I helped with this. 😅
"
you are great!
lol I noticed this as well and went straight to the comments but I didn't catch the first mistake he made, great eye!
2 mistakes and those 2 mistakes cancelled each other out.
one mistake actually but then he continued writing it correctly
Here? 4:19
@@wolfler_vii Yes then when je groups 2*x^2, he puts a minus instead of a +
Mistakes can be happen in maths especially at algebra but genuinely i fell in love with video and beautiful hand write really an appreciative video
The part of my personality that resonates as autistic wants to pixelate the "+" you write at 4:22. I paused a long time when I first saw it. The next step corrected the mistake, relieving most of my anxiety, but the mistake itself continued to make me nervous. Worth it for the magical solution, though. That solution is what I clicked the video for!
I love your passion
Thank you for this video!
As x is not 1 and 8-x² must be non-negative (as equal to a square), x must be between -2sqrt(2) and 2sqrt(2), 1 excluded. So if there is at least one solution in Z, it should be -2, -1, 0 or 2. Luckily 2 is solution, even twice actually so we are left with a second-degree polynom in the end. It doesn't work each time but it was worth the try here. 🙂
Hi. In second column, row 2, in the right side you have a little mistake with sign.
Beutiful way of answer.
Now x² + (x/(x - 1))² = 8.
So, x² and hence (x/(x - 1))²
must both be small, and in fact -3 < x < 3, such that x ≠ 1.
So if we try the possible integer solutions, we get lucky and find that x = 2 is a solution.
Now as x² + (x/(x - 1))² = 8, then
x²(x - 1)² + x² = 8(x - 1)²
So, x⁴ - 2x³ - 6x² + 16x - 8 = 0 = f(x) say.
Applying synthetic division to f(x) where x = 2,
we obtain f(x) = (x - 2)(x³ - 6x + 4) = 0.
Let's try and factorise x³ - 6x + 4 = 0 using the rational roots theorem.
So possible rational roots are x = ±1, ±2 or ±4.
Trying x = 2, we see it is a root.
Using synthetic division on x³ - 6x + 4 = 0 with x = 2, we can factorise x³ - 6x + 4 = 0 as (x - 2)(x² + 2x +2)= 0
So f(x) = (x - 2)²(x² + 2x + 2) = 0 and it's easy to find the other two roots (solution) using this equation.
Good way of solving
Guess x = 2, then factorize x-2, x^3-6x+4 = 0, factorize x-2 again, x^2+2x-2=0, then get irrational roots
Love from India 🇮🇳
By inspection x=2 is one root. There must be a second real root.
x^4-2x^3+x^2-8x^2+16x-8+x^2=x^4-2x^3-6x^2+16x-8=0
Divide by (x-2): x^3-6x+4=0; By inspection x=2 is a second real root.
Divide by (x-2): x^2+2x-2=0; x=(-2±2√3)/2=-1±√3
There is a mistake in terms of the positive sign concerning the second term of the numerator of the second fraction at the second line exactly beforre time point 4:30.We should put -2X unstead of +2X as the two fration members of the numerator share the same negative outside sign similar they are between brakets.
This man's smile is so beautiful 😊
The equation to solve is
8 − x² = (x/(x − 1))²
These types of equations seem to be tremendously popular on TH-cam lately, but the proposed solution methods are often awkward at best.
This equation can be solved elegantly (1) without having to mess with fractions and (2) without substitutions. The key here is to use an ancient technique which is often useful in solving algebraic equations, i.e. converting a product of quantities into a difference of squares.
If we first bring x² over to the right hand side and then multiply both sides by (x − 1)² to eliminate the fraction (which is allowed since x ≠ 1) and then swap sides we can write the equation as
x²(x − 1)² + x² = 8(x − 1)²
I ultimately want to create a difference of two squares, and to do so I first incorporate the term x² at the left hand side into the product x²(x − 1)² which can be done since they both have a common factor x². Expanding (x − 1)² at the left hand side we have
x²(x² − 2x + 1) + x² = 8(x − 1)²
and taking out the common factor x² at the left hand side this gives
x²((x² − 2x + 1) + 1) = 8(x − 1)²
which is
x²(x² − 2x + 2) = 8(x − 1)²
We now have a product of two quantities x² and x² − 2x + 2 at the left hand side which we can turn into a difference of squares. The average (arithmetic mean) of x² and x² − 2x + 2 is half their sum which is x² − x + 1 and the difference between x² and x² − 2x + 2 is x² − (x² − 2x + 2) = 2x − 2 = 2(x − 1) so half their difference is (x − 1). So, we have x² = (x² − x + 1) + (x − 1) and x² − 2x + 2 = (x² − x + 1) − (x − 1) and we can therefore rewrite the equation as
((x² − x + 1) + (x − 1))((x² − x + 1) − (x − 1)) = 8(x − 1)²
and applying the difference of two squares identity (a + b)(a − b) = a² − b² to the left hand side this can be written as
(x² − x + 1)² − (x − 1)² = 8(x − 1)²
and bringing the term 8(x − 1)² from the right hand side over to the left hand side this gives
(x² − x + 1)² − 9(x − 1)² = 0
We now again have a difference of two squares on the left hand side and again applying the difference of two squares identity a² − b² = (a + b)(a − b) to the left hand side we have
(x² − x + 1 + 3(x − 1))(x² − x + 1 − 3(x − 1)) = 0
which is
(x² + 2x − 2)(x² − 4x + 4) = 0
and applying the zero product property this means
x² + 2x − 2 = 0 ⋁ x² − 4x + 4 = 0
and solving these quadratic equations by completing the square we have
(x + 1)² = 3 ⋁ (x − 2)² = 0
x = −1 + √3 ⋁ x = −1 − √3 ⋁ x = 2
and the equation is solved. Note that x = 2 is a double root which is why we have only three different solutions.
Can you solve 4×a^2+4×b^2+3=4c+4 sqrt of a+b-c
5:04 .... Shouldn't it be (x+1)? Can you please tell me how it is (x-1)? (I love your videos BTW... They r great!)
I've replied to this in another comment just now. 😊
this is due to the previous line he made a mistake grouping the numbers with a minus, -(2x^3+2x^2) here should be -(2x^3-2x^2) so that we could take (x-1) out.
really that is just as hard if not harder to find than the straight forward working out. move 8-x^2 to other side as x^2-8. Next common denominator of (x-1)^2. simplify the numerator and you should get something nice.
Brilliant 😊
Great work ❤❤❤❤
Excellent. Please continue your algebra series
4:22 it's 2 (times) cube of x minus(-) 2 (times) square of x (2x^3-2x^2).
Acctually this reminds me of a classical math problem:
th-cam.com/video/CZKD7rffF0M/w-d-xo.html
which can be solved both in geometry way and in algebra way.
@PrimeNewtons
I have left a comment in that linked video showing how to solve this problem with algebras,just do some transformation to the equation and set something to a new "a" (in that case it was (x²+1)/x=a),and such that we get a new equations concerning a.
but most of the times,I don't think people could easily find that "a" for a substitution,and then we usually get the standard power of 4th equation,in this case it's:
x^4-2x^3-6x^2+16x-8=0
well,that doesn't mean your algebra way of solving this problem is going to end here,what I recommend is to do "the double cross" trick:
a1 b1 c1
a2 b2 c2
a1a2=1(x^4);c1c2=-8(x^0);a1b2+a2b1=-2(x^3);b1c2+b2c1=16(x);a1c2+a2c1+b1b2=-6(x^2),with all the above condition satified,we get the double cross looking like the following:
1 -4 4
1 2 -2
then it's literally the exact factorization of the polynomium on the left
(x^4-4x+4)(x^2+2x-2)=0
okay,it is a much easier case now,(x-2)^2(x^2+2x-2)=0
x={2,√3-1,-√3-1}
in that linked video of @MindYourDecisions,he also made a little mistake that the other shorter length should not be considered a solution,well it's not true,I think.if you imagine the ground was the wall,and take the wall as the ground,you will find that shorter length as one of the solutions.
A simple slip: @ 4.23, +2x^2 instead of the correct -2x^2 ; as noted elsewhere.
Magic! 😊
To be honest even if we multiply and distribute everything out, we'll get an biquadratic equation ( a degree 4 but you can treat x² as a new variable) you'll just have to use the quadratic formula 2 times, nothing much
I don 't believe it....calculate..
.
Awesome.
Line infront of equal sign should be in the center.
Thank you
Way cool 😎
👍🏻
Nice!
First method comes to my mind is quadratic formula and （a-b)^2 but obviously both of them are not the anser😢 I haven’t noticed it can be ax^-bx-C=0
Thanks for an other video master
two errors at about 5:00 - it should be 2x^3-2x^2 but you wrote 2x^3+2x^2 ...
then you wrote 2x^2(x-1) which is actually correct, but does not follow from the previous step
ارجوالمعذره لكن عندما استخرجت x2عامل مشترك في الحدالثاني غيرت الاشاره من الموجب الئ السالب دون ان تغيراشارة الكسر ارجو االتوضيح
Sir, Would you please let us know how to solve the Korean National Entance Exam to Universitily problems, since they are notorious for its difficulties?
you made a sign error at 4:24 and corrected it in the next line xD
En la 2da. columna, en la fila 2, el signo en el numerador de la fracción es negativo no positivo !!!, en la 3ra. fila cuando factorizas, corriges el signo, esto me da a entender que el ejercicio lo haces de " memoria " NO ???
I have to say the more I look closely at you, the more I like your smile. It is, nevertheless, a symbol of pleasure and satisfaction. Once again, thanks.
The answer is right ✅
Because: -2X^3 + 2X^2 = -2(X^3 - X^2)
= -2X^2( X - 1 )
I wish you a happy christmas and new year!
So I somehow made a silly mistake by trying to distribute the denominator and still got the correct answer somehow. Math is magic and I shouldn’t be doing mental algebra at midnight
The (first) mistake was already in 4:24. The numerator of the second fraction should have already been 2x³-2x². But with the transcription error in 5:06 (second mistake) he compensated for his first mistake.
As nice as the solution is, it unfortunately only works for very specific number constellations. Therefore, the method presented is unfortunately not generally usable, but only possible for specially tailored equations.
I think there's an error at the second line in the right board an error at sign it has to be
Yes
Can you please make a video on log and natural log.
TED-Ed has a great video on logarithms. If you've already watched videos on logarithms and natural logarithms I would recommend you to play around with them in a graphing calculators to develop intuition.
Thanks, but i already watched it.
Two mistakes in the 4:24 and 5:06 mae one correct! 😉
Ive noticed that at 5:02, why when factoring 2x^3+2x^2 become 2x^2(x-1) but not 2x^2(x+1)? Any explanations ?
I've seen that too, and it's actually a "fortunate mistake". What I mean with this, the mistake carries from the previous step. He had xˆ4-2xˆ3+xˆ2+xˆ2 on the numerator. This equals to xˆ4-2xˆ3+2xˆ2.
When he splits this in two parts, he writes xˆ4 on one side, and then the other side, he writes it as -(2xˆ3+2xˆ2)
This is where the first mistake is done. The numerator grouping should equal to -(2xˆ3-2xˆ2), which is then equal to -2xˆ2(x-1)
He writes, incorrectly, that grouping as -(2xˆ3+2xˆ2), but then he actually factors it correctly, fixing inadvertently the previous mistake.
I was about to write about this too. I hope I helped with this. 😅
@@AurynBeorn Wow.. I see then. Thanks for the help!
There is an intermediiate mistake. 2nd column, 2nd row. instead of +2x^2, must be minus 2x^2
R u on Instagram too??
Plz reply sir
Foi cometido um erro de sinal na sétima linha.
i got that right away and im in 7th grade
Error!
Hello
I think he puts something wrong in every video
Mistake
Your answers are wrong.
The answer is right ✅
Because: -2X^3 + 2X^2 = -2(X^3 - X^2)
= -2X^2( X - 1 )
Here you are some considerations, my dear friend: isn't it obvious at a simple glance that x=2 is a solution, right?...Okey dokey! Then, you have a natural solution to the equation, haven't you? ... Well, what if you apply Ruffini to the polinomial of 4 degree, yes? You'd get rapidly x=2 is a doble solution of the equation, thus you get a residual polinomial of degree 2, which is easily solvable. Best regards, man!
There's so much knowledge in the comments. Thank you
asnwer=1x
It is interesting and amazing. The solution is not difficult more than expected. but you made a mistake twice. so as a result your solution is right. please consider the comment of @AurynBeorn. thanks~~