2:21 - the mistake here is that you cannot just simply "cancel" out the square on both the sides as - (-2²) =2² but -2≠2 This is where you just mistaken if two things have the equal square, you cannot just simply cancel it out as we have seen in the example above
The mistake is when he removed the powers incorrectly because he did not put this kind of rule √x²=|x| 'The square of that number squared is equal to the absolute value of that number" so the correct cancellation is (4-6)²=(8-6)² |4-6| = |8-6| |-2|=|2| 2=2
You cannot simply remove the squares on both sides without considering the fact that both are true for positive and negative values within their respective brackets.
no you actually take square root of (4-6)^2 = (8-6)^2 but square root of something can't be negative so either put modulus or do like this (6-4)^2 = (8-6)^2
The error occurs when you assume that the equation simplifies directly to 4−6=8−6. In reality, the square roots of both sides represent absolute values, so you can't drop the absolute value symbols. The proper step should be: ∣4−6∣=∣8−6∣ which simplifies to: -(4 - 6) = +(8 - 6) {Rule for > or < than 0} -(-2) = 2 2 = 2
When you take the square root on both sides, you need to include + or - on both sides (leaving us with + or - 2 = + or - 2, disallowing his arithmetic tangent. This is because sqrt(x^2) will always be equal to +x AND -x
Mistake is in cancelling the square on both side because (4-6)²= (-2)²=4=(2²)=(8-6)² But if the square cancelled then (4-6)=(-2)≠(2)=(8-6) Thank me later😎😎😎😁😁 Pls pin my comment so that it will be useful ...😎😎😎
in this solution, x got two values ,not one , x=4 in the left side and x = 8 in the right side ,where y =6 in both sides , so the two values of x made everything else built on it wrong mathematically without logic
4-6= -2 while, 8-6= 2 -2 ≠ 2 (that is the mistake!) in fact, (-2)²=(2)² but, -2≠2 ; thats it! Then, you cannot cancel the "squares" squares as you will...
Okay. Then İf x=y x²=xy x²-y²=xy-y² (In here a²-b² = (a+b) × (a-b) and ax-bx = x(a-b)) (x+y)(x-y) = y(x-y) And if we divide this 2 equations with x-y x+y=y. Remember at the first x=y and now x+y=y. İf we say x=1, 1+1 =1. 2=1. WOW
The critical error in this proof occurs in Dividing by (𝑎2−𝑎𝑏)(a 2 −ab) is invalid if 𝑎2−𝑎𝑏=0a 2 −ab=0, which is true since 𝑎=𝑏a=b. This makes the denominator undefined, rendering the entire proof incorrect.
First comment please pin it
2:21 - the mistake here is that you cannot just simply "cancel" out the square on both the sides as -
(-2²) =2² but
-2≠2
This is where you just mistaken if two things have the equal square, you cannot just simply cancel it out as we have seen in the example above
The mistake is he removed the exponents from (4-6)² and (8-6)²
The mistake is when he removed the powers incorrectly because he did not put this kind of rule √x²=|x|
'The square of that number squared is equal to the absolute value of that number" so the correct cancellation is
(4-6)²=(8-6)²
|4-6| = |8-6|
|-2|=|2|
2=2
This makes sense
2:24 the absolute value of /4-6/ = /8-6/ here 's the mistake
Yes, you cant just cancel the squares
You cannot simply remove the squares on both sides without considering the fact that both are true for positive and negative values within their respective brackets.
No we cant cancel square that way because by doing square root both sides we get modulus or absolute value function of x, i.e. by definition ✓x² = |x|
no you actually take square root of (4-6)^2 = (8-6)^2
but square root of something can't be negative so either put modulus or
do like this (6-4)^2 = (8-6)^2
The error occurs when you assume that the equation simplifies directly to
4−6=8−6. In reality, the square roots of both sides represent absolute values, so you can't drop the absolute value symbols.
The proper step should be:
∣4−6∣=∣8−6∣
which simplifies to:
-(4 - 6) = +(8 - 6) {Rule for > or < than 0}
-(-2) = 2
2 = 2
When you take the square root on both sides, you need to include + or - on both sides (leaving us with + or - 2 = + or - 2, disallowing his arithmetic tangent. This is because sqrt(x^2) will always be equal to +x AND -x
This man is Quite genius woth numbers , unorthodo methods
Removing the squares from (4-6)²=(8-6)² just means -2=2.
Mistake is in cancelling the square on both side because (4-6)²= (-2)²=4=(2²)=(8-6)²
But if the square cancelled then (4-6)=(-2)≠(2)=(8-6)
Thank me later😎😎😎😁😁
Pls pin my comment so that it will be useful ...😎😎😎
the wrong spot is that you cannot just square root both sides when (4-6)^2=(8-6)^2
because you need to add plus or minus!
in this solution, x got two values ,not one , x=4 in the left side and x = 8 in the right side ,where y =6 in both sides , so the two values of x made everything else built on it wrong mathematically without logic
4-6= -2
while,
8-6= 2
-2 ≠ 2 (that is the mistake!)
in fact,
(-2)²=(2)²
but, -2≠2 ; thats it!
Then, you cannot cancel the "squares" squares as you will...
Happy new year....🎉🎉🎉🥳🥳🥳😎😎😎🍪🎂🎂🎂🧁🧁
4=4
52-48=100-96
16+36-48=64+36-96
4^2+6^2+2x4x6=8^2+6^2-2x8x6
(4-6)^2=(8-6)^2
(-2)^2=(2)^2 *[**2:21** is the mistake]*
4=4
4/4=4/4
1=1
1+1=
2
Excellent comments❤
When You Canceled The squares, There must be +-(4-6)=+-(8-6)
You can't cancel the square because (4-6)²≠(4-6)
I would first check if it was in a different base vs base 10
At 2:40 you Definitely made a mistake
0:28 it was true cause me, you and our children😋
Okay. Then
İf x=y
x²=xy
x²-y²=xy-y²
(In here a²-b² = (a+b) × (a-b) and ax-bx = x(a-b))
(x+y)(x-y) = y(x-y)
And if we divide this 2 equations with x-y
x+y=y. Remember at the first x=y and now x+y=y. İf we say x=1, 1+1 =1. 2=1. WOW
😅
No it is x=0
How you can assume a value until you didn't solve whole equation x+y = y =⟩ x=0 , so how you can take x=1
-2=2 hmm so accurate
2:40
OBJECTIOM
YOU ARE CHEAT
you are not supposed to say it out loud 🤫😉
No haha 😂😂
First comment
The critical error in this proof occurs in Dividing by (𝑎2−𝑎𝑏)(a 2 −ab) is invalid if 𝑎2−𝑎𝑏=0a 2 −ab=0, which is true since 𝑎=𝑏a=b. This makes the denominator undefined, rendering the entire proof incorrect.
what? this never happened, you probably didn't even watch the video