This is just the quadratic formula x=(-b±√(b²-4ac))/2a Which can be broken down into 2 part The x value of the turning point: -b/2a And the distance to the roots: √(b²-4ac)/2a So the quadratic formula basically wants you to find the turning point and add a distance on both sides of the turning point Hence the formula Fun fact: The turning point variation of the quadratic equation: y=a(x-p)²+q Can be written as y=a(x-(-b/2a))²-(b²-4ac)/4a The y value of the turning point(q) = -(b²-4ac)/4a !!!!!
How does this "method" supposedly "not depend on memorization"? You want us to be juggling with coefficients without knowing what we're actually doing. Personally, I usually use the "Completing the square" method, unless factoring or using the quadratic formula is faster. Your method is basically simply a version of the quadratic formula: x = m ± √(m² - c/a) where m = -b/(2a) By the way, please learn the difference between "(polynomial) expression", "(polynomial) function", and "equation".
Sorry, but how does "completing the square" require memorization or guessing? *Example: x² − 5x + 6 = 0* Po-Shen Lo Method M = −(−5)/(2*1) = 5/2, P = 6/1 = 6 (5/2 − u) (5/2 + u) = 6 25/4 − u² = 6 u² = 25/4 − 6 = 1/4 u = 1/2 x = 5/2 ± u = 5/2 ± 1/2 x = 3 or 2 Completing the square Method x² − 5x + 6 = 0 x² − 5x + (5/2)² = −6 + 25/4 (x − 5/2)² = 1/4 x − 5/2 = ± 1/2 x = 5/2 ± 1/2 x = 3 or 2 My main problem with the Po-Shen Lo method is that it disconnects you from the original equation and the variable x, while you go on a side quest to find u, which you then use to calculate x. All this without really being any simpler than completing the square
2x² + 11x + 23 = 0 I always first check the determinant D = b² - 4ac = (11)² - 4 . (2) . (23) = 121 - 184 = - 63 D is negative so no solutions for y = f(x) = 0 you joker ! But I will play with you: x² + 11/2 . x + 23/2 = 0 (x + m) . (x + n) = 0 with m + n = 11/2 and m . n = 23/2 M = 11/2 /2 = 11/4 (11/4 + d) . (11/4 - d) = 23/2 so (11/4)² - d² = 23/2 121/16 - d² = 23/2 d² = 121/16 - 184/16 = - 63 so no solutions for y = f(x) = 0
This is so much harder...
Maybe we are not used to it yet 😅 Give it some practice
@BrainStationAdvanced there are way easier methods
@@BrainStationAdvanced Say "ZDTF is very cool"
And I'll sub to you
I will not say "ZDTF is very cool" 😎 because I want genuine subscribers
@@BrainStationAdvanced Owlman
This is just the quadratic formula
x=(-b±√(b²-4ac))/2a
Which can be broken down into 2 part
The x value of the turning point: -b/2a
And the distance to the roots: √(b²-4ac)/2a
So the quadratic formula basically wants you to find the turning point and add a distance on both sides of the turning point
Hence the formula
Fun fact: The turning point variation of the quadratic equation:
y=a(x-p)²+q
Can be written as
y=a(x-(-b/2a))²-(b²-4ac)/4a
The y value of the turning point(q) = -(b²-4ac)/4a !!!!!
How does this "method" supposedly "not depend on memorization"? You want us to be juggling with coefficients without knowing what we're actually doing.
Personally, I usually use the "Completing the square" method, unless factoring or using the quadratic formula is faster.
Your method is basically simply a version of the quadratic formula:
x = m ± √(m² - c/a)
where m = -b/(2a)
By the way, please learn the difference between "(polynomial) expression", "(polynomial) function", and "equation".
That's computing Brahmagupta's formula without saying it.
woah. as someone who is able to count quickly, this is actually helpful. thanks!
You deserve more subs.
looks exactly like the formula
He used simplified formula of quadratic formula.
x= {-b+/-(b^2-4ac)}/2a= -b/2a+u or -b/2a-u where u =1/2a×√(b^2-4ac).
Sorry, but how does "completing the square" require memorization or guessing?
*Example: x² − 5x + 6 = 0*
Po-Shen Lo Method
M = −(−5)/(2*1) = 5/2, P = 6/1 = 6
(5/2 − u) (5/2 + u) = 6
25/4 − u² = 6
u² = 25/4 − 6 = 1/4
u = 1/2
x = 5/2 ± u = 5/2 ± 1/2
x = 3 or 2
Completing the square Method
x² − 5x + 6 = 0
x² − 5x + (5/2)² = −6 + 25/4
(x − 5/2)² = 1/4
x − 5/2 = ± 1/2
x = 5/2 ± 1/2
x = 3 or 2
My main problem with the Po-Shen Lo method is that it disconnects you from the original equation and the variable x, while you go on a side quest to find u, which you then use to calculate x. All this without really being any simpler than completing the square
Sorry mate, i am already married to the quadratic formula.
Divorce her and marry my method (Just Kidding 😂)...btw it’s simpler, and faster
x^2-22x+117=0 x=9 x=13 It’s in my head.
The quality of your teaching is out of tis world! Thank you so much ❤❤
Completing the square is objectively better
X^2 -5X+6=0 X=2 X=3
x^2-4x-7=0 x=2±Sqrt[11]
The last "so GOOOD"
Was funni
Great solutions my fellow Mathematician ❤ I have subscribed
Awesome! Thank you!
Take 117 =121-4=11^2-2^2=(11+2)(11-2)
=13×9
Sum =13+9=22.
So the roots are 9,13.
im practically married to factorisation dawg 😭
Looks like vieta's formula
Yeah, we were shown this, but you were thaught to calculate the discriminant at all times anyway, so that's what's stuck in my head rn
The only thing worse than this A. I. voice is the semi literate person who typed up the transcript for it.
2x^2+11x+23=0 x=(-11±3Sqrt[7])/4=-2.75±0.75Sqrt[7]
Woah
Never heard of this method
Really cool
So cool
x^2-5x+6=0 , (x-3)(x-2)=0 , x^2-22+117=0 , (x-9)(x-13)=0 ,
1 -3 1 -9
-2 6 -13 117
It's no good if the roots turn out to be complex.
Why not?
Think it’s fine. u becomes imaginary.
make trick on multiplying and dividing big no.
Lemme guess
Area method?
Nope
117=9*13=> x=9 or 13
Needs even more memorisation.
Am gonna stick to quadratic formula because i fear i will foget to substitute u in final answer😅. Still really great
9 13
For the question in tambnail
No memorization??? LOL!!!
X,2×+5=8
2x² + 11x + 23 = 0 I always first check the determinant D = b² - 4ac = (11)² - 4 . (2) . (23) = 121 - 184 = - 63
D is negative so no solutions for y = f(x) = 0 you joker !
But I will play with you:
x² + 11/2 . x + 23/2 = 0 (x + m) . (x + n) = 0 with m + n = 11/2 and m . n = 23/2
M = 11/2 /2 = 11/4 (11/4 + d) . (11/4 - d) = 23/2 so (11/4)² - d² = 23/2
121/16 - d² = 23/2 d² = 121/16 - 184/16 = - 63 so no solutions for y = f(x) = 0
No solutions? Who said we could not enter the complex realm?