@@user-hp9qh1cy6o He's finding the x-coordinate of the vertex of the parabola made by the equation, which is always -b/2a, and since he made a=1, it's just -b/2 or b/-2.
I was hired to teach middle school history but was asked to teach math instead( I had 18 hr of undergrad math and computer science). When solving a question sumilar to this, the bells and whistles came out. Look how beautiful it is! I hadn't done thatwork in 30 years but it all came back.
This method is basically a change of coordinates that move the origin to the vertex of the parabola so that the quadratic equation turns to have the b value equal zero
This is interesting and I've never seen it before. Is this something relatively new, or did it just escape my awareness since my first algebra class in junior high school? I don't see any compelling advantage of it over using the quadratic formula, completing the square "manually," or even factoring, but I haven't tried any problems with it yet. Thank you for sharing this (and all of your other excellent content). I wish I could have learned from you when I was still teaching algebra because I think much of what you present would make me a more effective teacher.
This is comlpleting the square. He has chose numbers that make integers when he divides by the a value. If division here leads to a fraction, the result is more challenging, and the quadratic formula might be easier.
It is not something new. It is called Viete's Theorem. (Mr. H, has simply modified it to suit his own purposes). Sum of Roots = - b/a, and Product of Roots = c/a It is exactly what you do when solving a quadratic equation by the method of factoring. What are the factors of 'c' that will sum up to 'b'? Viete's Theorem is not mentioned in high school Algebra when treating Quadratic Equations. The concept is a bit high to be mentioned in high school Algebra. However, when dealing with quadratic functions related to the parabola in conic sections (or in Analytic Geometry), some tutors might bring it up. Mr. H is simply flexing his mathematical skills here without telling anybody that he is just reinterpreting Viete's Theorem.
This is just modified quad formula. At the start he divides b by -2(a) and then he solves for t using (b^2 - 4ac)/4. Still pretty cool but it can get a little messy when solving free fall and area problems.
Mr. H, you should inform your subscribers that what you are doing is simply a modified Viete's Theorem. Given: 2x^2 - 24x - 216 = 0 a = 2, b = - 24, and c = - 216 According to Viete's Theorem; Sum of Roots = -b/a And; Product of Roots = c/a Sum of Roots = - (-24)/2 = 24/2 = 12 Product of Roots = - 216/2 = - 108 The factors of - 108 that sum up to 12 are + 18 and - 6 (-6) × (18) = - 108 (-6) + (18) = 12 Therefore, the roots of the given quadratic function are: x = - 6 and/or x = 18.
I feel like the quadratic formula, completing the square, and whatever one where you multiply c by a and divid a by a in the beginning, are much faster. I can do the above in less than 30 secs without a calc and this seems like much more work.
It is not something new. It is called Viete's Theorem. (Mr. H has simply modified it to suit his own purposes). Sum of Roots = - b/a, and Product of Roots = c/a It is exactly what you do when solving a quadratic equation by the method of factoring. What are the factors of 'c' that will sum up to 'b'? Viete's Theorem is not mentioned in high school Algebra when treating Quadratic Equations. The concept is a bit high to be mentioned in high school Algebra. However, when dealing with quadratic functions related to the parabola in conic sections (or in Analytic Geometry), some tutors might bring it up. Mr. H is simply flexing his mathematical skills here without telling anybody that he is just reinterpreting Viete's Theorem.
Did statistics for my undergrad and now post grad. in analytics, and i'm just learning about this trick. Thank you so much. I'm dealing with complex numbers, and this is just a genius short -cut
This is simply a reinterpretation of Viete's Theorem. This is nothing novel or a reflection of genius. Our tutor here is simply flexing without telling anybody that what he is doing is derived from Viete's Theorem.
Step 1 dived every thing by the x^2 coefficient to make a=1 Step 2 This the special case of the quadratic formula when a =1 Solution =b/-2 +/- sqrt(b^2/4-c)
U definetely got a sub.This is great for exercises w long numbers.However, does this work w any equation of the form: ax^2+bx+c? Or just for some and if so for which exact ones does it, id really like to know!
Interesting. Because 1st step is to set a = 1 Then magically all the formula from (-b +- √(b^2 - 4ac))/2a Becomes (-b +- √(b^2 -4c))/2 Then b/-2 is just the 1st part (in video -12/-2) √b^2 -4c got divided all by 4 (from the 2 outside the root) becomes √((b/2)^ - c) thus becoming √(6^2 - 108). Very easy, just makes sure to make a = 1
this is the same as doing it on a slide rule. with a KL-1 circular rule it's only 3 steps. Find Roots to Quadratic of the form x^2 -px +q = 0: 1) rotate the black knob so q on C scale is on hairline 2) rotate the red knob until values on C and CI sum to p 3) these values are the roots to the quadratic
Because x² need to have value 1 ~> 1x²=x² and more in an equation if you dividing all the first term by 2, then you must divide by 2 the second term too ~> (.........)÷2 = 0÷2 , but 0÷2 is always 0.
Essentially what we are doing, is shifting horizontally (a distance h) to make the b-term disappear, so that the vertex is at the y-axis. You are rephrasing the parabola in terms or capital X, such that the vertex occurs when capital X = 0. Once you do that, you will isolate X^2, and we can solve for X directly. Then, we undo the shift, to find the original x's we wanted.
This is just alternate way to derive the quadratic formula. Just for fun, i had derived the formula from using middle term split on unknowns. I am sure many students who know basic algebra would have done that
Where the heck did the "t" come from? There are much simpler methods to do this. Find two numbers that multiply to a•c and add to b. Then split the b term into those two values. Factor each pair separately. Then factor the result. (Can't really show that in a comment.)
Isn’t this just the quadratic formula done in pieces? Dividing b by a and then -2 is the same as -b/2a which is the first term in the quadratic formula.
He made a new equation using the variable t to find the value of t so that he can plug it into x= 6+t, x=6-t . It's not transferred from the original equation.
Yes, however the difference is that the original quadratic equation is difficult to factor, whereas the 2nd quadratic equation is designed to solved easily.
Bro why is everyone using dumbest way ever exists to solve a quadratic equation, let me show you the easiest way- 2x²-24x-216=0 2(x²-12x-108)=0 Taking 2 on the other side, will give x²-12x-108=0 This can be also written as x²-18x+6x-108=0 x(x-18)+6(x-18)=0 (x-18)(x+6)=0 x=18,-6 This method is called the Middle term splitting term method, easiest way to solve most of the quadratic problems you can ask me if you are confused anywhere in my solution 🙂
It's a variable he made up. What he is doing, is finding the x-coordinate of the parabola's vertex from the b-coefficient, and then establishing that the parabola has reflectional symmetry about the vertical line through the vertex. This means the two roots are equidistant from x = -24/(-2), or x = +12. He assigned the variable t, to be the distance from x=+12, to find the roots. One root will be at x = 12 + t, the other root will be at x = 12 - t.
What he's doing is using the b-term to locate the vertex. For parabolas with a=1, the vertex will be located at -b/2. For parabolas in general, the vertex will be located at -b/(2*a). Play around with a graphing calculator to confirm that this is true. The t-variable is something he made up. He's using the property of reflectional symmetry to find the roots, once he locates the vertex. The two roots will be a distance of t, away from the vertex's x-position. Once he finds the vertex occurs at x=h, then the two roots will occur at x=h+t, and x=h-t. We know h=-b/(2*a), so once we solve for t, we know the two solutions for x.
Here's yet another way to solve a quadratic equation.
Is there a particular reason for always dividing by -2?
@@user-hp9qh1cy6o He's finding the x-coordinate of the vertex of the parabola made by the equation, which is always -b/2a, and since he made a=1, it's just -b/2 or b/-2.
@@prismo5459 Nope, see my above comment
Middle term splitting left mathematics 💀
@@itx_nitin The title says "without factoring".
Your handwriting and the sound of the marker are so soothing
I took piano lessons, same thing, totally addictive
His arms are well built.
you're well focusing 😂😂
Bro are you gay?😂
bro just hits the gym its not that deep
Why was that also the first thing i noticed
Stop objectifying us men. We don’t appreciate being… oh wait. We do appreciate it. 😂
this guy does things outside the box, amazingly refreshing
Well, he is high I.Q. (member of MENSA).
This is suspiciously close to completing the square. Keep up the good work.
Good eye!
Thank you
There is need of a save button in YT Shorts man.😭
I 100% agree.
Like button left TH-cam after seeing this comment
save it to a playlist
Share it to your email - easy find in the future
Playlist:
The famous Professor Po-Shen Loh's method
I was hired to teach middle school history but was asked to teach math instead( I had 18 hr of undergrad math and computer science). When solving a question sumilar to this, the bells and whistles came out. Look how beautiful it is! I hadn't done thatwork in 30 years but it all came back.
I always wondered if you could find solutions using the vertex formula, this is amazing!
I'm gonna stick to old trusty quadratic. She's never steered me wrong.
yup this one involves more opportunities for mistakes.
amazing job sir
This method is basically a change of coordinates that move the origin to the vertex of the parabola so that the quadratic equation turns to have the b value equal zero
It's just a reference i guess just to solve the problem
This is interesting and I've never seen it before. Is this something relatively new, or did it just escape my awareness since my first algebra class in junior high school?
I don't see any compelling advantage of it over using the quadratic formula, completing the square "manually," or even factoring, but I haven't tried any problems with it yet.
Thank you for sharing this (and all of your other excellent content). I wish I could have learned from you when I was still teaching algebra because I think much of what you present would make me a more effective teacher.
This is comlpleting the square. He has chose numbers that make integers when he divides by the a value. If division here leads to a fraction, the result is more challenging, and the quadratic formula might be easier.
@@markrobinson9956 Thanks! That's why I included "manually" in my original comment re: completing the square.
It is not something new. It is called Viete's Theorem. (Mr. H, has simply modified it to suit his own purposes).
Sum of Roots = - b/a, and
Product of Roots = c/a
It is exactly what you do when solving a quadratic equation by the method of factoring. What are the factors of 'c' that will sum up to 'b'?
Viete's Theorem is not mentioned in high school Algebra when treating Quadratic Equations. The concept is a bit high to be mentioned in high school Algebra.
However, when dealing with quadratic functions related to the parabola in conic sections (or in Analytic Geometry), some tutors might bring it up.
Mr. H is simply flexing his mathematical skills here without telling anybody that he is just reinterpreting Viete's Theorem.
This is just modified quad formula. At the start he divides b by -2(a) and then he solves for t using (b^2 - 4ac)/4. Still pretty cool but it can get a little messy when solving free fall and area problems.
Mr. H, you should inform your subscribers that what you are doing is simply a modified Viete's Theorem.
Given:
2x^2 - 24x - 216 = 0
a = 2, b = - 24, and c = - 216
According to Viete's Theorem;
Sum of Roots = -b/a
And;
Product of Roots = c/a
Sum of Roots = - (-24)/2 = 24/2 = 12
Product of Roots = - 216/2 = - 108
The factors of - 108 that sum up to 12 are + 18 and - 6
(-6) × (18) = - 108
(-6) + (18) = 12
Therefore, the roots of the given quadratic function are:
x = - 6 and/or x = 18.
I feel like the quadratic formula, completing the square, and whatever one where you multiply c by a and divid a by a in the beginning, are much faster. I can do the above in less than 30 secs without a calc and this seems like much more work.
Wow, this is new and amazingly working!
So simpler than factoring when there are so many factors for c and much easier than using quadratic formula
It is not something new. It is called Viete's Theorem. (Mr. H has simply modified it to suit his own purposes).
Sum of Roots = - b/a, and
Product of Roots = c/a
It is exactly what you do when solving a quadratic equation by the method of factoring. What are the factors of 'c' that will sum up to 'b'?
Viete's Theorem is not mentioned in high school Algebra when treating Quadratic Equations. The concept is a bit high to be mentioned in high school Algebra.
However, when dealing with quadratic functions related to the parabola in conic sections (or in Analytic Geometry), some tutors might bring it up.
Mr. H is simply flexing his mathematical skills here without telling anybody that he is just reinterpreting Viete's Theorem.
It is a good equations. Would love a more details on explanation with a quadratic graph. Still interesting regardless.
Nice simplification of the quadratic root formula. Love it.
How did you get the t=+-12?
And I have been using long and tedious quadratic equation my whole life 😢.
More trick math!!!! I like it
2x^2 - 24x - 216 = 0 I dividing by 2
x^2 - 12x - 108 = 0 I adding and subtracting 36 to the left side
x^2 - 2*x*6 + 36 - 36 - 108 = 0
(x - 6)^2 - 144 = 0
(x - 6)^2 = 144
1. x - 6 = 12 -> x = 18
2. x - 6 = -12 -> x = -6
Man u are a great mathematician this took like 2 sec while the basic method takes 2 min
Good trick niceeeeeee
Thank you for teaching me something new again today.
making a quadratic equation from a quadratic equation is another level of crazy
This is so helpful, thank you so much Sir
Out of curiosity, why did you decide to use the variable "t" to represent the distance from the midpoint to either extreme of the parabola?
Crazy... This makes it easier for me 💯
Did statistics for my undergrad and now post grad. in analytics, and i'm just learning about this trick. Thank you so much. I'm dealing with complex numbers, and this is just a genius short -cut
This is simply a reinterpretation of Viete's Theorem.
This is nothing novel or a reflection of genius.
Our tutor here is simply flexing without telling anybody that what he is doing is derived from Viete's Theorem.
What kind of method is this one?
x^2 - 12x - 108 = 0
x^2 = 12x + 108
Fórmula *X = b/2 + - ^[ (b/2)2 + (c) ]*
X = 6 + - ^[ 36 + 108 ]
X = 6 + - ^[ 144 ]
*X' = 6 + 12= 18*
*X" = 6 - 12 = - 6*
Parabéns Professor, sucesso sempre. 🎉
It’s an interesting brain exercise by combining vieda’s formulas and quadratic roots
Thats a nice question.
Step 1 dived every thing by the x^2 coefficient to make a=1
Step 2
This the special case of the quadratic formula when a =1
Solution =b/-2 +/- sqrt(b^2/4-c)
Mr. H.....I've never seen this method of solving the quadratics before. Can you do a video explaining your method?
U definetely got a sub.This is great for exercises w long numbers.However, does this work w any equation of the form: ax^2+bx+c? Or just for some and if so for which exact ones does it, id really like to know!
Yes, this will work with any quadratic equation in the form of ax^2+bx+c.
Thank you for the sub.
@@mrhtutoring its my pleasure, well deserved and thank u for the easier time w maths
Interesting.
Because 1st step is to set a = 1
Then magically all the formula from
(-b +- √(b^2 - 4ac))/2a
Becomes
(-b +- √(b^2 -4c))/2
Then b/-2 is just the 1st part (in video -12/-2)
√b^2 -4c got divided all by 4 (from the 2 outside the root) becomes √((b/2)^ - c) thus becoming √(6^2 - 108).
Very easy, just makes sure to make a = 1
I forgot it in class but now I understand it again 😅
Try this one
x^2 -12x - 108 = 0
x^2 -12x +36 = 144
(x-6)^2= 144
x-6 = plusminus 12
x= 18 or x = -6
Very old and forgetten but nice method
Po Shen Loh
😢😢😢 that's so unfair but cool after knowing this better than using a quadratic formula
Good. There’s a way of doing it without any algebra too. You can just write the answer down. 👍
Looks interesting, but on small numbers it's better to solve with Vieta
This is nuts
He's definitely a gym rat💪🏻😂
Perfection👏👏👏
this is the same as doing it on a slide rule.
with a KL-1 circular rule it's only 3 steps.
Find Roots to Quadratic of the form x^2 -px +q = 0:
1) rotate the black knob so q on C scale is on hairline
2) rotate the red knob until values on C and CI sum to p
3) these values are the roots to the quadratic
He’s good at math and jacked??? 😍😍
You should see papa flammy
How was 6 derived? It wasn't given in the question. Is it because it's a common multiple of 12 and 108?
Is it applicable for all quadratic equations
Sir but we could have also done this by middle term splitting right ?
Yes, definitely
The title should be "It's Always Negative Two".
Why do you divide by -2?
Because x² need to have value 1 ~> 1x²=x² and more in an equation if you dividing all the first term by 2, then you must divide by 2 the second term too ~> (.........)÷2 = 0÷2 , but 0÷2 is always 0.
Essentially what we are doing, is shifting horizontally (a distance h) to make the b-term disappear, so that the vertex is at the y-axis. You are rephrasing the parabola in terms or capital X, such that the vertex occurs when capital X = 0.
Once you do that, you will isolate X^2, and we can solve for X directly. Then, we undo the shift, to find the original x's we wanted.
Explain, please, why it works
This is just alternate way to derive the quadratic formula.
Just for fun, i had derived the formula from using middle term split on unknowns. I am sure many students who know basic algebra would have done that
Thanks mn
This is much faster than the ac method.
👏👏👏
Moi je vois, en dérivant, 2x-12=0
Donc x=6?
Est-ce exact comme raisonnement ou est-ce que je me plante?
just to ask , does is it really that the whites not goodat math ?
🤭
You could just do the delta mode
"Solving without factoring or quadratic formula"
Wait, don't they teach you Vieta's theorem at American school?
You can do it easily by using quadratic equations methods
this is the po shen loh method right?
Yes
Can you do partial fraction decomp videos?
I'll do it in a long video.
Look up the Heaviside cover-up method. It's a very handy shortcut for partial fractions.
Teacher ❌ Bodybuilder ✅
Haha, thanks.
I think. 🤔
😅you can make fitness videos also ....lol
👍
What if there is no a value?
Where the heck did the "t" come from?
There are much simpler methods to do this.
Find two numbers that multiply to a•c and add to b.
Then split the b term into those two values.
Factor each pair separately.
Then factor the result.
(Can't really show that in a comment.)
The "t" came from recognition of symmetry of the roots on either side of the vertex.
How did you get x=6-t and 6+t? Can you please explain that? Thanks
I'll try in a regular video.
The explanation is too long for the TH-cam shorts.
Why divided by -2
Reason?
Isn’t this just the quadratic formula done in pieces? Dividing b by a and then -2 is the same as -b/2a which is the first term in the quadratic formula.
Middle term splitting method left mathematics 💀
You have -108 equals 0, when you transfer -108 to other side - minus sign -108 should change into positive plus sign. Pls check.
He never transferred anything to the other side
He made a new equation using the variable t to find the value of t so that he can plug it into x= 6+t, x=6-t . It's not transferred from the original equation.
This is possible with numbers divisible by two only
It actually works for all numbers
Wow its a short cut i think👍
Is it working for quadratic equations?
Yes, any quadratic equation
So i guess it uses the product and sum of roots to find X
You still need to know the t formula thing...
after dividing 12 by -2 i got lost
🎉hello how do u arrive at 12 pls explian
Ok, and how did you solve your auxiliary quadratic without factoring or the quadratic formula? How did you get t=+-12? you skipped that part...
It's somewhat trivial. Isolate t^2 and take the square root. The second quadratic is easier since it removes the "bx" term.
Does it work when b and or c is an odd number?
Yes, it does!
@@mrhtutoring make a video on it😅
known as Po Shen Lo method!
Loh's method.
So what if b is not divisible by 2?
Still works. It will just be a fraction
+12/-2 always -2bro
you…formed another quadratic equation to solve a quadratic equation…
Yes, however the difference is that the original quadratic equation is difficult to factor, whereas the 2nd quadratic equation is designed to solved easily.
What happens if b value is odd?
You would use the same method.
If the b value is odd, you'd get a fraction after dividing by -2.
Always been asian
Bro why is everyone using dumbest way ever exists to solve a quadratic equation, let me show you the easiest way-
2x²-24x-216=0
2(x²-12x-108)=0
Taking 2 on the other side, will give
x²-12x-108=0
This can be also written as
x²-18x+6x-108=0
x(x-18)+6(x-18)=0
(x-18)(x+6)=0
x=18,-6
This method is called the Middle term splitting term method, easiest way to solve most of the quadratic problems you can ask me if you are confused anywhere in my solution 🙂
It's just another method.
Depending on the equation, one method may be better than the other.
@itx_nitin he said he would do it without factoring, read the title
This lost me.
Where did a “T” come from?
Wait wait
Pls did he got the answer 12
Why is it always -2
Huh, Where did the T come from ?
It's a variable he made up. What he is doing, is finding the x-coordinate of the parabola's vertex from the b-coefficient, and then establishing that the parabola has reflectional symmetry about the vertical line through the vertex.
This means the two roots are equidistant from x = -24/(-2), or x = +12. He assigned the variable t, to be the distance from x=+12, to find the roots. One root will be at x = 12 + t, the other root will be at x = 12 - t.
this is just proving the qf
Yes, and also the completing the square method.
Thanks for noticing.
I got lost when u said about b why divide and how got t
What he's doing is using the b-term to locate the vertex. For parabolas with a=1, the vertex will be located at -b/2. For parabolas in general, the vertex will be located at -b/(2*a). Play around with a graphing calculator to confirm that this is true.
The t-variable is something he made up. He's using the property of reflectional symmetry to find the roots, once he locates the vertex. The two roots will be a distance of t, away from the vertex's x-position. Once he finds the vertex occurs at x=h, then the two roots will occur at x=h+t, and x=h-t. We know h=-b/(2*a), so once we solve for t, we know the two solutions for x.