Factor ANY Quadratic Equation Without Guessing | Outlier.org

What if you could solve any quadratic function without any guess-and-check or memorization? This method was actually known by the Babylonians nearly 2,000-3,000 years ago but it's hardly taught in schools. Po-Shen Loh shares how to use the technique, the moment he discovered it, and why it's still such a secret.
Po-Shen Loh is a social entrepreneur, working in the fields of mathematics, education, and healthcare. He founded expii.com, a free personalized learning platform that gives the middle school math curriculum a focus on creative thinking. He has created KZbin videos totaling over 10 million views. Po-Shen earned his Bachelors of Science (B.S.) in Mathematics from Caltech, his Master of Advanced Study (MASt) in Mathematics from the University of Cambridge, and his Ph.D. in Mathematics from Princeton before becoming a math professor at Carnegie Mellon University. He is also the national coach of the USA International Mathematical Olympiad team. In 2019, he received the prestigious United States Presidential Early Career Award for Scientists and Engineers.
This lesson is excerpted from Outlier’s Precalculus course. Learn more about this course and check out our full catalog at www.outlier.org.
About @OutlierOrg :
Outlier.org (from the co-founder of MasterClass) has developed the world's best online, university-level courses, taught by some of the most celebrated educators in the world. Our courses employ engaging content and cutting-edge educational technology to create an immersive, student-friendly learning experience complete with transferable credits. Outlier is on a mission to make higher education more accessible, equitable, and affordable for everyone.
#Precalculus #Po-ShenLoh #FactoringMethod #Outlier

[@chocolateangel8743]

"I think that's how you multiply." 🤣 That's especially funny coming from a math professor -- but I get it. Ever since I learned about the area model of multiplication (and variations of it), I no longer use the standard algorithm. . As for this method of solving Quadratics, it took me seeing it demonstrated, using a graph, before I was able to understand it. Now, it's easy, and I like it. I'm able to visualize a graph in my head now.

[@lakshya5946]

I'm literally crying 😭 right now my biggest fear in maths is quadratic equations and professor just turned suspence into mist thanks outlier ❤️

[@lakshya5946]

When I saw this my mathematics search was over truely mathematics is a very cool think and more cool with outlier ❤️♥️❤️

[@agrocassiano]

(2x2 - 16x + 26 = 0) : 2 a b c *x² = 8x - 13* a b c Fórmula _ *X = b/2 + - ^[ (b/2)² + (c) ]* X = 8/2 + - ^[ (8/2)² + (- 13) X = 4 + - ^[ 16 - 13 ] *X = 4 + - ^[ 3 ]* Parabéns Outlier, sucesso sempre.

[@anonyone8834]

Thatnk you, it helped me to understand how dors quadratuc equations work.

[@fineganw]

Outstanding Professor! Wished I knew this method years ago, it would have made the difference between a B and an A for sure! (smile). More than ever, I shall subscribe to your channel. A statement that learning should always be continuous! Blessings Prof! Thanks for this method!

[@primefactor5417]

awesome technique .....

[@snow-108]

thank you pro shen loh

[@dustinvidrine3954]

im screaming. Thank you

[@RiverHeights]

This seems very confusing at first, but I'm going to learn it, and keep trying so I get good at it

[@mr.radium1202]

Hey , You really have great courses.

[@mr.radium1202]

Can I use it for JEE mains and advanced??

[@OutlierOrg]

Glad you like them!

[@xx1its___me1xx14]

@@mr.radium1202 can u ?

[@jaysonbalanday6054]

Thanks Sir. Hmm. Sir i have also an alternative method of Solving Quadratic equation with my own.

[@lyingcat9022]

Ax^2 + Bx + C == A(x+a)(x+b) To avoid a mess of fractions sometimes. Instead of factoring out A and having to do (B/A)/2 and having C/A… you can… Leave B alone and let C equal C•A then do B/2 and the rest as usual and factor A back out of the solution at the end…. A(x + r/A)(x + s/A) giving you back A(x+a)(x+b)

[@HerimosConstruction]

What happens when an odd number is involved, for example 2x squared-17x+26=0

[@chrishall9107]

It still works. divide by the x^2 coefficient: x^2 - 17/2 x + 13 = 0 Look for u: (17/4 + u)(17/4 - u) = 13 => 289/16 - u^2 = 13 => u^2 = 81/16 => u = 9/4 17/4 + 9/4 = 26/4 = 13/2 17/4 - 9/4 = 8/4 = 2 Factorisation is (x - 2)(x - 13/2) = 0

[@oldtrendz2848]

This is a great method but I think it only work, when the coefficient of ( x ) is an even number but for odd it will not work. ❤

[@oasisarah]

yes it does. take x^2+3x+2. then (3/2+a)(3/2-a)=9/4-a^2=2. subtract 9/4 then multiply by -1, you get a^2=1/4. a=+/- 1/2, sub back into the beginning, and you get the solutions 1 and 2.

[@quiddam]

this is great! but it's still useful to memorise the quadratic formula,... if nothing else... to quickly know how many solutions (0, 2 real or 2 complex) you're gonna have, based on the discriminant.