How to Solve Quadratic Equations - Using 3 Different Methods

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Thank you! You helped a lot of people by making this video. I know that you helped me at least!

kaimirrekt

Within nine minutes I learned everything and understand perfectly. My math teacher has been trying to teach this in my math class for monthss. Thank you so much 😩.

Indicaa

I’ve seen so many different videos on “How to Solve Quadratic Equations” and out all this is the one that I understood clearly. I love how you explained every step thoroughly, I understand and know how to solve quadratic equations because of this video.
Thank you. 💓

TheLivelyLeahRae

Nothing to do while in isolation, this really helped me to get up to speed again after many years of doing no maths. It's amazing how quickly it all comes back.
Sir, Thanks very much.

wimm

Proud to say. I'm 43 and (re) learning this to be prepared for my son's algebra. You sir have made it simplest to understand. I thank you.

RobertoGarcia-nivl

Thank you very much! I have a test tomorrow and i think this video helped me understand it better. I also like that you go a bit slower that helps me learn too. thank you!

sofiar

Thank you so very much!!
I was literally stressed out as I have an exam on 19th October. This video proved to be of great help!
Thank you once again!😊

imameatball

I just come to see the comment section if there is any one complaining after this extraordinary explanation

tehuvzf

I'm trying to help my son with his homework, and this video was perfect! Thanks so much for going to the time to help so many. Stay well!

katebrophy

my teacher teach this a week ago i still don't get it but now i watched this video i know how to do this

yaurora

OH WOW THE EXPLANATION ON ALL THREE METHODS IS AMAZING, NO MORE SEARCHING!

ybuouyy

Wonderfully explained 3 different ways solve an algebraic puzzle. Loved it very much. Three forms are;
1. Quadratic formula a, b, c
2. Factoring ( ) ( )
3. Root square

goopalg

These solutions are interesting in that the formula (--b +_sqr(b^2--4*a*c))2*a) is normally obtained by completing the squares. The factoring method is a guessing game, as while it is possible for small nice numbers it is not possible to guess the roots for larger numbers .
If a quadratic A*x^2 + B*X +C = 0 then if we plot y= A*x^2 + B*X +C we find that this is a summation of three individual functions as

(A*x^2) ( parabola) added to ( B*X)(straight line) and added to (C ) a shifting constant.
If they are plotted individually then we have
A will open and close the parabola X^2
B will shift the parabola to the left or the right and even up and down keeping the parabola A*X^2 in the same orientation.
C will shift the parabola A^X^2 up and down on the Y axis keeping its shape and its orientation.

It is because the curve keeps its orientation and its symmetry the following method works. Knowing of this symmetry, then as detectives we need to look for the minimum location of this function hence we need to differential the function and equate it to zero.
y= A*x^2 + B*X +C let us put it in the form of y = X^2 +K1*X + K2 K1 = sum of roots and K2= product of roots

y = X^2 +K1*X + K2 it is convenient to take all constants as being positive.
dy/dx= for minimum = 0= 2*X + K1 hence Y is a minimum at X= --(K1)/2 NOTE THE NEGATIVE SIGN ATTACHED TO IT.
Then proceed taking into account the negative sign. of where the minimum is located. It is because of this symmetry that the average of the roots is located in the middle of the roots!
Let us solve for the equation ( X+4)*(X+2) =0= X^2 + (2+4)*x + 4*2 =0 = X^2 + 6*x + 8 = X^2 + ( sum of roots) *X + (products of roots) = 0
Note the negative sign hence average of roots is (--6/2) = --3
hence ( --3 --U)*( --3 +U) = 8
hence (--3)^2 -- U^2= 8
9 -- U^2=8
1 = U^2 hence U= + or - 1 and this in conjunction with the mid location of roots
one root = --3+1= --2 and other root is -- 3-1 = --4 which is the solution of equation ( X+4)*(X+2) =0

if a*x^2 +b^x +c = 0 then dividing by a, x^2 + (b/a)*x +c/a = 0 . The sum of roots is ( b/a) and product of roots is (c/a)
Average of roots is ( note negative sign --(b/2a) let U be the distance between the midpoint of the roots to the roots, hence
products of roots is ( ---(b/2a -- U) ( ---(b/2a+ U) = (c/a), hence b^2/4a^2 -- U^2 = c/a hence U^2 = b^2/4a^2 -- c/a hence
taking sqr of U^2 = b^2/4a^2 -- c/a we have U= sqr( b^2/4a^2 -- c/a) = + or -- (sqr( b^2 -- 4ac ))/2a Now we can find the roots.

one root is --(b/2a) + (sqr( b^2 -- 4ac ))/2a other root is --(b/2a) --(sqr( b^2 -- 4ac ))/2a
note the above is mean of roots plus distance of mean from root U to the sides of the mean.
Ths is the same as the formula found by completing the squares. --(b + or--(sqr( b^2 -- 4ac )))/2a

carmelpule

Thank you very much, I have final exams in December, I was so stressed, but I'm going to keep watching your videos to get more info, helped a lot! thanks!

daka

Most of the videos of other channels show how to use different methods of solving quadratic equations using different equations, which is helpful but not as helpful (at least for beginners like me) as seing the same equation being solved by different methods. Thank you.

deltakid