hello guys this is Pratiksha Mandal from
Mathematics Buzz welcome to my channel
and today we are going to learn about
quadratic equation what is a quadratic
equation and how to solve a quadratic
equation so if you like this video
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and comment let's start now quadratic
equation what is a quadratic equation
now equations of the form y = ax^2 +bx+c
where a is not
equal to 0 and a,b,c are the real numbers
are called quadratic equation so any
equation which is of this form is a
quadratic equation now to solve a
quadratic equation we equate the
equation ax^2 +bx+c equal
to 0 and is taking y equal to 0 we can
solve the equation now for example let
us see an example here I have wrote down
an example it is x^2 - 3x +2
equal to 0 in the quadratic equation
where a equal to 1 be equal to minus 3 and
c equal to 2 now see there are these
solutions or the values of x which we
get by solving the quadratic equations
so if we take 2 roots of the quadratic
equation as alpha and another as beta of
the equation ax^2 + bx+c
equal to 0 these are the two roots and
now how to solve a quadratic equation
now first is a factorization method I
have done one now number one you can see
that I have written down the example it
is 3y^2 - 20 equal to 160
minus 2y square so in
the next step we are taking  2y squared to the left that is
this becoming 3y squared plus two Y
Square and minus 20 to the right so it
will become 160 plus 20 the next step 5
is for equal to 180 and then and then it
becomes 5 y squared minus 180 equal to 0
the next step right over y squared - 36
equal to 0 so now now you can see that
it is becoming y square minus 36 equal
to 0 and Y square minus 6 whole square
equal to 0 so it will it can be applied
to this sum and it will become y equal
to 6 or - it took y equal to 6 and -6
are the two roots of the quadric way of
this quadratic equation so in a similar
way you can solve more other problems
and I will be solving another example so
let's write down the example so here the
example
x by x + 1 plus x+ 1 by x equal 2
one by 12 so how can we solve this
problem so it is a very simple method
let us take x by x + 1 as a where a
is some constant value which is equal
to x plus 1 so then we can write or x
plus 1 by x equal to 1 by a if we can
reverse this  we'll get this now in
the next step we will write it will
become a plus 1 by a we will solve this
so it will become a +1 /a equal to  25 / 12 so we will
now take the LCM it will become a square
plus 1 by a = 25 by 12 or a squared
plus 1 we will just
multiplied 12 to  to the left
and 25 to the right hand side
now the equation will become 12 a
squared plus 12  minus 25 a equal to 0
or 12 a squared minus 25 a plus 12 equal
to 0 now we will use factorization
method and so we can factorize it in
this method 12 a squared minus 16 a
minus 9 plus 12 equal to 0 or we can
take or a as common
-
Or
we'll take three are common and then it
will become 3a minus 4
 
minus 4 equal to zero
there will be four three
3a minus 4
is 4 by 3 so replacing a by X by X
plus 1 will get
 
the answer is
x equal to three comma minus four so
here is the solution of this problem I
solved it and I will be giving two more
problems number one is
two x plus one
Plus 3 by 2 x plus 1 equal to 4 where X
is not equal to minus 1/2 and another
problem I will be given here this number
it's a very important for probably one
by a plus  b plus x equal to 1 by a plus
1 by b plus 1 by x where it's true that
X is not equal to 0 and it's not equal
to minus a plus b now here is the end
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