**Shortcut Method To Solve Questions Related to Quadratic Equations**

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Hello Friends, Today we
will learn about how to solve Quadratic equations. We hope this article will be
helpful for you. Thank You J Keep
Learning !!

What is Quadratic Equation?

Any equation having in the form of

**ax**

^{2}+bx+c=0,
is a quadratic equation in where x represents
an unknown. a, b & c represents the real numbers such that a is
not equal to 0.

In this Equation:

a is
the coefficient of x

^{2}
b is the coefficient of x

c is the constant term

How to solve Single Variable Quadratic Equation ?

In single
variable quadratic equation we will be given one quadratic equation in the form
of

**ax**& we have to find the value of that unknown variable x.^{2}+bx+c=0
Let
us take an Example:

**Example:**Let we have given the equation: 4 x

^{2}+12x+8=0

*Solution:*
In the above given equation, we have

+4 is the coefficient of x

^{2}
+12 is the coefficient of x

+8 is a constant value

Step-1: Multiply
the coefficient of x & the given constant value

i.e (+4) * (+8) = +32

Step-2: Now we have
to split the number 32 into two numbers in such a way that after adding them we
get the number equal to coefficient of x i.e 12 & also their multiplication
results into 32.

Therfore we get, +12= (+4) + (+8)

Also, (+32)= (+4) * (+8)

Step 3: Now change
the signs of both the factors i.e 4 & 8

After changing the signs we get, (-4) & (-8)

Step 4: Divide
these factors by the coefficient of x

^{2}
Therefore, we get

-4/4=-1

& -8/4= -2

Thus, x= -1, -2

*Therefore -1 & -2 are the required values of x.*
Now
Let us take one more Example:

Q) 3x

^{2}+19x+28=0

*Solution:*
In the above given equation, we have

+3 is the coefficient of x2

+19 is the coefficient of x

+28 is the constant value

Step-1: Multiply
the coefficient of x & the constant value

i.e (+3) * (+28)
= +84

Step-2: Now we have
to split the number 84 into two numbers in such a way that after adding both
numbers we get the number equal to coefficient of x i.e 19 & also their
multiplication results to 84

Therefore, +19=
(+12) + (+7)

Also, +84 =
(+12) * (+7)

Step
3: Now change
the signs of both the factors

After
changing the signs we get, (-12) & (-7)

Step
5: Now Divide
these factors by the coefficient of x2

Therefore,
we get

-12/3=-4

& -7/3=
-7/3

Thus, x= -4,
-7/3

*Therefore -4 & -7/3 are the required values of x.*
How to Solve More than One Quadratic Equation?

In this type
of Quadratic Equation, more than one quadratic equation will be given. First we
have to solve the given equations individually & find the values of unknown
variables given in the equation & then have to find the relation between
these variables of the equation as given below:

1. x > y

2. x < y

3. x ≥ y

4. x ≤ y

5. x = y or relation cannot be determined

Example:

x

^{2}+13x+40=0 ----(1)
y

^{2}+7y+12=0 ----- (2)
First Solve
the Equation (1)

Therefore, +13 = (+5) + (+8)

Also, +40 = (+5) * (+8)

Now Change the signs & Divide them with
the coefficient of x

^{2}i.e with +1
Therefore, we get

-5/1= -5 & -8/1= -8

*Thus, -5 & -8 are the required values of x.*
Now similarly Solve the 2nd Equation,

+7 = (+3) + (+4)

Also +12 = (+3) * (+4)

Now Change the signs & Divide them with
the coefficient of y

^{2}i.e with +1
-3/1=-3 & -4/1=-4

*Thus, -3 & -4 are the required values of y.*
Therefore
Now we have x= -5, -8

y= -3, -4

Therefore, by
comparing both the values we can say that

**x < y.**
Thank You Friends !! J

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