Choose Your Language (अपनी भाषा चुनें)

Aug 17, 2015

SHORTCUT METHOD TO SOLVE QUESTIONS RELATED TO QUADRATIC EQUATIONS

Shortcut Method To Solve Questions Related to Quadratic Equations
Shared by: WWW.GovtJobsPortal.IN

Shortcut Method To Solve Questions Related to Quadratic Equations
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 
ax2+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 x2
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 ax2+bx+c=0 & we have to find the value of that unknown variable x.

Let us take an Example:

Example: Let we have given the equation: 4 x2+12x+8=0

Solution:
In the above given equation, we have
+4 is the coefficient of x2
+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 x2
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)  3x2+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:

x2+13x+40=0 ----(1)
y2+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 x2 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 y2 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
Related Government Jobs

0 comments:

Post a Comment

Video Bar

Loading...