Short Tricks to Solve Quadratic Equations in Seconds for Bank Exam

Short Tricks to Solve Quadratic Equations in Seconds for IBPS, SBI, RBI and Other Bank PO and Clerk Exam In every banking exam there are about 4-5 questions from Quadratic Equation. Here in this Article we will learn to solve quadratic equation questions quickly

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.

Therefore 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 4: 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

Short Tricks to Solve the Quadratic Equation :

+      +	-          -
-       +	+        +
+       -	-         +
-        -	+        -

Remember this diagram friends to solve the Questions of quadratic equation shortly in the exam but before exam you will need to practice more and more question to get speed and also to get familiar with the tricks.

1. I.5x2 – 87x + 378 = 0                 II.3y2 – 49y + 200 = 0

I.5x2 – 45x – 42x + 378 = 0

or,5x(x – 9) – 42(x – 9) = 0

or,(5x – 42) (x – 9) = 0

 x = 9,5

II.3y2 – 24y – 25y + 200 = 0

or,3y(y – 8) –25(y – 8) = 0

or,(y – 8) (3y – 25) = 0

y = 8,3

So , Answers : x > y

Now for this question we can solve this in less than thirty second how lets see,

Step-1. First take the Equation-I means equation with  variable x.

It is in the form of   ax2  +   (-)b x + c.

Step-2. See the Sign of b & c means sign of coefficient of x variable and sign of last one,

It is first negative(-) and second positive(+)

Step -3.From above table we can say that both the roots of x will be positive.

Step-4.Make Calculation 378*5=1890 first divide by 2 then 3 and then by 5 and follow above rule, so for this it will be 45 and 42 and now you don't need to write another two to three lines of taking common and then making in the forms of roots. This is your answer because we know from table that both root have positive sign.

Step-5 For those question in which coefficient of X2 is not one like in this question you will need to divide both the roots by (a).

So answer will be :-9,5

Follow same step for Equation-II also and find root in 20-30 seconds but to get this in short time you will need to practice lots of question then in exam you can solve these questions easily.

Practice Questions :-

1. I.x2 – 8x + 12 = 0                              II. y2 – 14y + 49  = 0

2. I.10x2 – x – 24 = 0                            II. y2 – 2y + 24 = 0

3. I.x2 – 5x + 6 = 0                                II.2y2 – 15y + 27 = 0

4. I.3x + 2y = 301                                  II.7x – 5y = 74

5. I.14x2 – 37x + 24 = 0                        II.28y2 – 53y + 24 = 0

6. I.11x + 5y = 117                                 II.7x + 13y = 153

7. I.6x2 + 51x + 105 = 0                        II.2y2 + 25y + 78 = 0

8. I.6x + 7y = 52                                    II.14x + 4y = 35

9. I.x2 + 11x + 30 = 0                            II.y2 + 12y + 36 = 0

10. I.2x2 + x – 1 = 0                              II.2y2 – 3y + 1 = 0

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