    Directions: In each question two equations numbered I and II are given. You have to solve both the equations and mark the answer.
Important for :
1
I.  x2 – 37x + 140 = 0

II.  y2 + 13 y + 120 = 0
» Explain it
A
I.  x2 – 37x + 140 = 0

Step 1: Find the square of the root part of middle cofficient of the given equation:

⇒ (2)2 = 2

Step 2: Divide the constant part of the equation by the number we get at step 1:

 ⇒ 140 = 70 2

Step 3: Find such factors of 70 that can give us the integer value of the middle cofficient; – 37

Two such factors are –35 & –2.

Step 4: The equation, therefore, can be written as

x2 – 35 2x – 2 2x + 140 = 0

Step 5: Value of x, hence will be

either +352 or +22

Similarly, value of y will be

either –83 or –53

Now, we can observe that both the values of x are positive while those of y negative.

Therefore, x > y.

Hence, option A is correct.

2
I.  x2 – 52 x = 48

II.  y2 – 2 2 y = 30
» Explain it
E
I. x2 – 52 x = 48

= x2 – 52 x – 48 = 0

Step 1: Find the square of the root part of middle cofficient of the given equation:

⇒ (2)2 = 2

Step 2: Divide the constant part of the equation by the number we get at step 1:

 ⇒ 48 = 24 2

Step 3: Find such factors of 24 that can give us the integer value of the middle cofficient; – 5

Two such factors are –8 & +3.

Step 4: The equation, therefore, can be written as

x2 – 8 2 x + 32 x – 48 = 0

Step 5: Value of x, hence will be

either +82 or –32

Similarly, value of y will be

either +52 or –32

Now, in approximation we can assume the values of 2 to be 1.

Applying the comparision rule, we find that one of the values of y is lying between the value of x. So, we can't find the relation between them.

Hence, option E is correct.

3
I.  2x2 – 8x – 24 = 0

II.  9y2 – 12y + 4 = 0
» Explain it
E
I.  2x2 – 8x – 24 = 0

∴  x2 – 4x – 12 = 0

∴  x2 − 6x + 2x − 12 = 0

∴   (x + 2)(x – 6) = 0

∴   x = – 2 or x = 6

II.  9y2 – 12y + 4 = 0

∴   9y2 – 6y – 6y + 4 = 0

∴   (3y – 2)(3y – 2) = 0

 ∴   y = 2 3

When x = 6, x > y and when x = −2, x < y

Thus, the relationship between x and y can't be established.

Hence, option E is correct.

4
I.  6x2 + 11x – 35 = 0

II.  5y2 – 2y – 9 = 0
» Explain it
E
Since both equations are of the form ax2 ± bx − c = 0, both equations have one positive and one negative root.

Hence, the relation between x and y can't be established.

Hence, option E is correct.

5
I. 5x2 − 6x − 63 = 0

II. 4y2 + y − 39 = 0
» Explain it
E
I. 5x2 − 6x − 63 = 0

∴   5x2 + 15x − 21x − 63 = 0

∴   (5x − 21)(x + 3) = 0

∴   x = –3 or x = 21/5

II. 4y2 + y − 39 = 0

∴  4y2 – 12y + 13y − 39 = 0

∴  (4y + 13)(y – 3) = 0

∴  y = 3 or y = −13/4

When x = 21/5, x > y

When x = −3 and y = 3, x < y

Hence, the relation between x and y cannot be established.

Hence, option E is correct.

## Quadratic Equation Questions for SBI Clerk Pre 2021, IBPS Clerk Pre

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Folks, here is a another quiz of Quadratic Equation, so we are going to discuss different kind of Quadratic Equation questions which are frequently asked in SBI clerk pre 2021 Pre Exam, So Quadratic Equation have a lot of chance to appear in Sbi clerk pre 2021.

First of all we discuss different kind of Quadratic Equation for Sbi clerk pre 2021 like New pattern based Quadratic Equation Question and Answer that are now being asked in the exams are high level and it can be based on root based quadratic equation and general Quadratic Equation, find out the root values and then compare them and last but not the least Shri Dharcharya Rule based Quadratic Equation, Quadratic Equation is one of the most favorite topics of Sbi clerk pre 2021 Pre 2021. Questions on Quadratic Equations are asked in Quantitative Aptitude section. Generally, two quadratic equations in two different variables are given.

We have to solve both of the Quadratic equations to get to know the relation between both the variables.

Suppose we have two variables ‘x’ and ‘y’. The relationship between the variables can be any one of the following:

x > y
x < y
x = y or relation can’t be established between x & y
x ≥ y
x ≤ y

Meaning of different symbols, Before getting deep into the quadratic equations, lets try to understand the meaning of the basic operations used in finding the relationship between the variables –

(1) ‘>’ symbol: This symbol indicates that variable on the left side is definitely greater than the variable on the right side of the symbol.

For example:  x > y means x is definitely greater than y.
(2) ‘<’ symbol: This symbol indicates that variable on the left is definitely smaller than the variable on the right side of the symbol.
For example: x<y means x is definitely smaller than y.

(3) ‘=’ symbol: This symbol indicates that variable on the left side is equal to the variable on the right side of the symbol.
For example: x = y means x is definitely equal to y.

(4) ‘≥’ symbol: This symbol indicates that variable on the left side is either greater than or equal to the variable on the right side of the symbol.
For example:  x ≥ y means x is either greater than y or equal to y.
(5) ‘≤’ symbol: This symbol indicates that variable on the left side is either smaller than or equal to the variable on the right side of the symbol.

For example: x≤y means x is either smaller than y or equal to y.

General form of a Quadratic Equation
ax2 + bx + c = 0

Quadratic equation means that it will definitely have the maximum clerk pre 2021wer of the variable as ‘2’ which means we will always see ax2 term in a quadratic equation.

Or we can say that b can be 0, c can be 0 but a will never be 0.

While solving quadratic equation, you will always get 2 values of the equation. These 2 values are called roots of the equation. The roots of the equation always satisfy the equation. So in case of doubt, we can check the solution by putting the values back into the equation. If the equation turns out to be zero then our roots are correct.

Here are different kind of example of Quadratic Equation , so that we will get a very clear concept of the basic formation of a quadratic equation to get prepared for Sbi clerk pre 2021 Pre.

Type 1: This is basic of quadratic equation for bank exams like Sbi clerk Pre, IBPS RRB Clerk Pre as shown below

I. x3 – 4913 = 0                       II. y2 – 361 = 0

Here is the link to practice a proper quiz for quadratic equation for bank clerk pre exams like, Sbi clerk Pre 2021, SBI Clerk pre  and NIACL Assistant Pre

Type 2: Root based Quadratic Equation, these type of quadratic equation is being asked in Sbi clerk pre 2021, SBI clerk pre 2021 Pre exams

I. x2 – 3√3x – 54 = 0               II. y2 – 7√2y – 36 = 0

Type 3: High level Quadratic Equation for Sbi clerk pre 2021, this kind of Quadratic has been shown in many bank clerk pre 2021 exams

I. 20x2 – 108x + 144 = 0

II. 8y2 + 18y + 4 = 0

Here is link for sbi clerk pre 2021 mock test

Free Mock Test of SBI clerk Pre 2021

I hope these Quadratic equations will help you in upcoming Sbi clerk pre 2021 exams, if you want to practice a Mock Test for Sbi clerk pre 2021 Based on SBI clerk pre 2021   link is below:

Testzone Provides Best test series for Sbi clerk pre 2021 and other Bank clerk pre 2021 Exams, you will practice here all different kind of difficult level mock test with brilliant analytics. You will get 1 test of Previous year paper and other test of Parallel exam’s memory based test rest are with Difficult, Moderate and Easy Pattern based Test for all Bank and SSC Exams it Covers RBI Assistant Pre 2021, Sbi clerk pre 2021, SBI clerk pre 2021, RBI Grade B, SSC CGL, SSC 10 plus 2, LIC, NIACL Assistant and NIACL Mains.