Quadratic Equation Question for IBPS Clerk 2017. Welcome to the letsstudytogether.co online Reasoning section. If you are preparing for IBPS PO/Clerk 2017 exam, you will come across a section on Reasoning Ability Section.Quadratic Equation Quiz Set -3 IBPS Clerk.

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## Quadratic Equation Question for IBPS Clerk 2017

1.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 + 17x + 52 = 0
II. y2 + 16y + 39 = 0

A. x ≥ y

B. x > y

C. x = y or no specific relation can be established

D. x ≤ y

E. x < y

C. x = y or no specific relation can be established

I. x2 + 17x + 52 = 0
⇒ x2 + 13x + 4x + 52 = 0
Splitting the middle term
⇒ x (x + 13) + 4 (x + 13) = 0
⇒ (x + 4) (x + 13) = 0
So, (x + 4) = 0
⇒ x = – 4
Or, (x + 13) = 0
⇒ x = – 13
II. y2 + 16y + 39 = 0
⇒ y2 + 13y + 3y + 39 = 0
Splitting the middle term
⇒ y (y + 13) + 3 (y + 13) = 0
⇒ (y + 3) (y + 13) = 0
So, (y + 3) = 0
⇒ y = – 3
Or, (y + 13) = 0
⇒ y = – 13
So, when x = – 4, x < y for y = – 3 and x > y for y = – 13
And when x = – 13, x < y for y = – 3 and x = y for y = – 13
∴ so, we can observe that no clear relationship cannot be determined between x and y.

2.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. 5x2 + 1 = 6x
II. 16y2 + 1 = 8y

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

C. x = y or no specific relation can be established

I. 5x2 + 1 = 6x
⇒ 5x2 – 6x + 1 = 0
⇒ 5x2 – 5x – x + 1 = 0
Now splitting middle term,
⇒ 5x (x – 1) – 1 (x – 1) = 0
⇒ (5x – 1) (x – 1) = 0
So, (5x – 1) = 0
⇒ x = 1/5
Or, (x – 1) = 0
⇒ x = 1
II. 16y2 + 1 = 8y
⇒ 16y2 – 8y + 1 = 0
Now splitting middle term
⇒ 16y2 – 4y – 4y + 1 = 0
⇒ 4y (4y – 1) – 1 (4y – 1) = 0
⇒ (4y – 1) (4y – 1) = 0
Then, y = ¼
So, when x = 1/5, x < y for y = ¼
And when x = 1, x > y for y = ¼
∴ So, we can observe that no clear relationship cannot be determined between x and y

3.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x3 – 120 = 392
II. y3 – 217 = 512

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. y > x

E. y > x

I. x3 – 120 = 392
⇒ x3 = 392 + 120
⇒ x3 = 512
⇒ x = 8
II. y3 – 217 = 512
⇒ y3 = 512 + 217
⇒ y3 = 729
⇒ y = 9
∴ We can observe that y > x.

4.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

C. x = y or no specific relation can be established

5.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 – 26x = – 5x – 68
II. y2 + 18y = 2y – 48

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

D. x > y

I. x2 – 26x = – 5x – 68
⇒ x2 – 26x + 5x + 68 = 0
⇒ x2 – 21x + 68 = 0
Splitting middle term
⇒ x2 – 17x – 4x + 68 = 0
⇒ x (x – 17) – 4 (x – 17) = 0
⇒ (x – 4) (x – 17) = 0
So, x = + 4 or x = + 17
II. y2  + 18y = 2y – 48
⇒ y2 + 18y – 2y + 48 = 0
⇒ y2 + 16y + 48 = 0
⇒ y2 + 12y + 4y + 48 = 0
⇒ y (y + 12) + 4 (y + 12) = 0
⇒ (y + 4) (y + 12) = 0
So, y = – 4 or y = – 12
∴ As, we can observe that both of the values of x are positive and both of the values of y are negative then we can clearly say that x > y

6.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 + 96 = 321
II. y + 1296 = 1312

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

E. x < y

I. x2 + 96 = 321
⇒ x2 = 321 – 96 = 225
⇒ x = √225
⇒ x = ± 15
II. y +1296 = 1312
⇒ y = 1312 – 1296
⇒ y = 16
So, when x = + 15, x < y for y = 16
And when x = – 15, x < y for y = 16
Hence, we can conclude x < y.

7.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 – 16x + 28 = 0
II. y2  – 14y + 13 = 0

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

C. x = y or no specific relation can be established

I. x2 – 16x + 28 = 0
⇒ x2 – 14x – 2x + 28 = 0
⇒ x(x – 14) – 2(x – 14) = 0
⇒ (x – 14)(x – 2) = 0
Then, x = + 14 or x = + 2
II.y2 – 14y + 13 = 0
⇒ y2 – 13y – y + 13 = 0
⇒ y(y – 13) – 1(y – 13) = 0
⇒ (y – 13)(y – 1)
Then, y = + 13 or y = + 1
So, when x = + 14, x > y for y = + 13 and x > y for y = + 1
And when x = + 2, x < y for y = + 13 and x > y for y = + 1
∴ So, we can observe that no clear relationship cannot be determined between x and y.

8.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 + 6x + 8 = 0
II. y2 + 13y +42 = 0

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

D. x > y

I. x2 + 6x + 8 = 0
⇒ x2 + 4x + 2x + 8 = 0
⇒ x(x + 4) + 2(x + 4) = 0
⇒ (x + 4)(x + 2) = 0
Then, x = – 4 or x = – 2
II.y2 + 13y + 42 = 0
⇒ y2 + 6y + 7y + 42 = 0
⇒ y(y + 6) + 7(y + 6) = 0
⇒ (y + 7)(y + 6) = 0
Then, y = – 7 or y = – 6
So, when x = – 4, x > y for y = – 7 and x > y for y = – 6
And when x = – 2, x > y for y = – 7 and x > y for y = – 6
∴ We can clearly observe that x > y.

9.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. x2 – 17x + 72 = 0
II. y2 – 23y + 130 = 0

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

E. x < y

I. x2 – 17x + 72 = 0
⇒ x2 – 9x – 8x + 72 = 0
⇒ x(x – 9) – 8(x – 9) = 0
⇒ (x – 8)(x – 9) = 0
Then, x = 8 or x = 9
II. y2 – 23y + 130 = 0
⇒ y2 – 13y – 10y + 130 = 0
⇒ y(y – 13) – 10(y – 13) = 0
⇒ (y – 10)(y – 13) = 0
Then, y = 10 or y = 13
As, we can observe that both the values of x are less than both values of y then we can conclude x < y.

10.In the following question, two equations are given. You have to solve these equations and find out the values of x and y and determine the relation between them.
I. 3x2 – 51x + 180 = 0
II. 2y2 – 24y + 64 = 0

A. x ≥ y

B. y ≥ x

C. x = y or no specific relation can be established

D. x > y

E. x < y

C. x = y or no specific relation can be established

I. 3x2 – 51x + 180 = 0
⇒ x2 – 17x + 60 = 0 [Dividing both sides by 3] ⇒ x2 – 12x – 5x + 60 = 0
⇒ x(x – 12) – 5(x – 12) = 0
⇒ (x – 12)(x – 5) = 0
Then, x = 12 or x = 5
II. 2y2 – 24y + 64 = 0
⇒ y2 – 12y + 32 = 0 [Dividing both sides by 2] ⇒ y2 – 8y – 4y + 32 = 0
⇒ y(y – 8) – 4(y – 8) = 0
⇒ (y – 8)(y – 4) = 0
Then, y = 8 or y = 4
So, when x = 12, x > y for y = 8 and x > y for y = 4
And when x = 5, x < y for y = 8 and x > y for y = 4
∴ So, we can observe that no clear relationship cannot be determined between x and y.

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