 # Quadratic Equation Questions PDF for IBPS Clerk, IBPS PO Pre, NIACL, SBI Clerk, SBI PO Pre, 2019  Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer.
Important for :
1
I. 5x2 + 11x – 12 = 0,          II. 4y2 – 13y – 12 = 0
» Explain it
E
I. 5x2 + 11x – 12 = 0

⇒ 5x2 + 15x – 4x – 12 = 0

⇒ 5x (x + 3) – 4(x + 3) = 0

⇒  (5x – 4) (x + 3) = 0

 ⇒ x = 4 , – 3 5

II. 4y2 – 13y – 12 = 0

⇒ 4y2 –  16y + 3y – 12 = 0

⇒ 4y(y –  4) + 3 (y – 4) = 0

⇒  (4y + 3) (y – 4) = 0

 ⇒ y = – 3 , 4 4

While comparing the root values of x and y, we find that one root value of y lies between the root values of x.

Therefore, relationship between x and y can't be determined.

Hence, option E is correct.

2
I. 3x2 + 19x + 30 = 0,          II. 3y2 – 20y – 32 = 0
» Explain it
D
I. 3x2 + 19x + 30 = 0

⇒  3x2 + 9x + 10x + 30 = 0

⇒  3x2 + 9x + 10x + 30 = 0

⇒ 3x (x + 3) + 10 (x + 3) = 0

⇒ (3x + 10)(x + 3) = 0

 ⇒ x = – 10 , – 3 3

II. 3y2 – 20y – 32 = 0

⇒ 3y2 – 24y + 4y – 32 = 0

⇒ 3y(y – 8) + 4(y – 8) = 0

⇒ (3y + 4) (y – 8) = 0
 ⇒ y = – 4 , 8 3

While comparing the root values x and y, we find that root values x is less than y's.

Therefore, x < y

Hence, option D is correct.

3
I. x2 – 4√7x + 21 = 0,          II. 2y2 – 8√5y – 50 = 0
» Explain it
E
I. x2 – 4√7x + 21 = 0

⇒ x2 – √7x – 3√7x + 21 = 0

⇒ x (x – √7) – 3√7 (x – √7) = 0

⇒ (x – √7)(x – 3√7) = 0

⇒ x = √7, 3√7

II. 2y2 – 8√5y – 50 = 0

⇒ 2y2 – 8√5y – 50 = 0

Taking 2 as a common term, we get

⇒ y2 – 4√5y – 25 = 0

⇒ y2 + √5y – 5√5y – 25 = 0

⇒ y( y + √5) – 5√5 (y + √5) = 0

⇒ (y + √5) (y – 5√5) = 0

⇒ y = – √5, 5√5

While comparing the root values of x and y, we find that root values of y lies between the x's root values.

Therefore, relationship between x and y can't be determined.

Hence, option E is correct.

4
I. x2– 52x + 667 = 0,          II. y+ 8y – 33 = 0
» Explain it
A
x– 52x + 667 = 0

⇒ x2 – 23x – 29x + 667 = 0

⇒ x (x – 23) – 29 (x – 23) = 0

⇒ (x – 23) (x – 29) = 0

⇒ x = 23, 29

y+ 8y – 33 = 0

⇒ y2 – 3y + 11y – 33 = 0

⇒ y (y – 3) + 11 (y – 3) = 0

⇒ (y – 3) (y + 11) = 0

⇒ y = 3, – 11

Therefore, x > y

Hence, option A is correct.

5
I. x– 13 √2 x + 60 = 0,          II. y+ 3√5 y – 20 = 0
» Explain it
A
I.  x2 – 13√2 x + 60 = 0

⇒ x2 – 10√2x – 3√2 x + 60 = 0

⇒ x(x – 10√2) – 3√2 (x – 10√2) = 0

⇒ (x – 3√2) (x – 10√2) = 0

x = 3√2, 10√2

II.  y+ 3√5 y – 20 = 0

⇒  y2  + 4√5 y  – √5 y – 20 = 0

⇒  y(y  + 4√5)  – √5 (y + 4√5) = 0

⇒  (y – √5)(y + 4√5) = 0

⇒ y = – 4√5,√5

While comparing the root values of x and y, we find that the x's root values are greater than y's.

Hence, option A is correct.