# Set-2 Quadratic Equations (Inequalities) For SBI PO and SBI Clerk 2019 | Must Go Through These Questions

Dear Aspirants,
We are providing the most important Quadratic Equations (Inequalities) for SBI PO 2019, SBI Clerk 2019 and all other competitive bank and insurance exams. These questions have very high chances to be asked in SBI PO 2019, SBI Clerk 2019.

Directions:1-5) In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
1.

I.x2 +11x +18 = 0
II.y2 +16y+ 48 = 0

2.

I.3x2 +10x +7 = 0
II.3y2 +4y+ 1 = 0

3.

I.5x2 -7x +2 = 0
II.2y2 -7y+ 3 = 0

4.

I.2x2 -3x +1 = 0
II.y2-8 y+ 15 = 0

5.

I)x = (-4)2
II)y2=256

Directions:6-10):Two equations (I) and (II) are given in each question. On the basis of these questions, you have to decide the relation between p and q and give answer:

6.

1. 5p2– 87p + 378 = 0
2. 3q2– 49q + 200 = 0

7.

1. 10p2– p – 24 = 0
2. q2– 2q = 0

8.

1. p2– 5p + 6 = 0
2. 2q2– 15q + 27 = 0

9.

1. 3p + 2q = 301
2. 7p – 5q = 74

10.

1. 14p2– 37p + 24 = 0
2. 28q2– 53q + 24 = 0

• Directions:1-5) In the given question two equations are given. You have to Solve both the equations and given answer.

##### 1. Question

I.x2 +11x +18 = 0
II.y2 +16y+ 48 = 0

Ans:5
I)x2 +11x +18 = 0
x2 +2x+9x+18=0
x(x+2)+9(x+2)=0
(x+2) (x+9)=0
x=-2,-9y2 +16 y+ 48 = 0
y2+4y+12y+48=0
y(y+4) + 12(y+4)=0
(y+4)(y+12)=0
y=-4,-12
So relationship between x and y cannot be established

• ##### 2. Question

I.3x2 +10x +7 = 0
II.3y2 +4y+ 1 = 0

Ans:4
I.3x2 +10x +7 = 0
3x2 +3x+7x+7=0
3x(x+1) + 7(x+1)=0
(x+1) (3x+7)=0
x=-1,-7/3II.3y2 +4y+ 1 = 0
3y2 + 3y+y+1=0
3y(y+1) + 1(y+1) =0
(y+1) (3y+1)=0
y=-1,-1/3
So x ≤ y

• ##### 3. Question

I.5x2 -7x +2 = 0
II.2y2 -7y+ 3 = 0

Ans:5
I.5x2 -7x +2 = 0
5x2-5x-2x+2=0
5x(x-1) -2(x-1)=0
(x-1) (5x-2)=0
x=1,2/5II.2y2 -7 y+ 3 = 0
2y2-6y-y+3=0
2y(y-3) -1(y-3)=0
(y-3) (2y-1)=0
y=3,1/2
Relationship between x and y cannot be established

• ##### 4. Question

I.2x2 -3x +1 = 0
II.y2-8 y+ 15 = 0

Ans:2
I.2x2 -3x +1 = 0
2x2 -2x-x +1=0
2x(x-1)-1(x-1)=0
(x-1) (2x-1)=0
x=1,1/2II.y2-8 y+ 15 = 0
y2-3y-5y+15=0
y(y-3)-5(y-3)=0
(y-3)(y-5)=0
y=3,5
So x <y

• ##### 5. Question

I)x = (-4)2
II)y2=256

Ans:3
I)x = (-4)2x=16

II)y2=256

y=16 and -16

Hence x ≥ y

Directions:6-10):Two equations (I) and (II) are given in each question. On the basis of these questions, you have to decide the relation between p and q and give answer

##### 6. Question
1. 5p2– 87p + 378 = 0
2. 3q2– 49q + 200 = 0
Ans:1
I.5p– 45p – 42p + 378 = 05p (p – 9) – 42(p- 9) = 0

(5p – 42)(p -9) = 0

p = 9, 42/5 = 9, 8.4

II.3q– 24q – 25q + 200=0

3q (q – 8) -25(q – 8) = 0

(q – 8)(3q-25)=0

q = 8, 25/3 = 8, 8.33

Hence, p>q

• ##### 7. Question
1. 10p2– p – 24 = 0
2. q2– 2q = 0
Ans:5
I.10p+15p – 16p – 24 = 05p(2p + 3) – 8(2p+ 3)=0

(2p + 3)(5p – 8) = 0

p = -3/2, 8/5 = -1.5, 1.6

II.q2– 2q = 0

q(q -2) = 0

q = 0, 2

(i.e.) no relationship exists between p and q.

• ##### 8. Question
1. p2– 5p + 6 = 0
2. 2q2– 15q + 27 = 0
Ans:4
p2 – 2p – 3p + 6 =0p(p -2) – 3(p -2) = 0

(p -2) (p -3) = 0

p =2, 3

2q – 6q – 9q + 27 = 0

2q(q – 3)-9(q -3) = 0

(q- 3) (2q -9) = 0

q = 3, 9/2 = 3, 4.5

Hence, p ≤ q

• ##### 9. Question
1. 3p + 2q = 301
2. 7p – 5q = 74
Ans:2
I.eqn (I) × 5 + eqn (II) × 2[15p + 10q = 1505] + [14p – 10q = 148] = 29p = 1653

p = (1653/29) = 57

And q = 65

Hence, p< q

• ##### 10. Question
1. 14p2– 37p + 24 = 0
2. 28q2– 53q + 24 = 0
Ans:3
14p2 – 37p + 24 = 014p– 21p- 16p + 24 = 0

7p(2p -3) -8(2p -3) = 0

(2p – 3)(7p – 8) = 0

p = (3/2), (8/7)

II.28q– 53q + 24 = 0

28q -21q – 32q + 24 =0

7q(4q – 3) -8(4q – 3) = 0

(7q – 8) (4q – 3) = 0

q = 8/7, 3/4

p ≥ q