# Set-21 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 the following questions, two equations I and II are given. You have to solve both the equations.

1.
I. 15a2 – 10a – 5 = 0;
II. 20b2 + 18b + 4 = 0

2.
I. 26a2 – 88 = 16;
II. 3√b/8 – √b/4 = 1/√b

3.
I. 2a + 3b = 20;
II. 4a + 2b = 24

4.
I. 10a2 + 9a -7 = 0;
II. 12b + 84 = 122

5.
I. a2 = 196;
II. b2 + 11b + 24 = 0

Directions:6-10) In each of the following questions, two equations (i) and (ii) are given. You have to solve them and find the correct option.

6.
i) 2x2+11x+12=0
ii) x2+2x-35=0

7.
i) x2-20x+99=0
ii) 2y2-15y+28=0

8.
i) 2x+5y-√2601=0
ii) 7x+3y-77=0

9.
i) x2+19y+90=0
ii) 14y2-37y+24=0

10.
i) 5x2+36√3x+192=0
ii) 2y2+9√3y+27=0

• Directions:1-5) In the following questions, two equations I and II are given. You have to solve both the equations.

##### 1. Question

I. 15a2 – 10a – 5 = 0;
II. 20b2 + 18b + 4 = 0

Ans:5
15 a2 + 15a – 5a –  5 = 0

15a (a + 1) – 5 (a + 1) = 0

(15a – 5) (a + 1) = 0

a = 1/3, -1

20b2 + 10b + 8b + 4 = 0

10b (2b +1) + 4 (2b + 1) = 0

(10b + 4) (2b + 1) = 0

b = -2/5,    -1/2

Hence there is no relation.

• ##### 2. Question

I. 26a2 – 88 = 16;
II. 3√b/8 – √b/4 = 1/√b

Ans:3
26a2 = 16 + 88 = 104

a2 = 104/26 = 4

a = ±2

3√b/8 – 2√b/8 = 1/√b

√b/8 = 1/√b

b = 8

a < b.

• ##### 3. Question

I. 2a + 3b = 20;
II. 4a + 2b = 24

Ans:5
4a + 6b = 40 —-I

4a + 2b = 24 —-II

Solve I and II we get

4b = 16

b = 4

Apply the b = 4 in eqn I

a = 4

a= b.

• ##### 4. Question

I. 10a2 + 9a -7 = 0;
II. 12b + 84 = 122

Ans:3
10a2 + 14a – 5a – 7 = 0

2a (5a + 7) – 1 (5a + 7) = 0

(2a – 1) (5a + 7) = 0

a = 1/2, -7/5

12b = 144 – 84 = 60

b = 60/12 = 5

a < b.

• ##### 5. Question

I. a2 = 196;
II. b2 + 11b + 24 = 0

Ans:5
a = ±14

b2 + 8b + 3b + 24 = 0

b (b + 8) + 3 (b + 8) = 0

(b + 3) (b + 8) = 0

b = -3, -8

Hence there is no relation.

Directions:6-10) In each of the following questions, two equations (i) and (ii) are given. You have to solve them and find the correct option.

##### 6. Question

i) 2x2+11x+12=0
ii) x2+2x-35=0

Ans:5
a) 2x2+11x+12=0

2x2+8x+3x+12=0

(x+4)(2x+3)=0

x= -4 and -3/2

b) x2+2x-35=0

x2+7x-5x-35=0

(x+7)(x-5)=0

x=-7 or 5

Hence the relationship cannot be determined

• ##### 7. Question

i) x2-20x+99=0
ii) 2y2-15y+28=0

Ans:1
a) x2-20x+99=0

x2-9x-11x+99=0

(x-9)(x-11)=0

x=9 and 11

b) 2y2-15y+28=0

2y2-8y-7y+28=0

(y-4)(2y-7)=0

y= 7/2 and 4

Thus x>y

• ##### 8. Question

i) 2x+5y-√2601=0
ii) 7x+3y-77=0

Ans:1
After solving above equations we have x= 8 and y=7

Hence x>y

• ##### 9. Question

i) x2+19y+90=0
ii) 14y2-37y+24=0

Ans:2
a) x2+19y+90=0

x2+9y +10y+90=0

(x+9)(x+10)=0

x= -9 and -10

b) 14y2-37y+24=0

14y2-21y-16y+24=0

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

y= 3/2 or 8/7

Hence x<y

• ##### 10. Question

i) 5x2+36√3x+192=0
ii) 2y2+9√3y+27=0

Ans:2
a)5x2+36√3x+192=0

5x2+20√3x+16√3x+192=0

(5x+16√3)(x+4√3)=0

x=-4√3 or -16√3/5

b) 2y2+9√3y+27=0

2y2+6√3y+3√3y +27=0

(2y+3√3)(y+3√3)=0

y= -3√3 or -3√3/2

Thus x<y