# Set-20 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 numbered are given in variables a and b. you have to solve both the equations and find out the relationship between a and b. Then give answer accordingly-

1.

I.5a2 – 18a + 9 =0
II.3b2 + 5b – 2 = 0

2.

I.√a – √6 / √a = 0
II. b3 – 63/2 = 0

3.

I.(625)1/4a + √1225 = 155
II.√196b + 13 = 279

4.

I. 3a2 – 17a + 24 =0
II. 4b2 – 15b + 14 = 0

5.

I. a2 – 2a – √5 a + 2√5 = 0
II. b2 – √3 b – √2 b + √6 =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.

a)3+1/x-10/x2=0
b)3y2-5y=12

7.

a)3y-4x+5=0
b) 6x-5y=11

8.

a)7x2+21x+14=0
b)6y2-35y-26=0

9.

a)15x2-61x+56=0
b)20y2+51y+28=0

10.

a)4x2-17x+18=0
b)2y2+13y+21=0

• Directions:1-5) In the following questions, two equations numbered are given in variables a and b. you have to solve both the equations and find out the relationship between a and b. Then give answer accordingly-

##### 1. Question

I.5a2 – 18a + 9 =0
II.3b2 + 5b – 2 = 0

Ans: 1
5a2 – 18a + 9  = 0
=> 5a2 – 15a – 3a  + 9 =0
=> (5a – 3 )(a-3 )= 0
=> a=  3/ 5 or  a= 3
3b2 + 5b – 2 = 0
=> 3b2 + 6b – b -2 = 0
=> (3b-1)(b + 2) = 0
=> b = 1/3  or  -2
• ##### 2. Question

I.√a – √6 / √a = 0
II. b3 – 63/2 = 0

Ans: 5
√a –   √6 / √a = 0
a – √6 = 0
a = √6
b3 – 6(3/2) = 0
=>b3 = (√6)3
=> b = √6There is no relation between a and b

• ##### 3. Question

.(625)1/4a + √1225 = 155
II.√196b + 13 = 279

Ans: 1
5a + 35 = 155
=> 5a = 155 – 35
=> a = 120/ 5 = 24√196 b + 13 = 279
=> 14b = 279 -13
=> b = 266/14 =19

From I and II, a>b

• ##### 4. Question

I. 3a2 – 17a + 24 =0
II. 4b2 – 15b + 14 = 0

Ans: 1
3a2 -17a  + 24 =0
=> 3a2 – 9a  – 8a +24 = 0
=> (3a- 8)(a-3) = 0
=>  a=  8/3 or 3
4b2 – 15b + 14 = 0
=> 4b2 – 8b -7b +14 = 0
=> (4b -7 )(b-2)=0
=> b = 7 /4 or  2From I and II, a>b

• ##### 5. Question

I. a2 – 2a – √5 a + 2√5 = 0
II. b2 – √3 b – √2 b + √6 =0

Ans: 1
a2 – 2a- √5 a + 2√5 = 0
=> a(a-2 ) – √5 (a-2 )= 0
=> (a-2)(a-√5)=0
=> a= 2 or √5
b2 – √3 b  – √2 b + √6 =0
=> b(b-√3) – √2(b – √3) = 0
=> (b – √2)(b- √3) =0
=> b = √2 or √3From I and II, a>b

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

a)3+1/x-10/x2=0
b)3y2-5y=12

Ans: 5
a)3+1/x-10/x2=03+1/x-10/x2=0

3x2+x-10=0

(3x-5)(x+2)=0

x=5/3 and -2

b)3y2-5y=12

3y2-5y-12=0

(3y+4)(y-3)=0

y= 3 and -4/3

Hence, relationship cannot be determined

• ##### 7. Question

a)3y-4x+5=0
b) 6x-5y=11

Ans: 1
3y-4x+5=0………………….. (1)6x-5y=11…………………… (2)

Multiply equation (i) by3

9y-12x+15=0

And multiply equation (ii) by 2

12x-10y=22

Solving these equations, we get

x=-4 and y=-7

Hence x>y

• ##### 8. Question

a)7x2+21x+14=0
b)6y2-35y-26=0

Ans: 2
a)7x2+21x+14=07x2+14x+7x+14=0

(x+1)(7x+14)=0

x=-1 and -2

b)6y2-35y-26=0

6y2+4y-39y-26=0

2y(3y+2)-13(3y+2)=0

(2y-13)(3y+2)=0

y=13 /2 and -2/3

y= 6.5 and -0.6

Hence x<y

• ##### 9. Question

a)15x2-61x+56=0
b)20y2+51y+28=0

Ans: 1
a)15x2-61x+56=015x2-40x-21x+56=0

(5x-7)(3x-8)=0

x=8/3 and 7/5

b)20y2+51y+28=0

20y2+16y+35y+28=0

(4y+7)(5y+4)=0

y=-4/5 and -7/4

Hence, x>y

• ##### 10. Question

a)4x2-17x+18=0
b)2y2+13y+21=0

Ans: 1
a)4x2-17x+18=04x2-8x-9x+18=0

(4x-9)(x-2)=0

x=9/4 and 2

b)2y2+13y+21=0

2y2+7y+6y+21=0

(y+3)(2y+7)=0

y=-3 and -7/2

Hence x>y