# Set-6 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.
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Directions:1-10) In each of these questions, two equations I and II are given. You have to solve both the equations and given answer

1.

I)a7– {(45)15/2 /√a} =0

II) 44/ √b + √b =89 / √b

2.

1. (a+189)1/2+(5184)1/2 =86
2. b2-11b+30 =0

3.

1. a=3√512
2. b=√289

4.

I.15a2-8a+1=0

II.18b2-11b+1=0

5.

1. a2-9a=0
2. 3a2+5a+2 =0

6.

1. 4x2 – 15x + 9 = 0
2. 2y2 – 13y + 21 = 0

7.

1. 3x2 – 17x + 20 = 0
2. 3y2 + y – 10 = 0

8.

1. 3x2 – 4x – 32 = 0
2. 3y2 + 14y + 16 = 0

9.

1. 4x2 + 4x – 3 = 0
2. 2y2 – 13y + 21 = 0

10.

1. 3x2 – 7x – 20 = 0
2. 3y2 + 20y + 25 = 0

Check your Answers below:

• Directions:1-10) In each of these questions, two equations I and II are given. You have to solve both the equations and given answer

##### 1. Question

I)a7– {(45)15/2 /√a} =0

II) 44/ √b + √b =89 / √b

Ans:5

1. I)a7– {(45)15/2 / √a} =0

a15/2 =(45)15/2

a=45

1. II)(44+b) / √b = 89/√b

44+b=89

b=89-44=45

• ##### 2. Question
1. (a+189)1/2+(5184)1/2 =86
2. b2-11b+30 =0
Ans:1

1. (a+189)1/2+(5184)1/2=86

Or,(a+189)1/2=86-72=14

Now, squaring both sides, we get

a+189=196

Or, a=196-189=7

II.b2-11b+30=0

Or,b2-6b-5b+30=0

Or, b(b-6)-5(b-6)=0

Or,(b-5)(b-6)=0

∴ b= 5, 6

Hence a>b

• ##### 3. Question
1. a=3√512
2. b=√289
Ans:2
a = 3√512

a = 8

b = √289 = 17

a<b

• ##### 4. Question

I.15a2-8a+1=0

II.18b2-11b+1=0

Ans:4
I.15a2-8a+1 =0

Or,15a2-5a-3a+1=0

Or,5a(3a-1)-1(3a-1) =0

Or,(5a-1) (3a-1) =0

∴a=1/5,1/3

II.18b2-11b+1=0

or,18b2-9b-2b+1=0

or,9b(2b-1)-1(2b-1) =0

or,(9b-1) (2b-1) =0

∴b =1/9,1/2

Hence, no relation can be established between a and b.

• ##### 5. Question
1. a2-9a=0
2. 3a2+5a+2 =0
Ans:1
I.a2-9a =0

Or,a2=9a

∴a=9

II.3a2+5a+2 =0

Or,3a2+3a+2a+2=0

Or,3a(a+1)+2(a+1) =0

Or,(3a+2)(a+1) =0

∴a= -2/3, -1

Hence a>b

• ##### 6. Question
1. 4x2 – 15x + 9 = 0
2. 2y2 – 13y + 21 = 0
Ans:4
I.4x2 – 15x + 9 = 0

=> 4x2 – 12x – 3x+ 9 = 0

=> 4x(x – 3) – 3(x – 3) = 0

=> (4x – 3) (x – 3) = 0

=> x = 3, 4/3

II. 2y2 – 13y + 21 = 0

=> 2y2 – 7y – 6y + 21 = 0

=> y(2y – 7) – 3(2y – 7) = 0

=> y= 3, 7/2

Y≥X

• ##### 7. Question
1. 3x2 – 17x + 20 = 0
2. 3y2 + y – 10 = 0
Ans:3
I.3x2 – 17x + 20 = 0

=> 3x2 – 12x – 5x + 20 = 0

=> 3x(x – 4) – 5(x – 4) = 0

=> (3x – 5)(x – 4) = 0

X = 5/3, 4

II. 3y2 + y – 10 = 0

=> 3y2 + 6y – 5y – 10 = 0

=> 3y( y + 2) – 5( y + 2) = 0

=> (y + 2)(3y – 5) = 0

Y = -2, 5/3

X≥Y

• ##### 8. Question
1. 3x2 – 4x – 32 = 0
2. 3y2 + 14y + 16 = 0
Ans:5
I. 3x2 – 4x – 32 = 0

=> 3x2 – 12x + 8x – 32 = 0

=> 3x(x – 4) + 8(x – 4) = 0

=> (x – 4) (3x + 8) = 0

X = 4, -8/3

II. 3y2 + 14y + 16 = 0

=> 3y2 + 8y + 6y + 16 = 0

=> Y(3y + 8) + 2(3y + 8) = 0

=> (y + 2)(3y + 8) = 0

Y= -8/3, -2

X=Y or No relation

• ##### 9. Question
1. 4x2 + 4x – 3 = 0
2. 2y2 – 13y + 21 = 0
Ans:1
I.4x2 + 4x – 3 = 0

=> 4x2 + 6x – 2x – 3 = 0

=> 2x(2x + 3) – 1(2x + 3) = 0

=> (2x – 1)(2x + 3) = 0

X = ½ , – 3/2

II. 2y2 – 13y + 21 = 0

=> 2y2 – 7y – 6y + 21 = 0

=> Y(2y– 7) – 3(2y – 7) = 0

=> (y – 3)(2y – 7)= 0

Y = 3, 7/2

Y>X

• ##### 10. Question
1. 3x2 – 7x – 20 = 0
2. 3y2 + 20y + 25 = 0
Ans:3
I. 3x2 – 7x – 20 = 0

=> 3x2 – 12x + 5x – 20 = 0

=> 3x(x – 4) + 5(x – 4) = 0

=> (x – 4)(3x + 5)= 0

X= 4, -5/3

II. 3y2 + 20y + 25 = 0

=> 3y2 + 15y + 5y + 25 = 0

=> 3y(y + 5) + 5(y + 5) = 0

=> (y + 5)(3y + 5) = 0

Y = – 5, -5/3

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