# Set-14 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:1-5) 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. 6a2 + 7a -3 = 0

II. b (10b – 1) = 2

2.

I. 4a2 + 3a – 27 = 0

II. 15b2 – 38b – 21 = 0

3.

I. 8a2 + 5a – 13 = 0

II. 2b2 + 23b + 63 = 0

4.

I. 4a2 + 19a + 21 = 0

II. 2b2 – 25b – 27 = 0

5.

I. a2 – 9a + 20 = 0

II. b2 – 11b + 30 = 0

Directions:6-10) In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,

6.

I. 4x2 + 17x + 18 = 0

II. 4y2 – 4y  –  15 = 0

7.

I. x3 = 2197

II. y2 + 26y + 169 = 0

8.

I. 3x = 5y – 5

II. x – 4y – 3 = 0

9.

I. 3x2 – 16x + 21 = 0

II. 3y2– 18y + 27 = 0

10.

I. x + √289 = √1089

II. y1/2 = 64 ÷ y

Check your Answers below:

• 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. 6a2 + 7a -3 = 0

II. b (10b – 1) = 2

Ans:5
6 a2 + 7a -3 = 0
6 a2 + 9a – 2a – 3 = 0
So a = -1.5, 0.3
b (10b – 1) = 2
10b2 – b – 2 = 0
10b2 – 5b + 4b – 2 = 0
So b = 0.5, -0.4
-1.5… -0.4… 0.3… 0.5.
• ##### 2. Question

I. 4a2 + 3a – 27 = 0

II. 15b2 – 38b – 21 = 0

Ans:5
4a2 + 3a – 27 = 0
4a2 + 12a – 9a – 27 = 0
So a =2.25, -3
15b2 – 38b – 21 = 0
15b2 – 45b + 7b – 21 = 0
So b = 3, – 0.46
-3… -0.46… 2.25…. 3.
• ##### 3. Question

I. 8a2 + 5a – 13 = 0

II. 2b2 + 23b + 63 = 0

Ans:1
8a2 + 5a – 13 = 0
8a2 + 13a – 8a – 13 = 0
So a = -1.625, 1
2b2 + 23b + 63 = 0
2b2 + 14b + 9b + 63 = 0
So b = -7, -4.5
-7…. -4.5 ….-1.625…. 1.
• ##### 4. Question

I. 4a2 + 19a + 21 = 0

II. 2b2 – 25b – 27 = 0

Ans:2
4a2 + 19a + 21 = 0
4a2 + 12a + 7a + 21 = 0,
So a = -3, – 1.75
2b2 – 25b – 27 = 0
2b2 – 27b + 2b – 27 = 0
So b = 13.5, -1
-3…. -1.75….. -1…..13.5.
• ##### 5. Question

I. a2 – 9a + 20 = 0

II. b2 – 11b + 30 = 0

Ans:4
a2 – 9a + 20 = 0
a2 – 4a – 5a + 20 = 0
So a = 4, 5
b2 – 11b + 30 = 0
b2 – 5b -6b + 30 = 0
So b = 5, 6
4 …. 5….5……6.

Directions:6-10) In the following questions, two equations I and II are given. You have to solve both the equations and give Answer as,

##### 6. Question

I. 4x2 + 17x + 18 = 0

II. 4y2 – 4y  –  15 = 0

Ans:3
I. 4x2 + 17x + 18 = 04x2 + 8x + 9x + 18 = 0

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

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

X = -9/4, -2 = -2.25, -2

II. 4y2 – 4y – 15 = 0

4y2 + 6y – 10y – 15 = 0

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

(2y – 5) (2y + 3) = 0

Y = 5/2, -3/2 = 2.5, -1.5

X < y

• ##### 7. Question

I. x3 = 2197

II. y2 + 26y + 169 = 0

Ans:1
x3 = 2197X = 13

II. y2 + 26y + 169 = 0

(y + 13) (y + 13) = 0

Y = -13, -13

X > y

• ##### 8. Question

I. 3x = 5y – 5

II. x – 4y – 3 = 0

Ans:3
3x = 5y – 5 —- > (1)x – 4y – 3 = 0 –à (2)

By substituting (1) and (2), we get,

X = -5, y = -2

X < y

• ##### 9. Question

I. 3x2 – 16x + 21 = 0

II. 3y2– 18y + 27 = 0

Ans:4
I. 3x2 – 16x + 21 = 03x2 – 9x – 7x + 21 = 0

3x(x – 3) -7(x – 3) = 0

(3x – 7) (x – 3) = 0

X = 7/3, 3 = 3, 2.33

II. 3y2– 18y + 27 = 0

3y2– 9y – 9y + 27 = 0

3y(y -3) -9(y – 3) = 0

(3y – 9) (y – 3) = 0

Y = 9/3, 3 = 3, 3

x ≤ y

• ##### 10. Question

I. x + √289 = √1089

II. y1/2 = 64 ÷ y

Ans:5
I. x + √289 = √1089X + 17 = 33

X = 33 – 17 = 16

II. y1/2 = 64 ÷ y

Y1/2 * y = 64

Y3/2 = 64

Y = (64)2/3 = (641/3)2

Y = 42 = 16

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