# Set-19 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. Solve both the equations and give Answer

1.

I.(25 / p2) – (12 / p) + (9 / p2) = (4 / p2)
II.9.84 – 2.64 = 0.95 + q2

2.

I. √(900)p + √(1296) = 0
II.(256)1/4q + (216)1/3 = 0

3.

I.[(3)5 + (7)3]/ 3 = p3
II.7q3 = – (15 × 2) + 17q3

4.

I.(p1/4 / 16)2 = 144 / p3/2
II.q1/3  × q2/3 × 3104 = 16 × q2

5.

I.3p2 – 19p +28 = 0
II.5q2 – 18q + 16 = 0

Directions:6-10) In each question, two equations are given. Find p and q and give the answer:

6.

a)(p³ – 12p2 – p + 12)/(p+1) = 0
b)(q³ + 5q² – 2q – 24)/(q-2) = 0

7.

a)q = 2p + 1
b)2q = 3p – 1

8.

a)9p² – 29p + 22 = 0
b)q² – 7q + 12 = 0

9.

a)3p² – 4p – 32 = 0
b)12q² – 109q + 247 = 0

10.

a)4p + 7q = 42
b)3p – 11q = -1

Directions:1-5) In the following questions, two equations I and II are given. Solve both the equations and give Answer

##### 1. Question

I.(25 / p2) – (12 / p) + (9 / p2) = (4 / p2)
II.9.84 – 2.64 = 0.95 + q2

Ans: 2
I.(25/p2) + (9/p) – (4/p2 )  = (12/p)

(25 + 9 – 4) / p2 = 12/p = 30/p2 = 12/p

12p = 30

p = 30 / 12 = 5/2 = 2.5

II.9.84 -2.64 = 0.95 + q2

7.2 – 0.95 = q2

q = √(6.25) = ± (2.5)

Clearly p ≥ q

##### 2. Question
1.  √(900)p + √(1296) = 0
2. (256)1/4q + (216)1/3 = 0
Ans: 1
I.√(900)p + √(1296) =0

√(900)p = -√(1296)

30p = -36

p = -36 / 30 = -1.2

II.(256)1/4 q = (216)1/3

(44)1/4 q = – (63)1/3 à 4q = -6

Q = -(6/4) = -1.5

Clearly, p > q

##### 3. Question

I.[(3)5 + (7)3]/ 3 = p3
II.7q3 = – (15 × 2) + 17q3

Ans: 1
I.[(3)5 + (7)3] / 3 = p3

(243 + 343) / 3 = p3

(586 / 3) = p3 à p = 5.8

II.7q3 = -30 + 17q3 = 10q3 = 30

q= 30/10 = 3 à q = 1.44

Clearly, p > q

##### 4. Question
1. (p1/4 / 16)2 = 144 / p3/2
2. q1/3  × q2/3 × 3104 = 16 × q2
Ans: 3
I.(p1/4 / 16)2 = (144 / p3/2 )  = (p1/2 / 256) = (144 / p3/2 )

(p1/2 )  × (p3/2 ) = 256 × 144

p2 = (256 × 144)

p = √(256 × 144)

p = ± (16 × 12) = ±192

II.q1/3 × q2/3 × 3104 = 16q2

q × 3104 = 16q2

3104 = 16q

Q = 3104 / 16 = 194

Clearly, p < q

##### 5. Question

I.3p2 – 19p +28 = 0
II.5q2 – 18q + 16 = 0

Ans: 1
I.3p2 – 9p + 28 = 0

3p2 – 12p – 7p + 28 = 0

3p (p – 4) – 7 (p – 4) = 0

(p – 4) (3p – 7) = 0

p = 4, 7/3

II.5q2 – 18q + 16 = 0

5q2 – 10q – 8q + 16 = 0

5q (q – 2) – 8 (q – 2) = 0

(q – 2) (5q – 8) = 0

Q = 2, 8/5

Clearly, p > q

Directions:6-10) In each question, two equations are given. Find p and q and give the answer:

##### 6. Question

a)(p³ – 12p2 – p + 12)/(p+1) = 0
b)(q³ + 5q² – 2q – 24)/(q-2) = 0

Ans: 1
a)(p³ – 12p2 – p + 12)/(p+1) = 0

[p2 (p-12) -1(p-12)]/(p+1) = 0

[(p-1)(p+1)(p-12)]/(p+1) = 0

p² – 13p + 12 = 0

Solving, p = 12, 1

b)(q³ + 5q² – 2q – 24)/(q-2) = 0

(q³ + 7q² – 2q² -14q + 12q – 24)/(q-2) = 0

(q-2)(q² + 7q + 12)/(q-2) = 0

Solving, q = -4, -3

Thus, p>q

##### 7. Question

a)q = 2p + 1
b)2q = 3p – 1

Ans: 1
Solving the two using substitution method,

p = -3

q = -5

Thus, p>q

##### 8. Question

a)9p² – 29p + 22 = 0
b)q² – 7q + 12 = 0

Ans: 2
a)9p² – 18p – 11p + 22 = 0

9p(p-2) – 11(p-2) = 0

(9p-11)(p-2) = 0

p = 11/9, 2.

b)q² – 4q – 3q + 12 = 0

q(q-4) – 3(q-4) = 0

(q-4)(q-3) = 0

q = 3, 4

Thus, p<q

##### 9. Question

a)3p² – 4p – 32 = 0
b)12q² – 109q + 247 = 0

Ans: 2
a)3p² – 12p + 8p – 32 = 0

3p(p-4) + 8(p-4) = 0

(3p+8)(p-4) = 0

p = -8/3, 4

b)12q² – 52q – 57q + 247 = 0

4q (3q – 13) – 19(3q – 13) = 0

(4q-19)(3q-13) = 0

q = 19/4, 13/3

Thus, p<q

##### 10. Question

a)4p + 7q = 42
b)3p – 11q = -1

Ans: 1
Sol. Solving the two equations using the substitution method,

p = 7

q = 2

Thus, p>q