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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 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
Check your Answers below:
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
I.(25/p2) + (9/p2 ) – (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
- √(900)p + √(1296) = 0
- (256)1/4q + (216)1/3 = 0
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
I.[(3)5 + (7)3] / 3 = p3
(243 + 343) / 3 = p3
(586 / 3) = p3 à p = 5.8
II.7q3 = -30 + 17q3 = 10q3 = 30
q3 = 30/10 = 3 à q = 1.44
Clearly, p > q
4. Question
- (p1/4 / 16)2 = 144 / p3/2
- q1/3 × q2/3 × 3104 = 16 × q2
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
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
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
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
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
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
Sol. Solving the two equations using the substitution method,
p = 7
q = 2
Thus, p>q
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