All Exams > UPPSC (UP) > 6 Months Preparation Course for UPPSC > MCQ Questions
Quadratic Equations MCQs for UPPSC (UP) Exam
It covers all Important Questions with answers on Quadratic Equations for the UPPSC (UP) exam. The questions are based on important topics. Details about the questions:- Topic: Quadratic Equations
- Type of Questions: MCQs with solutions
- Number of Questions: 20
- You can attempt them on EduRev to score high in UPPSC (UP) exam.
I. x2 + 5x + 6 = 0,
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?- a)If x < y
- b)If x > y
- c)If x ≤ y
- d)If x = y or the relationship between x and y cannot be established.
- e)If x ≥ y
Correct answer is option 'D'. Can you explain this answer?
I. x2 + 5x + 6 = 0,
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?
II. y2 + 9y +14 = 0 to solve both the equations to find the values of x and y?
a)
If x < y
b)
If x > y
c)
If x ≤ y
d)
If x = y or the relationship between x and y cannot be established.
e)
If x ≥ y
 | Kavya Saxena answered |
I. x2 + 3x + 2x + 6 = 0
=> (x + 3)(x + 2) = 0 => x = -3 or -2
II. y2 + 7y + 2y + 14 = 0
=> (y + 7)(y + 2) = 0 => y = -7 or -2
No relationship can be established between x and y.
=> (x + 3)(x + 2) = 0 => x = -3 or -2
II. y2 + 7y + 2y + 14 = 0
=> (y + 7)(y + 2) = 0 => y = -7 or -2
No relationship can be established between x and y.
Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for "Quadratic Equations" under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations. Q. Find the roots of the quadratic equation: x2 + 2x - 15 = 0?- a)-5, 3
- b)3, 5
- c)-3, 5
- d) -3, -5
- e)5, 2
Correct answer is option 'A'. Can you explain this answer?
Practice Quiz or MCQ (Multiple Choice Questions) with solutions are available for Practice, which would help you prepare for "Quadratic Equations" under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.
Q. Find the roots of the quadratic equation: x2 + 2x - 15 = 0?
a)
-5, 3
b)
3, 5
c)
-3, 5
d)
-3, -5
e)
5, 2
 | Manoj Ghosh answered |
Rules for factorising:
ax^2 + bx + c = 0
m x n = c
m + n = b
∴(x+m)(x+n)=0
In this case: 5 x −3 = −15 = c
5 + −3 = 2 = b
∴ x^2 + 2x −15 = 0 → (x+5)(x−3) = 0
Either of the brackets must be equal to 0.
Assuming (x+5) = 0:
x = −5
Assuming (x-3) = 0:
x= 3
I. a2 - 2a - 8 = 0,
II. b2 = 9 to solve both the equations to find the values of a and b?- a)If a < b
- b)If a ≤ b
- c)If the relationship between a and b cannot be established
- d)If a > b
- e)If a ≥ b
Correct answer is option 'C'. Can you explain this answer?
I. a2 - 2a - 8 = 0,
II. b2 = 9 to solve both the equations to find the values of a and b?
II. b2 = 9 to solve both the equations to find the values of a and b?
a)
If a < b
b)
If a ≤ b
c)
If the relationship between a and b cannot be established
d)
If a > b
e)
If a ≥ b
| | Kavya Saxena answered |
Explanation:
I. (a - 4)(a + 2) = 0
=> a = 4, -2
II. b2 = 9
=> b = ± 3
-2 < 3, -2 > -3, 4 > 3, 4 > -3,
No relation can be established between a and b.
=> a = 4, -2
II. b2 = 9
=> b = ± 3
-2 < 3, -2 > -3, 4 > 3, 4 > -3,
No relation can be established between a and b.
The sum and the product of the roots of the quadratic equation x2 + 20x + 3 = 0 are?- a)10, 3
- b) -10, 3
- c)20, -3
- d)-10, -3
- e)None of these
Correct answer is option 'E'. Can you explain this answer?
The sum and the product of the roots of the quadratic equation x2 + 20x + 3 = 0 are?
a)
10, 3
b)
-10, 3
c)
20, -3
d)
-10, -3
e)
None of these
 | Niharika Dey answered |
Explanation:
Sum of the roots and the product of the roots are -20 and 3 respectively.
I. a2 + 11a + 30 = 0,
II. b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?- a)If a < b
- b)If a ≤ b
- c)If the relationship between a and b cannot be established
- d)If a > b
- e)If a ≥ b
Correct answer is option 'B'. Can you explain this answer?
I. a2 + 11a + 30 = 0,
II. b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?
II. b2 + 6b + 5 = 0 to solve both the equations to find the values of a and b?
a)
If a < b
b)
If a ≤ b
c)
If the relationship between a and b cannot be established
d)
If a > b
e)
If a ≥ b
 | Yash Patel answered |
Explanation:
I. (a + 6)(a + 5) = 0
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b
=> a = -6, -5
II. (b + 5)(b + 1) = 0
=> b = -5, -1 => a ≤ b
I. a2 - 9a + 20 = 0,
II. 2b2 - 5b - 12 = 0 to solve both the equations to find the values of a and b?- a)If a < b
- b)If a ≤ b
- c)If the relationship between a and b cannot be established
- d)If a > b
- e)If a ≥ b
Correct answer is option 'E'. Can you explain this answer?
I. a2 - 9a + 20 = 0,
II. 2b2 - 5b - 12 = 0 to solve both the equations to find the values of a and b?
II. 2b2 - 5b - 12 = 0 to solve both the equations to find the values of a and b?
a)
If a < b
b)
If a ≤ b
c)
If the relationship between a and b cannot be established
d)
If a > b
e)
If a ≥ b
 | Anaya Patel answered |
Explanation:
I. (a - 5)(a - 4) = 0
=> a = 5, 4
II. (2b + 3)(b - 4) = 0
=> b = 4, -3/2 => a ≥ b
=> a = 5, 4
II. (2b + 3)(b - 4) = 0
=> b = 4, -3/2 => a ≥ b
Find the quadratic equations whose roots are the reciprocals of the roots of 2x2 + 5x + 3 = 0?- a)3x2 + 5x - 2 = 0
- b)3x2 + 5x + 2 = 0
- c)3x2 - 5x + 2 = 0
- d)3x2 - 5x - 2 = 0
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
Find the quadratic equations whose roots are the reciprocals of the roots of 2x2 + 5x + 3 = 0?
a)
3x2 + 5x - 2 = 0
b)
3x2 + 5x + 2 = 0
c)
3x2 - 5x + 2 = 0
d)
3x2 - 5x - 2 = 0
e)
None of these
| | Yash Patel answered |
Explanation:
The quadratic equation whose roots are reciprocal of 2x2 + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3x2 + 5x + 2 = 0
Hence, 2(1/x)2 + 5(1/x) + 3 = 0
=> 3x2 + 5x + 2 = 0
The roots of the equation 3x2 - 12x + 10 = 0 are?- a)rational and unequal
- b) complex
- c)real and equal
- d)irrational and unequal
- e)rational and equal
Correct answer is option 'D'. Can you explain this answer?
The roots of the equation 3x2 - 12x + 10 = 0 are?
a)
rational and unequal
b)
complex
c)
real and equal
d)
irrational and unequal
e)
rational and equal
 | Gowri Chakraborty answered |
The discriminant of the quadratic equation is (-12)2 - 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?- a)10
- b)8
- c)15
- d)7.50
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
a)
10
b)
8
c)
15
d)
7.50
e)
None of these
 | Nikita Singh answered |
Explanation:
Let the price of each note book be Rs.x.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.
Let the number of note books which can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.
The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?- a)9, 10
- b) 10, 11
- c)11, 12
- d)12, 13
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?
a)
9, 10
b)
10, 11
c)
11, 12
d)
12, 13
e)
None of these
 | Dhruv Mehra answered |
Let the two consecutive positive integers be x and x + 1
x2 + (x + 1)2 - x(x + 1) = 91
x2 + x - 90 = 0
(x + 10)(x - 9) = 0 => x = -10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10.
Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?- a)15
- b)14
- c)24
- d)26
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
Find the value of a/b + b/a, if a and b are the roots of the quadratic equation x2 + 8x + 4 = 0?
a)
15
b)
14
c)
24
d)
26
e)
None of these
| | Manoj Ghosh answered |
a/b + b/a = (a2 + b2)/ab = (a2 + b2 + a + b)/ab
= [(a + b)2 - 2ab]/ab
a + b = -8/1 = -8
ab = 4/1 = 4
Hence a/b + b/a = [(-8)2 - 2(4)]/4 = 56/4 = 14.
I. a2 - 7a + 12 = 0,
II. b2 - 3b + 2 = 0 to solve both the equations to find the values of a and b?- a)if a < b
- b)if a ≤ b
- c)if the relationship between a and b cannot be established.
- d)if a > b
- e) if a ≥ b
Correct answer is option 'D'. Can you explain this answer?
I. a2 - 7a + 12 = 0,
II. b2 - 3b + 2 = 0 to solve both the equations to find the values of a and b?
II. b2 - 3b + 2 = 0 to solve both the equations to find the values of a and b?
a)
if a < b
b)
if a ≤ b
c)
if the relationship between a and b cannot be established.
d)
if a > b
e)
if a ≥ b
| | Gowri Chakraborty answered |
I.(a - 3)(a - 4) = 0
=> a = 3, 4
II. (b - 2)(b - 1) = 0
=> b = 1, 2
=> a > b
I. a2 + 8a + 16 = 0,
II. b2 - 4b + 3 = 0 to solve both the equations to find the values of a and b?- a)If a < b
- b)If a ≤ b
- c)If the relationship between a and b cannot be established
- d)If a > b
- e)If a ≥ b
Correct answer is option 'A'. Can you explain this answer?
I. a2 + 8a + 16 = 0,
II. b2 - 4b + 3 = 0 to solve both the equations to find the values of a and b?
II. b2 - 4b + 3 = 0 to solve both the equations to find the values of a and b?
a)
If a < b
b)
If a ≤ b
c)
If the relationship between a and b cannot be established
d)
If a > b
e)
If a ≥ b
 | Ravi Singh answered |
Explanation:
I. (a + 4)2 = 0 => a = -4
II.(b - 3)(b - 1) = 0
=> b = 1, 3 => a < b
II.(b - 3)(b - 1) = 0
=> b = 1, 3 => a < b
If the roots of a quadratic equation are 20 and -7, then find the equation?- a)x2 + 13x - 140 = 0
- b)x2 - 13x + 140 = 0
- c)x2 - 13x - 140 = 0
- d)x2 + 13x + 140 = 0
- e)None of these
Correct answer is option 'C'. Can you explain this answer?
If the roots of a quadratic equation are 20 and -7, then find the equation?
a)
x2 + 13x - 140 = 0
b)
x2 - 13x + 140 = 0
c)
x2 - 13x - 140 = 0
d)
x2 + 13x + 140 = 0
e)
None of these
| | Deepika Banerjee answered |
Explanation:
Any quadratic equation is of the form
x2 - (sum of the roots)x + (product of the roots) = 0 ---- (1)
where x is a real variable. As sum of the roots is 13 and product of the roots is -140, the quadratic equation with roots as 20 and -7 is: x2 - 13x - 140 = 0.
x2 - (sum of the roots)x + (product of the roots) = 0 ---- (1)
where x is a real variable. As sum of the roots is 13 and product of the roots is -140, the quadratic equation with roots as 20 and -7 is: x2 - 13x - 140 = 0.
One root of the quadratic equation x2 - 12x + a = 0, is thrice the other. Find the value of a?- a)29
- b)-27
- c)28
- d)7
- e)None of these
Correct answer is option 'E'. Can you explain this answer?
One root of the quadratic equation x2 - 12x + a = 0, is thrice the other. Find the value of a?
a)
29
b)
-27
c)
28
d)
7
e)
None of these
| | Surbhi Sen answered |
Explanation:
Let the roots of the quadratic equation be x and 3x.
Sum of roots = -(-12) = 12
a + 3a = 4a = 12 => a = 3
Product of the roots = 3a2 = 3(3)2 = 27.
Sum of roots = -(-12) = 12
a + 3a = 4a = 12 => a = 3
Product of the roots = 3a2 = 3(3)2 = 27.
If the roots of the equation 2x2 - 5x + b = 0 are in the ratio of 2:3, then find the value of b?- a)3
- b)4
- c)5
- d)6
- e)None of these
Correct answer is option 'A'. Can you explain this answer?
If the roots of the equation 2x2 - 5x + b = 0 are in the ratio of 2:3, then find the value of b?
a)
3
b)
4
c)
5
d)
6
e)
None of these
| | Aarav Sharma answered |
To find the value of b, we need to use the fact that the roots of the equation are in the ratio of 2:3.
Let's assume the roots of the equation are 2k and 3k, where k is a constant.
Using the sum and product of roots formulas, we can write the equation as follows:
Sum of roots: 2k + 3k = -(-5/2) = 5/2
Product of roots: (2k)(3k) = b/2
Simplifying the equations, we get:
5k = 5/2
6k^2 = b/2
Now, let's solve for k:
5k = 5/2
k = 1/2
Substituting the value of k in the second equation, we get:
6(1/2)^2 = b/2
6(1/4) = b/2
6/4 = b/2
3/2 = b/2
b = 3
Therefore, the value of b is 3, which corresponds to option A.
Let's assume the roots of the equation are 2k and 3k, where k is a constant.
Using the sum and product of roots formulas, we can write the equation as follows:
Sum of roots: 2k + 3k = -(-5/2) = 5/2
Product of roots: (2k)(3k) = b/2
Simplifying the equations, we get:
5k = 5/2
6k^2 = b/2
Now, let's solve for k:
5k = 5/2
k = 1/2
Substituting the value of k in the second equation, we get:
6(1/2)^2 = b/2
6(1/4) = b/2
6/4 = b/2
3/2 = b/2
b = 3
Therefore, the value of b is 3, which corresponds to option A.
For all x, x2 + 2ax + (10 − 3a) > 0, then the interval in which a lies, is?- a)a < -5
- b)a > 5
- c)-5 < a < 2
- d)2 < a < 5
- e)a < -2
Correct answer is option 'C'. Can you explain this answer?
For all x, x2 + 2ax + (10 − 3a) > 0, then the interval in which a lies, is?
a)
a < -5
b)
a > 5
c)
-5 < a < 2
d)
2 < a < 5
e)
a < -2
| | Rahul Mehta answered |
In f(x) = ax2 + bx + c
When a > 0 and D < 0
Then f(x) is always positive.
x2 + 2ax + 10 − 3a > 0, ∀x ∈ R
⇒ D < 0
⇒ 4a2 − 4(10 − 3a) < 0
⇒ a2 + 3a − 10 < 0
⇒ (a+5)(a−2) < 0
⇒ a ∈ (−5,2)
When a > 0 and D < 0
Then f(x) is always positive.
x2 + 2ax + 10 − 3a > 0, ∀x ∈ R
⇒ D < 0
⇒ 4a2 − 4(10 − 3a) < 0
⇒ a2 + 3a − 10 < 0
⇒ (a+5)(a−2) < 0
⇒ a ∈ (−5,2)
If the roots of the equation (a2 + b2)x2 − 2b(a + c)x + (b2+c2) = 0 are equal then - a)2b = ac
- b)b2 = ac
- c)b = 2ac/(a + c)
- d)b = ac
- e)b = 2ac
Correct answer is option 'B'. Can you explain this answer?
If the roots of the equation (a2 + b2)x2 − 2b(a + c)x + (b2+c2) = 0 are equal then
a)
2b = ac
b)
b2 = ac
c)
b = 2ac/(a + c)
d)
b = ac
e)
b = 2ac
| | Ritika Choudhury answered |
(a2 + b2)x2 − 2b(a + c)x + (b2+c2) = 0
Roots are real and equal ∴ D = 0
D = b2 − 4ac = 0
⇒ [−2b(a+c)]2 − 4(a2 + b2)(b2 + c2) = 0
⇒ b2(a2 + c2 + 2ac) −(a2b2 + a2c2 + b4 + c2c2) = 0
⇒ b2a2 + b2c2 + 2acb2 − a2b2 − a2c2 − b4 − b2c2 = 0
⇒ 2acb2 − a2c2 − 2acb2 = 0
⇒ (b2 − ac)2 = 0
⇒ b2 = ac
Roots are real and equal ∴ D = 0
D = b2 − 4ac = 0
⇒ [−2b(a+c)]2 − 4(a2 + b2)(b2 + c2) = 0
⇒ b2(a2 + c2 + 2ac) −(a2b2 + a2c2 + b4 + c2c2) = 0
⇒ b2a2 + b2c2 + 2acb2 − a2b2 − a2c2 − b4 − b2c2 = 0
⇒ 2acb2 − a2c2 − 2acb2 = 0
⇒ (b2 − ac)2 = 0
⇒ b2 = ac
Chapter doubts & questions for Quadratic Equations - 6 Months Preparation Course for UPPSC 2026 is part of UPPSC (UP) exam preparation. The chapters have been prepared according to the UPPSC (UP) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for UPPSC (UP) 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Quadratic Equations - 6 Months Preparation Course for UPPSC in English & Hindi are available as part of UPPSC (UP) exam. Download more important topics, notes, lectures and mock test series for UPPSC (UP) Exam by signing up for free.
6 Months Preparation Course for UPPSC774 videos|2785 docs|928 tests |
