Quadratic Equations Test Class 10 Maths — 40 Marks
Pandey Sir’s BEST Classes
PERIODIC TEST — Quadratic Equations (2026-27) | CLASS 10 | MATHEMATICS
Max Marks: 40 · Time: 1 hour 30 minutes (Time left: 90:00)
| Section | Marks | Questions | Time | Strategy |
|---|---|---|---|---|
| A | 12 | 12 MCQ | 15 min | Discriminant and root facts must be instant |
| B | 8 | 4 Short | 20 min | Factorise first; use the formula only if it will not factorise |
| C | 12 | 4 Long | 30 min | Write the formula, substitute, then simplify |
| D | 8 | 2 Case | 20 min | Turn the words into an equation before solving |
| Review | — | — | 5 min | Reject roots that do not fit the situation |
General Instructions: All questions are compulsory. Section A has 12 MCQs of 1 mark each. Internal choice is given in Sections B, C and D — attempt any ONE option. Where a word problem gives a physical quantity, reject the root that is not possible.
Section A — Multiple Choice (12 × 1 = 12 marks)
Choose the single best option.
Which of the following is a quadratic equation?
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Check the highest power
A quadratic equation has degree 2.
(a) degree 2 ✓
(b) degree 3
(c) degree 1
(d) degree 4
(a) degree 2 ✓
(b) degree 3
(c) degree 1
(d) degree 4
Answer: (a) x2 + 2x + 1 = 0
The roots of x2 − 5x + 6 = 0 are
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Factorise
x2 − 5x + 6 = x2 − 2x − 3x + 6
= x(x − 2) − 3(x − 2)
= (x − 2)(x − 3)
So x = 2 or x = 3
= x(x − 2) − 3(x − 2)
= (x − 2)(x − 3)
So x = 2 or x = 3
Answer: (a) 2, 3
The discriminant of 2x2 − 4x + 3 = 0 is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
D = b2 − 4ac
a = 2, b = −4, c = 3
D = (−4)2 − 4 × 2 × 3
= 16 − 24
= −8
D = (−4)2 − 4 × 2 × 3
= 16 − 24
= −8
Answer: (a) −8
If the discriminant of a quadratic equation is 0, the equation has
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Meaning of D
D > 0 → two distinct real roots
D = 0 → two equal real roots
D < 0 → no real roots
D = 0 → two equal real roots
D < 0 → no real roots
Answer: (b) two equal real roots
The value of k for which x2 + kx + 9 = 0 has equal roots is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Set D = 0
a = 1, b = k, c = 9
k2 − 4 × 1 × 9 = 0
k2 = 36
k = ±6
k2 − 4 × 1 × 9 = 0
k2 = 36
k = ±6
Answer: (b) ±6
The sum of the roots of 3x2 − 9x + 5 = 0 is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
α + β = −ba
a = 3, b = −9
α + β = 93
= 3
α + β = 93
= 3
Answer: (a) 3
The product of the roots of x2 + 7x + 10 = 0 is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
αβ = ca
a = 1, c = 10
αβ = 10
αβ = 10
Answer: (c) 10
Which equation has no real roots?
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Check each discriminant
(a) D = 16 − 16 = 0
(b) D = 16 − 20 = −4 → no real roots
(c) D = 25 − 24 = 1
(d) D = 9 − 8 = 1
(b) D = 16 − 20 = −4 → no real roots
(c) D = 25 − 24 = 1
(d) D = 9 − 8 = 1
Answer: (b) x2 + 4x + 5 = 0
The roots of x2 = 16 are
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Take both square roots
x2 − 16 = 0
(x − 4)(x + 4) = 0
x = 4 or x = −4
(x − 4)(x + 4) = 0
x = 4 or x = −4
Answer: (c) ±4
If one root of x2 − 3x + k = 0 is 2, then k is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Substitute x = 2
(2)2 − 3(2) + k = 0
4 − 6 + k = 0
k = 2
4 − 6 + k = 0
k = 2
Answer: (b) 2
The quadratic formula for ax2 + bx + c = 0 is x =
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Recall the formula
x = −b ± √(b2 − 4ac)2a
Both the sign in front of b and the denominator 2a must be correct.
Both the sign in front of b and the denominator 2a must be correct.
Answer: (a)
The quadratic equation whose roots are 3 and −2 is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Use sum and product
Sum = 3 + (−2) = 1
Product = 3 × (−2) = −6
x2 − (sum)x + product = 0
x2 − 1x − 6 = 0
Product = 3 × (−2) = −6
x2 − (sum)x + product = 0
x2 − 1x − 6 = 0
Answer: (a) x2 − x − 6 = 0
Section B — Short Answer (4 × 2 = 8 marks)
Solve x2 − 7x + 12 = 0 by factorisation.
Marking Scheme: Correct factors 1 mark · Roots 1 mark.
Total: 2 Marks
Total: 2 Marks
Split the middle term
x2 − 7x + 12
−7x = −3x − 4x and (−3)(−4) = 12
= x2 − 3x − 4x + 12
= x(x − 3) − 4(x − 3)
= (x − 3)(x − 4)
−7x = −3x − 4x and (−3)(−4) = 12
= x2 − 3x − 4x + 12
= x(x − 3) − 4(x − 3)
= (x − 3)(x − 4)
Roots
(x − 3)(x − 4) = 0
x = 3 or x = 4
x = 3 or x = 4
Answer: x = 3, 4
Find the discriminant of 3x2 − 2x + 13 = 0 and hence state the nature of its roots.
Marking Scheme: Discriminant 1 mark · Correct conclusion 1 mark.
Total: 2 Marks
Total: 2 Marks
Compute D
a = 3, b = −2, c = 1/3
D = b2 − 4ac
= 4 − 4 × 3 × 1/3
= 4 − 4
= 0
D = b2 − 4ac
= 4 − 4 × 3 × 1/3
= 4 − 4
= 0
Nature of roots
D = 0, so the roots are real and equal.
x = −b2a = 26 = 13
x = −b2a = 26 = 13
Answer: D = 0; roots are real and equal (each = 1/3)
Find the value of k for which x2 + kx + 4 = 0 has equal roots.
Marking Scheme: Setting D = 0 1 mark · Solving for k 1 mark.
Total: 2 Marks
Total: 2 Marks
Set D = 0
a = 1, b = k, c = 4
k2 − 4 × 1 × 4 = 0
k2 − 16 = 0
k2 − 4 × 1 × 4 = 0
k2 − 16 = 0
Solve
k2 = 16
k = 4 or k = −4
k = 4 or k = −4
Answer: k = ±4
Find the roots of 2x2 + x − 6 = 0.
Option 1 — 2x2 + x − 6 = 0
Marking Scheme: Correct factorisation or formula 1 mark · Both roots 1 mark.
Total: 2 Marks
Total: 2 Marks
Factorise
2x2 + x − 6
= 2x2 + 4x − 3x − 6
= 2x(x + 2) − 3(x + 2)
= (x + 2)(2x − 3)
= 2x2 + 4x − 3x − 6
= 2x(x + 2) − 3(x + 2)
= (x + 2)(2x − 3)
Roots
x + 2 = 0 → x = −2
2x − 3 = 0 → x = 3/2
2x − 3 = 0 → x = 3/2
Answer: x = −2 and x = 3/2
— OR —
Option 2
If the sum of the roots of kx2 + 2x + 3k = 0 is equal to their product, find k (k ≠ 0).
Marking Scheme: Sum and product written 1 mark · Equation solved for k 1 mark.
Total: 2 Marks
Total: 2 Marks
Sum and product
α + β = −ba = −2k
αβ = ca = 3kk = 3
αβ = ca = 3kk = 3
Apply the condition
−2k = 3
−2 = 3k
k = −23
−2 = 3k
k = −23
Answer: k = −2/3
[Attempt any ONE of the two options of Q16]
Section C — Long Answer (4 × 3 = 12 marks)
Solve 3x2 − 5x + 2 = 0 by the quadratic formula.
Marking Scheme: Formula written and values substituted 1 mark · Discriminant 1 mark · Both roots 1 mark.
Total: 3 Marks
Total: 3 Marks
Write the formula
a = 3, b = −5, c = 2
x = −b ± √(b2 − 4ac)2a
x = −b ± √(b2 − 4ac)2a
Discriminant
D = (−5)2 − 4 × 3 × 2
= 25 − 24
= 1
= 25 − 24
= 1
Substitute and simplify
x = 5 ± √16
x = 5 + 16 = 1 or x = 5 − 16 = 23
x = 5 + 16 = 1 or x = 5 − 16 = 23
Answer: x = 1 and x = 2/3
The product of two consecutive positive integers is 156. Find the integers.
Marking Scheme: Correct equation formed 1 mark · Solving the quadratic 1 mark · Rejecting the invalid value and answer 1 mark.
Total: 3 Marks
Total: 3 Marks
Let the integers be x and x + 1
x(x + 1) = 156
x2 + x − 156 = 0
x2 + x − 156 = 0
Solve by the formula
D = 12 − 4(1)(−156) = 1 + 624 = 625
√625 = 25
x = −1 ± 252
x = 12 or x = −13
√625 = 25
x = −1 ± 252
x = 12 or x = −13
Reject the invalid root
The integers are positive, so x = −13 is rejected.
x = 12, so the integers are 12 and 13.
Check: 12 × 13 = 156 ✓
x = 12, so the integers are 12 and 13.
Check: 12 × 13 = 156 ✓
Answer: 12 and 13
Find the roots of x2 + 4x − 5 = 0 by the method of completing the square.
Option 1 — Completing the square
Marking Scheme: Completing the square 1 mark · Taking square roots 1 mark · Both roots 1 mark.
Total: 3 Marks
Total: 3 Marks
Make a perfect square
x2 + 4x = 5
Add (4/2)2 = 4 to both sides:
x2 + 4x + 4 = 5 + 4
(x + 2)2 = 9
Add (4/2)2 = 4 to both sides:
x2 + 4x + 4 = 5 + 4
(x + 2)2 = 9
Take square roots
x + 2 = ±3
x = −2 + 3 = 1 or x = −2 − 3 = −5
x = −2 + 3 = 1 or x = −2 − 3 = −5
Answer: x = 1 and x = −5
— OR —
Option 2
A rectangular garden is 4 m longer than it is wide and its area is 96 m2. Find its length and breadth.
Marking Scheme: Equation formed 1 mark · Solved 1 mark · Rejecting the invalid width and stating both dimensions 1 mark.
Total: 3 Marks
Total: 3 Marks
Form the equation
Let the breadth be x m, so the length is (x + 4) m.
x(x + 4) = 96
x2 + 4x − 96 = 0
x(x + 4) = 96
x2 + 4x − 96 = 0
Solve
D = 16 + 384 = 400, √400 = 20
x = −4 + 202 = 8 (the negative root is rejected as a length)
Length = 8 + 4 = 12
x = −4 + 202 = 8 (the negative root is rejected as a length)
Length = 8 + 4 = 12
Answer: breadth 8 m, length 12 m
[Attempt any ONE of the two options of Q19]
Find the value of k for which 2x2 + kx + 8 = 0 has equal roots.
Option 1 — Equal roots
Marking Scheme: Condition D = 0 set up 1 mark · Simplifying 1 mark · Both values of k 1 mark.
Total: 3 Marks
Total: 3 Marks
Equal roots means D = 0
a = 2, b = k, c = 8
k2 − 4 × 2 × 8 = 0
k2 − 64 = 0
k2 − 4 × 2 × 8 = 0
k2 − 64 = 0
Solve
k2 = 64
k = 8 or k = −8
k = 8 or k = −8
Answer: k = ±8
— OR —
Option 2
If α and β are the roots of x2 − 5x + 6 = 0, find α2 + β2.
Marking Scheme: Sum and product 1 mark · Correct identity 1 mark · Answer 1 mark.
Total: 3 Marks
Total: 3 Marks
Sum and product
α + β = −ba = 5
αβ = ca = 6
αβ = ca = 6
Use the identity
α2 + β2 = (α + β)2 − 2αβ
= 52 − 2 × 6
= 25 − 12
= 13
= 52 − 2 × 6
= 25 − 12
= 13
Answer: α2 + β2 = 13
[Attempt any ONE of the two options of Q20]
Section D — Case Study (2 × 4 = 8 marks)
A train has to cover a distance of 360 km at a uniform speed. If its speed were 5 km/h more, the journey would take 1 hour less.
Marking Scheme: Part (a) 1.5 marks · Part (b) 1.5 marks · Part (c) 1 mark.
Total: 4 Marks
Total: 4 Marks
(a) Form the equation
Let the usual speed be x km/h.
Time at x = 360x hours
Time at (x + 5) = 360x + 5 hours
360x − 360x + 5 = 1
Time at x = 360x hours
Time at (x + 5) = 360x + 5 hours
360x − 360x + 5 = 1
(b) Simplify to a quadratic
360(x + 5) − 360x = x(x + 5)
1800 = x2 + 5x
x2 + 5x − 1800 = 0
1800 = x2 + 5x
x2 + 5x − 1800 = 0
(c) Solve
D = 25 + 7200 = 7225, √7225 = 85
x = −5 + 852 = 40 (the negative root is rejected as a speed)
x = −5 + 852 = 40 (the negative root is rejected as a speed)
Answer: (a) 360/x − 360/(x + 5) = 1 (b) x2 + 5x − 1800 = 0 (c) usual speed = 40 km/h
The area of a rectangular plot is 528 m2. Its length is one metre more than twice its breadth.
Option 1 — Rectangular plot
Marking Scheme: Part (a) 1.5 marks · Part (b) 1.5 marks · Part (c) 1 mark.
Total: 4 Marks
Total: 4 Marks
(a) Form the equation
Let the breadth be b m.
Length = (2b + 1) m
b(2b + 1) = 528
2b2 + b − 528 = 0
Length = (2b + 1) m
b(2b + 1) = 528
2b2 + b − 528 = 0
(b) Solve
D = 1 + 4 × 2 × 528 = 4225, √4225 = 65
b = −1 + 654 = 16 (negative root rejected)
Length = 2(16) + 1 = 33
b = −1 + 654 = 16 (negative root rejected)
Length = 2(16) + 1 = 33
(c) Check
16 × 33 = 528 m2 ✓
Answer: breadth 16 m, length 33 m
— OR —
Option 2
A motorboat whose speed in still water is 18 km/h takes 1 hour more to travel 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Marking Scheme: Part (a) 1.5 marks · Part (b) 1.5 marks · Part (c) 1 mark.
Total: 4 Marks
Total: 4 Marks
(a) Form the equation
Let the speed of the stream be x km/h.
Upstream speed = (18 − x), downstream = (18 + x)
2418 − x − 2418 + x = 1
Upstream speed = (18 − x), downstream = (18 + x)
2418 − x − 2418 + x = 1
(b) Simplify to a quadratic
24(18 + x) − 24(18 − x) = (18 − x)(18 + x)
48x = 324 − x2
x2 + 48x − 324 = 0
48x = 324 − x2
x2 + 48x − 324 = 0
(c) Solve
D = 2304 + 1296 = 3600, √3600 = 60
x = −48 + 602 = 6
(x must be less than 18, and 6 < 18 ✓)
x = −48 + 602 = 6
(x must be less than 18, and 6 < 18 ✓)
Answer: speed of the stream = 6 km/h
[Attempt any ONE of the two options of Q22]
Self-assessment for Sections B, C and D (28 marks)
The MCQ section is scored for you. For the written questions, tick each one you got right out of the marks shown, and add them up honestly.
Written total: 0 / 28
Written and checked by Pandey Sir · Pandey Sir’s BEST Classes, Dammaiguda, Hyderabad. Every answer on this paper is worked out and checked before publishing.
This is board-pattern practice for Quadratic Equations, built to the current CBSE Class 10 Maths syllabus 2026-27. It is not a prediction of any examination paper.
Curriculum scope checked against cbseacademic.nic.in. Spotted a mistake? Tell us here.