Department of Mathematics — Secondary Education
Quadratic Equations
Annual Examination — Class X
Name
Roll No.
Section
General Instructions
- This question paper consists of five sections. Read all instructions carefully.
- Section A has 20 MCQs of 1 mark each. No negative marking.
- Section B has 5 questions of 2 marks each. Attempt all.
- Section C has 6 questions of 3 marks each. Attempt all.
- Section D has 4 questions of 5 marks each. Attempt all.
- Section E has 3 case-study questions of 4 marks each.
- All workings must be shown clearly for Sections B–E.
- Use of calculator is not permitted.
Mark Distribution
Section A
20
20 × 1 mark
Section B
10
5 × 2 marks
Section C
18
6 × 3 marks
Section D
20
4 × 5 marks
Section E
12
3 × 4 marks
Total
80
All sections
Section A
Multiple Choice Questions — Choose the most appropriate option
20 × 1 = 20 Marks
1.
The standard form of a quadratic equation in variable x is:
(a) ax + b = 0
(b) ax² + bx + c = 0, a ≠ 0
(c) ax³ + bx² + c = 0
(d) ax² = 0
[1]
2.
The maximum number of roots a quadratic equation can have is:
(a) 1
(b) 2
(c) 3
(d) Infinitely many
[1]
3.
The discriminant of ax² + bx + c = 0 is defined as:
(a) b² + 4ac
(b) b² − 4ac
(c) −b² − 4ac
(d) 2b − 4ac
[1]
4.
If the discriminant D > 0, then the quadratic equation has:
(a) No real roots
(b) Two equal real roots
(c) Two distinct real roots
(d) One real root only
[1]
5.
If D = 0, the roots of ax² + bx + c = 0 are:
(a) Imaginary
(b) Two distinct reals
(c) Equal: x = −b/2a
(d) Zero
[1]
6.
The roots of x² − 5x + 6 = 0 are:
(a) 2 and 3
(b) −2 and −3
(c) 1 and 6
(d) −1 and 6
[1]
7.
Which of the following is NOT a quadratic equation?
(a) 2x² + 3x − 1 = 0
(b) x(x + 1) + 8 = (x + 2)(x − 2)
(c) (x + 2)³ = x³ − 4
(d) x² − 4x + 4 = 0
[1]
8.
The discriminant of 2x² − 4x + 3 = 0 is:
(a) 8
(b) −8
(c) 4
(d) −4
[1]
9.
The roots of 100x² − 20x + 1 = 0 are:
(a) 1/10 and 1/10
(b) 1/5 and 1/10
(c) 10 and −10
(d) −1/10 and −1/10
[1]
10.
For the equation kx² + 4x + 1 = 0 to have equal roots, the value of k is:
(a) 4
(b) 2
(c) 8
(d) 16
[1]
11.
The quadratic formula for roots of ax² + bx + c = 0 is:
(a) x = (b ± √D) / 2a
(b) x = (−b ± √D) / 2a
(c) x = (−b ± √D) / a
(d) x = (b ± √D) / a
[1]
12.
The roots of 6x² − x − 2 = 0 are:
(a) 2/3 and −1/2
(b) −2/3 and 1/2
(c) 1/3 and −2
(d) 2/3 and 1/2
[1]
13.
If the sum of roots of ax² + bx + c = 0 is −b/a, the product of roots is:
(a) b/a
(b) −b/a
(c) c/a
(d) −c/a
[1]
14.
A prayer hall has area 300 m². Its length is (2x + 1) m and breadth x m. The quadratic equation modelling this is:
(a) 2x² + x + 300 = 0
(b) 2x² + x − 300 = 0
(c) x² + 2x − 300 = 0
(d) 2x + 1 = 300
[1]
15.
The nature of roots of 2x² − 3x + 5 = 0 is:
(a) Two distinct real roots
(b) Two equal real roots
(c) No real roots
(d) One real root
[1]
16.
The zeroes of the polynomial ax² + bx + c and the roots of ax² + bx + c = 0 are:
(a) Always different
(b) The same
(c) Negatives of each other
(d) Reciprocals of each other
[1]
17.
For kx(x − 2) + 6 = 0 to have two equal roots (k ≠ 0), the value of k is:
(a) k = 3
(b) k = 6
(c) k = 4
(d) k = 12
[1]
18.
The equation x(x + 1) + 8 = (x + 2)(x − 2) is:
(a) A quadratic equation
(b) A cubic equation
(c) A linear equation
(d) Not an equation
[1]
19.
If one root of x² − 5x + k = 0 is 2, then k equals:
(a) 6
(b) −6
(c) 3
(d) −3
[1]
20.
The product of two consecutive positive integers is 306. The quadratic equation representing this is:
(a) x² + x + 306 = 0
(b) x² − x − 306 = 0
(c) x² + x − 306 = 0
(d) x² − x + 306 = 0
[1]
— Section B begins on this page — All questions are compulsory —
Section B
Short Answer Questions — Show all working clearly
5 × 2 = 10 Marks
21.
Check whether (x + 1)² = 2(x − 3) is a quadratic equation. Justify your answer by simplifying into standard form.
[2]
22.
Find the discriminant of 3x² − 4√3 x + 4 = 0 and state the nature of its roots.
[2]
23.
Find the roots of 2x² + x − 6 = 0 by factorisation.
[2]
24.
Represent the following situation as a quadratic equation: "The product of two consecutive positive integers is 306."
[2]
25.
For what value of k does 2x² + kx + 3 = 0 have two equal roots?
[2]
Section C
Short Answer (Extended) — Full working required
6 × 3 = 18 Marks
26.
Find the roots of √2 x² + 7x + 5√2 = 0 by factorisation.
[3]
27.
Find the roots of 2x² − 6x + 3 = 0 using the quadratic formula. Express answers in simplified surd form.
[3]
28.
Find two numbers whose sum is 27 and product is 182.
[3]
29.
Find two consecutive positive integers, the sum of whose squares is 365.
[3]
30.
Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m²? If so, find its length and breadth.
[3]
31.
A cottage industry produces pottery articles daily. The cost of production of each article (in ₹) is 3 more than twice the number of articles produced. If the total cost of production is ₹90, find the number of articles and the cost per article.
[3]
Section D
Long Answer Questions — All steps must be shown
4 × 5 = 20 Marks
32.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. Show all steps including formulation, factorisation, and verification.
[5]
33.
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more to cover the same distance. Find the speed of the train. Verify your answer.
[5]
34.
Rohan's mother is 26 years older than him. The product of their ages (in years), 3 years from now, will be 360. Find Rohan's present age. Also find how old his mother will be when Rohan is 18.
[5]
35.
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres, such that the difference of its distances from two diametrically opposite gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected? Draw a diagram and show full working.
[5]
Section E
Case Study Questions — Read each case carefully before answering
3 × 4 = 12 Marks
Case Study I — Architecture [Q.36 to Q.37]
A charity trust decides to build a prayer hall having a carpet area of 300 square metres, with its length one metre more than twice its breadth. The architect sets up the problem algebraically to find the dimensions of the hall. Let the breadth of the hall be x metres.
36.
Write the quadratic equation in x representing the situation. Using factorisation, find the breadth and length of the hall. [2 marks]
[2]
37.
If the area were changed to 528 m² with the same length–breadth relationship (length = 2×breadth + 1), what would the new dimensions be? Also, is it possible to have a hall with area 400 m² if perimeter is 80 m? Justify. [2 marks]
[2]
Case Study II — Number Theory [Q.38 to Q.39]
John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. This situation can be modelled using a quadratic equation.
38.
Letting x be the number of marbles John originally had, formulate and solve the quadratic equation to find how many marbles each of them had to start with. [2 marks]
[2]
39.
In a similar problem, two positive numbers have a sum of 20 and a product of 96. Using the discriminant, determine whether such numbers exist and find them. [2 marks]
[2]
Case Study III — History of Mathematics [Q.40]
Sridharacharya (C.E. 1025) derived the quadratic formula by the method of completing the square. The formula x = (−b ± √(b²−4ac)) / 2a is now universally used. Consider the equation 3x² − 2x + 1/3 = 0 (as quoted by Bhaskara II in historical texts).
40.
Using the quadratic formula, find the discriminant, state the nature of roots, and find the roots of 3x² − 2x + ⅓ = 0. Also verify the roots using the sum and product relationships (α + β = −b/a and αβ = c/a). [4 marks]
[4]
Official Answer Key & Marking Scheme
Quadratic Equations — Class X Examination — 80 Marks
STRICTLY CONFIDENTIAL
For Examiner Use Only
For Examiner Use Only
Section A — MCQ Answers (1 mark each)
20 × 1 = 20 marks
Section B — Short Answers (2 marks each)
5 × 2 = 10 marks
21.
(x + 1)² = 2(x − 3) — is it quadratic?
Solution [2 marks]
1.LHS: (x+1)² = x²+2x+1
2.RHS: 2(x−3) = 2x−6
3.x²+2x+1 = 2x−6 → x²+7 = 0
YES — it is a quadratic equation (degree 2, a=1≠0)
★ Award 1 mark for correct expansion, 1 mark for standard form and conclusion.
22.
Discriminant of 3x² − 4√3 x + 4 = 0
Solution [2 marks]
1.a=3, b=−4√3, c=4
2.D = (−4√3)² − 4(3)(4) = 48 − 48 = 0
D = 0 → Two equal real roots: x = −b/2a = 4√3/6 = 2/√3
23.
Roots of 2x² + x − 6 = 0 by factorisation
Solution [2 marks]
1.ac = −12. Find p,q: p×q=−12, p+q=1 → p=4, q=−3
2.2x²+4x−3x−6 = 2x(x+2)−3(x+2) = (2x−3)(x+2) = 0
x = 3/2 or x = −2
24.
Product of two consecutive positive integers = 306
Solution [2 marks]
1.Let integers be x and x+1.
2.x(x+1) = 306 → x²+x = 306
x² + x − 306 = 0
25.
Value of k for 2x² + kx + 3 = 0 to have equal roots
Solution [2 marks]
1.For equal roots: D = 0 → k²−4(2)(3) = 0 → k² = 24
k = ±2√6 (both values are valid)
★ Do NOT penalise students who write only the positive root if they show correct method.
Section C — Short Extended Answers (3 marks each)
6 × 3 = 18 marks
26.
Roots of √2 x² + 7x + 5√2 = 0
Solution [3 marks]
1.ac = √2 × 5√2 = 10. Find p,q: p×q=10, p+q=7 → p=5, q=2
2.√2 x² + 5x + 2x + 5√2 = x(√2 x+5) + √2(√2 x+5)
3.= (x+√2)(√2 x+5) = 0
x = −√2 or x = −5/√2 = −5√2/2
27.
Roots of 2x² − 6x + 3 = 0 by quadratic formula
Solution [3 marks]
1.a=2, b=−6, c=3. D = 36−24 = 12 > 0. √12 = 2√3
2.x = (6 ± 2√3)/4 = (3 ± √3)/2
x = (3+√3)/2 or x = (3−√3)/2
28.
Two numbers with sum 27 and product 182
Solution [3 marks]
1.Let numbers be x and 27−x. Then x(27−x) = 182
2.x²−27x+182 = 0 → (x−13)(x−14) = 0
The numbers are 13 and 14
29.
Two consecutive positive integers, sum of squares = 365
Solution [3 marks]
1.Let integers be x and x+1. x²+(x+1)² = 365
2.2x²+2x+1 = 365 → x²+x−182 = 0 → (x−13)(x+14) = 0
3.x=13 (discard x=−14, integers must be positive)
The integers are 13 and 14
30.
Mango grove: length = 2×breadth, area = 800 m²
Solution [3 marks]
1.Let breadth = x. Length = 2x. Area: 2x² = 800 → x² = 400
2.D = 1600 > 0. x = ±20. Reject x=−20 (breadth > 0).
YES — possible. Breadth = 20 m, Length = 40 m
31.
Pottery: cost per article = 2x+3, total cost = ₹90
Solution [3 marks]
1.x(2x+3) = 90 → 2x²+3x−90 = 0
2.D = 9+720 = 729 = 27². x = (−3+27)/4 = 6 or x = −7.5 (rejected)
6 articles; cost per article = 2(6)+3 = ₹15. Check: 6×15 = ₹90 ✓
Section D — Long Answer (5 marks each)
4 × 5 = 20 marks
32.
Right triangle: altitude = base − 7, hypotenuse = 13 cm
Solution [5 marks — scheme shown]
1.[1m] Let base = x cm. Altitude = x−7 cm.
2.[1m] Pythagoras: x² + (x−7)² = 13² = 169
3.[1m] Expand: 2x²−14x+49 = 169 → x²−7x−60 = 0
4.[1m] Factorise: (x−12)(x+5) = 0 → x=12 or x=−5
5.[1m] Reject x=−5 (negative length). Base=12 cm, Altitude=5 cm.
Base = 12 cm, Altitude = 5 cm. Verify: 12²+5² = 144+25 = 169 = 13² ✓
33.
Train: 480 km, speed 8 km/h less → 3 hours more
Solution [5 marks]
1.[1m] Let speed = x km/h. Time = 480/x hrs.
2.[1m] Reduced speed: x−8. Equation: 480/(x−8) − 480/x = 3
3.[1m] Cross-multiply: 480x − 480(x−8) = 3x(x−8) → 3840 = 3x²−24x
4.[1m] Divide by 3: x²−8x−1280 = 0. D = 64+5120 = 5184 = 72²
5.[1m] x = (8+72)/2 = 40. Reject x=−32 (speed > 0).
Speed = 40 km/h. Verify: 480/40=12 hrs, 480/32=15 hrs. 15−12=3 ✓
34.
Rohan's age (mother 26 years older; product of ages in 3 years = 360)
Solution [5 marks]
1.[1m] Let Rohan's age = x. Mother = x+26.
2.[1m] In 3 years: (x+3)(x+29) = 360
3.[1m] Expand: x²+32x+87=360 → x²+32x−273=0
4.[1m] (x+39)(x−7)=0 → x=7 or x=−39. Reject x=−39.
5.[1m] When Rohan is 18: years passed = 11. Mother = 33+11 = 44 years old.
Rohan's present age = 7 years; Mother's age when Rohan is 18 = 44 years
35.
Pole on circular park boundary (diameter 13 m, |AP−BP| = 7 m)
Solution [5 marks]
1.[1m] Let BP = x. Since AP−BP=7, AP = x+7. Diagram with angle APB=90° (angle in semicircle).
2.[1m] Pythagoras: x²+(x+7)² = 13² = 169
3.[1m] Expand: 2x²+14x+49=169 → x²+7x−60=0
4.[1m] D=49+240=289>0 → Placement IS possible. √289=17. x=(−7+17)/2=5.
5.[1m] Reject x=−12. BP=5 m, AP=5+7=12 m.
Possible. Pole is 5 m from gate B and 12 m from gate A. Verify: 5²+12²=25+144=169=13² ✓
Section E — Case Study Answers (4 marks each)
3 × 4 = 12 marks
36.
Prayer hall — area 300 m², length = 2x+1
Solution [2 marks]
1.[1m] x(2x+1)=300 → 2x²+x−300=0. Factor: (x−12)(2x+25)=0
2.[1m] x=12 (reject x=−12.5). Breadth=12 m, Length=25 m.
Breadth = 12 m, Length = 25 m
37.
Modified hall (area 528 m²) + perimeter 80 m, area 400 m²
Solution [2 marks]
1.[1m] 2x²+x−528=0. D=1+4224=4225=65². x=(−1+65)/4=16. L=33 m, B=16 m.
2.[1m] Park: l+b=40, lb=400 → b²−40b+400=0. D=1600−1600=0. b=20=l. YES — square park, 20m×20m.
Area 528 m²: L=33 m, B=16 m. | Area 400 m², perim 80 m: Possible — square 20 m×20 m.
38.
Marbles: John + Jivanti = 45, lose 5 each, product = 124
Solution [2 marks]
1.[1m] (x−5)(40−x)=124 → x²−45x+324=0 → (x−9)(x−36)=0
2.[1m] Both roots valid. John=36, Jivanti=9 OR John=9, Jivanti=36.
Two valid scenarios: (John=36, Jivanti=9) or (John=9, Jivanti=36)
39.
Sum=20, product=96 — do such numbers exist?
Solution [2 marks]
1.[1m] x(20−x)=96 → x²−20x+96=0. D=400−384=16>0 → exist!
2.[1m] x=(20±4)/2 → x=12 or x=8.
Yes. The two numbers are 8 and 12. (8+12=20 ✓, 8×12=96 ✓)
40.
3x² − 2x + 1/3 = 0 — discriminant, roots, verification
Solution [4 marks]
1.[1m] a=3, b=−2, c=1/3. D=(−2)²−4(3)(1/3)=4−4=0
2.[1m] D=0 → Two equal real roots.
3.[1m] x=−b/2a = 2/6 = 1/3. Both roots are 1/3.
4.[1m] Verify: α+β=1/3+1/3=2/3=−b/a=2/3 ✓. αβ=1/9=c/a=1/9 ✓
D = 0; Two equal roots: x = 1/3, 1/3. Sum & product conditions verified ✓
Grade Boundaries
A+
72–80
A
64–71
B+
56–63
B
48–55
C
33–47
D
0–32
