Quadratic Equations PYQ Bank | Class X Maths

Questions

2-Mark Qs

3-Mark Qs

Years Covered

📋 Key Topics Covered (As per CBSE 2025-26 Syllabus)

Factorisation Method — splitting middle term

Quadratic Formula — x = (–b ± √D) / 2a

Nature of Roots — discriminant D = b² – 4ac

Find k for given conditions — equal / real / no roots

Word Problems — speed, age, area, number, fraction

Sum & Product of Roots — α+β = –b/a, αβ = c/a

⚠️

DELETED for 2025-26: Completing the Square Method is removed from the syllabus. Focus only on Factorisation and Quadratic Formula methods. Also, situational problems based on daily activities (real-life word problems of 5 marks) may still appear.

📝 SECTION A — 2-MARK QUESTIONS

15 Questions • Very Short Answer (VSA) Type • Board PYQs 2018–2025

CBSE 2025

Which of the following quadratic equations has real and equal roots? (a) (x+1)² = 2x+1 (b) x² + x = 0 (c) x² – 4 = 0 (d) x² + x + 1 = 0. Justify your answer.

2 Marks Discriminant ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Step 1 (a) (x+1)² = 2x+1 ⟹ x² + 2x + 1 = 2x + 1 ⟹ x² = 0

Here a = 1, b = 0, c = 0 ⟹ D = 0 – 4(1)(0) = 0 ✓ Equal roots

Step 2 (b) x² + x = 0 ⟹ D = 1 – 0 = 1 > 0 (distinct roots)

Step 3 (c) x² – 4 = 0 ⟹ D = 0 + 16 = 16 > 0 (distinct roots)

Step 4 (d) x² + x + 1 = 0 ⟹ D = 1 – 4 = –3 < 0 (no real roots)

∴ Answer: (a) — Only (x+1)² = 2x+1 has real and equal roots (D = 0)

CBSE 2025

Find the smallest value of p for which the quadratic equation x² – 2(p+1)x + p² = 0 has real roots.

2 Marks Nature of Roots ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Step 1 a = 1, b = –2(p+1), c = p²

Step 2 D = b² – 4ac = [–2(p+1)]² – 4(1)(p²)

= 4(p+1)² – 4p² = 4p² + 8p + 4 – 4p² = 8p + 4

Step 3 For real roots: D ≥ 0 ⟹ 8p + 4 ≥ 0 ⟹ p ≥ –1/2

∴ Smallest value of p = –1/2

CBSE 2024

If the roots of quadratic equation 4x² – 5x + k = 0 are real and equal, then find the value of k.

2 Marks Equal Roots → k ⭐ MOST REPEATED TYPE

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✅ DETAILED SOLUTION

For equal roots: D = b² – 4ac = 0

Step 1 a = 4, b = –5, c = k

Step 2 D = (–5)² – 4(4)(k) = 25 – 16k = 0

Step 3 16k = 25 ⟹ k = 25/16

∴ k = 25/16

CBSE 2024

If the roots of equation ax² + bx + c = 0 (a ≠ 0) are real and equal, which relation is true? (a) a = b²/c (b) b² = ac (c) ac = b²/4 (d) c = b²/a

2 Marks Discriminant Relation

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✅ DETAILED SOLUTION

Step 1 For real and equal roots: D = 0 ⟹ b² – 4ac = 0

Step 2 b² = 4ac ⟹ ac = b²/4

∴ Answer: (c) ac = b²/4

CBSE 2023

Find the sum and product of the roots of the quadratic equation 2x² – 9x + 4 = 0.

2 Marks Sum & Product of Roots

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✅ DETAILED SOLUTION

Sum of roots (α+β) = –b/a ; Product of roots (αβ) = c/a

Step 1 a = 2, b = –9, c = 4

Step 2 Sum = –(–9)/2 = 9/2

Step 3 Product = 4/2 = 2

∴ Sum of roots = 9/2 ; Product of roots = 2

CBSE 2023

Find the roots of the equation x² + 3x – 10 = 0.

2 Marks Factorisation

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✅ DETAILED SOLUTION

Step 1 x² + 3x – 10 = 0. Find two numbers: product = –10, sum = 3 → (+5) and (–2)

Step 2 x² + 5x – 2x – 10 = 0 ⟹ x(x+5) – 2(x+5) = 0

Step 3 (x–2)(x+5) = 0 ⟹ x = 2 or x = –5

∴ Roots are 2 and –5

CBSE 2022

If the sum of the roots of quadratic equation ky² – 11y + (k–23) = 0 is 13/21 more than the product of roots, find the value of k.

2 Marks Sum–Product Relation ⭐ HOTS

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✅ DETAILED SOLUTION

Step 1 Sum of roots = 11/k ; Product of roots = (k–23)/k

Step 2 Given: Sum = Product + 13/21

11/k = (k–23)/k + 13/21

Step 3 11/k – (k–23)/k = 13/21 ⟹ (11 – k + 23)/k = 13/21 ⟹ (34–k)/k = 13/21

Step 4 21(34–k) = 13k ⟹ 714 – 21k = 13k ⟹ 34k = 714 ⟹ k = 21

∴ k = 21

CBSE 2021

Write the quadratic equation in x whose roots are 2 and –5.

2 Marks Form Equation from Roots

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✅ DETAILED SOLUTION

Equation: x² – (sum of roots)x + (product of roots) = 0

Step 1 Sum = 2 + (–5) = –3 ; Product = 2 × (–5) = –10

Step 2 Equation: x² – (–3)x + (–10) = 0 ⟹ x² + 3x – 10 = 0

∴ x² + 3x – 10 = 0

CBSE 2022

Find the discriminant of the equation 3x² – 2x + 1/3 = 0 and hence find the nature of its roots.

2 Marks Discriminant

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✅ DETAILED SOLUTION

Step 1 a = 3, b = –2, c = 1/3

Step 2 D = b² – 4ac = (–2)² – 4(3)(1/3) = 4 – 4 = 0

∴ D = 0 → The equation has two real and equal roots.

Q10

CBSE 2020

Find the roots of 2x² – 5x + 3 = 0 by factorisation method.

2 Marks Factorisation ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 2x² – 5x + 3 = 0. Product = 2×3 = 6, Sum = –5 → (–2) and (–3)

Step 2 2x² – 2x – 3x + 3 = 0 ⟹ 2x(x–1) – 3(x–1) = 0

Step 3 (2x–3)(x–1) = 0 ⟹ x = 3/2 or x = 1

∴ Roots are 3/2 and 1

Q11

CBSE 2021

Find the nature of roots of the quadratic equation 2x² – 4x + 3 = 0.

2 Marks Nature of Roots

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✅ DETAILED SOLUTION

Step 1 a = 2, b = –4, c = 3

Step 2 D = b² – 4ac = (–4)² – 4(2)(3) = 16 – 24 = –8

Step 3 Since D < 0

∴ The equation has no real roots.

Q12

CBSE 2019

Find the value of k for which the equation 2x² + kx + 3 = 0 has two equal roots.

2 Marks Find k ⭐ REPEATED TYPE

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✅ DETAILED SOLUTION

Step 1 a = 2, b = k, c = 3. For equal roots: D = 0

Step 2 k² – 4(2)(3) = 0 ⟹ k² = 24 ⟹ k = ±2√6

∴ k = ±2√6 (i.e., k = 2√6 or k = –2√6)

Q13

CBSE 2023

Find the value of p for which one root of px² – 14x + 8 = 0 is 6 times the other.

2 Marks Root Relation → k ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 Let roots be α and 6α

Step 2 Sum: α + 6α = 14/p ⟹ 7α = 14/p ⟹ α = 2/p

Step 3 Product: α × 6α = 8/p ⟹ 6α² = 8/p

Step 4 6 × (2/p)² = 8/p ⟹ 6 × 4/p² = 8/p ⟹ 24/p² = 8/p ⟹ p = 3

∴ p = 3

Q14

CBSE 2018

Solve: x² – 5x + 6 = 0 by factorisation.

2 Marks Factorisation

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✅ DETAILED SOLUTION

Step 1 Product = 6, Sum = –5 → (–2) and (–3)

Step 2 x² – 2x – 3x + 6 = 0 ⟹ x(x–2) – 3(x–2) = 0

Step 3 (x–2)(x–3) = 0 ⟹ x = 2 or x = 3

∴ Roots are 2 and 3

Q15

CBSE 2020

Write all the values of p for which x² + px + 16 = 0 has equal roots. Find the roots.

2 Marks Equal Roots + Solve

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✅ DETAILED SOLUTION

Step 1 For equal roots: D = 0 ⟹ p² – 4(1)(16) = 0 ⟹ p² = 64 ⟹ p = ±8

Step 2 When p = 8: x² + 8x + 16 = 0 ⟹ (x+4)² = 0 ⟹ x = –4, –4

Step 3 When p = –8: x² – 8x + 16 = 0 ⟹ (x–4)² = 0 ⟹ x = 4, 4

∴ p = 8 (roots: –4, –4) or p = –8 (roots: 4, 4)

📝 SECTION B — 3-MARK QUESTIONS

15 Questions • Short Answer (SA) Type • Board PYQs 2018–2025

Q16

CBSE 2024

The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2 16/21, find the fraction.

3 Marks Fraction Word Problem

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✅ DETAILED SOLUTION

Step 1 Let numerator = x, so denominator = 2x + 1. Fraction = x/(2x+1)

Step 2 x/(2x+1) + (2x+1)/x = 58/21

Step 3 [x² + (2x+1)²] / [x(2x+1)] = 58/21

[x² + 4x² + 4x + 1] / [2x² + x] = 58/21

[5x² + 4x + 1] / [2x² + x] = 58/21

Step 4 21(5x² + 4x + 1) = 58(2x² + x)

105x² + 84x + 21 = 116x² + 58x

11x² – 26x – 21 = 0

Step 5 11x² – 33x + 7x – 21 = 0 ⟹ 11x(x–3) + 7(x–3) = 0

(11x+7)(x–3) = 0 ⟹ x = 3 (reject x = –7/11)

∴ Fraction = 3/(2×3+1) = 3/7

Q17

CBSE 2024

A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles. Find the side length of original tile by factorisation.

3 Marks Area Word Problem ⭐ COMPETENCY BASED

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✅ DETAILED SOLUTION

Step 1 Let side of original tile = x units. Area of floor = 200x²

Step 2 New tile side = (x+1). Area of floor = 128(x+1)²

Step 3 Since floor area is same: 200x² = 128(x+1)²

200x² = 128x² + 256x + 128

72x² – 256x – 128 = 0 ⟹ 9x² – 32x – 16 = 0 (÷ 8)

Step 4 9x² – 36x + 4x – 16 = 0 ⟹ 9x(x–4) + 4(x–4) = 0

(9x+4)(x–4) = 0 ⟹ x = 4 (reject x = –4/9)

∴ Side of original tile = 4 units

Q18

CBSE 2024

In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100 km/h and by doing so the time of flight increased by 30 minutes. Find the original duration of the flight.

3 Marks Speed-Time Problem ⭐ MOST IMPORTANT TYPE

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✅ DETAILED SOLUTION

Step 1 Let original speed = x km/h. Time = 2800/x hours

Step 2 New speed = (x – 100). New time = 2800/(x–100) hours

Step 3 2800/(x–100) – 2800/x = 1/2 (30 min = 1/2 hr)

2800[x – (x–100)] / [x(x–100)] = 1/2

2800 × 100 / [x(x–100)] = 1/2

x² – 100x = 560000

Step 4 x² – 100x – 560000 = 0

x² – 800x + 700x – 560000 = 0

(x–800)(x+700) = 0 ⟹ x = 800 (reject –700)

Step 5 Original time = 2800/800 = 3.5 hours = 3 hours 30 minutes

∴ Original duration = 3 hours 30 minutes

Q19

CBSE 2025

A circular park has diameter 65 m. An entry gate is at point P on the boundary such that PA = PB + 35 where AB is the diameter. Find PA and PB.

3 Marks Geometry + Pythagoras ⭐ MOST RECENT

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✅ DETAILED SOLUTION

Step 1 AB = 65 m (diameter). Angle in semicircle ⟹ ∠APB = 90°

Step 2 Let PB = x, then PA = x + 35

Step 3 By Pythagoras: PA² + PB² = AB²

(x+35)² + x² = 65²

x² + 70x + 1225 + x² = 4225

2x² + 70x – 3000 = 0 ⟹ x² + 35x – 1500 = 0

Step 4 x² + 60x – 25x – 1500 = 0 ⟹ (x+60)(x–25) = 0

x = 25 (reject –60)

∴ PB = 25 m, PA = 25 + 35 = 60 m

Q20

CBSE 2023

Solve by factorisation: 2x² + x – 6 = 0

3 Marks Factorisation

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✅ DETAILED SOLUTION

Step 1 2x² + x – 6 = 0. Product = 2 × (–6) = –12, Sum = 1 → (+4) and (–3)

Step 2 2x² + 4x – 3x – 6 = 0 ⟹ 2x(x+2) – 3(x+2) = 0

Step 3 (2x–3)(x+2) = 0 ⟹ x = 3/2 or x = –2

∴ Roots are 3/2 and –2

Q21

CBSE 2022

If (1+a²)x² + 2abx + (b²–c²) = 0 has equal roots, find the value of c²(1+a²).

3 Marks Discriminant Relation ⭐ HOTS

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✅ DETAILED SOLUTION

Step 1 A = (1+a²), B = 2ab, C = (b²–c²)

Step 2 D = 0 ⟹ B² – 4AC = 0

(2ab)² – 4(1+a²)(b²–c²) = 0

4a²b² – 4(b²–c²+a²b²–a²c²) = 0

4a²b² – 4b² + 4c² – 4a²b² + 4a²c² = 0

Step 3 –4b² + 4c² + 4a²c² = 0 ⟹ c² + a²c² = b²

∴ c²(1 + a²) = b²

Q22

CBSE 2021

Find the values of k for which the equation kx(x–2) + 6 = 0 has two equal roots.

3 Marks Equal Roots → k

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✅ DETAILED SOLUTION

Step 1 Expand: kx² – 2kx + 6 = 0

Step 2 a = k, b = –2k, c = 6. For equal roots: D = 0

Step 3 (–2k)² – 4(k)(6) = 0 ⟹ 4k² – 24k = 0 ⟹ 4k(k–6) = 0

Step 4 k = 0 or k = 6. But k ≠ 0 (otherwise not quadratic)

∴ k = 6

Q23

CBSE 2020

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.

3 Marks Speed-Time Problem ⭐ CLASSIC

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✅ DETAILED SOLUTION

Step 1 Let speed = x km/h. Time = 480/x hours

Step 2 New speed = (x–8). New time = 480/(x–8)

Step 3 480/(x–8) – 480/x = 3

480[x – (x–8)] / [x(x–8)] = 3

480 × 8 = 3x(x–8) ⟹ 3840 = 3x² – 24x

Step 4 x² – 8x – 1280 = 0 ⟹ x² – 40x + 32x – 1280 = 0

(x–40)(x+32) = 0 ⟹ x = 40 (reject –32)

∴ Speed of the train = 40 km/h

Q24

CBSE 2022

Find the value of m for which (m–1)x² + 2(m–1)x + 1 = 0 has two equal roots.

3 Marks Equal Roots → m

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✅ DETAILED SOLUTION

Step 1 a = (m–1), b = 2(m–1), c = 1. For equal roots: D = 0

Step 2 [2(m–1)]² – 4(m–1)(1) = 0

4(m–1)² – 4(m–1) = 0 ⟹ 4(m–1)[(m–1)–1] = 0

4(m–1)(m–2) = 0 ⟹ m = 1 or m = 2

Step 3 If m = 1, a = 0 → not quadratic. So m ≠ 1

∴ m = 2

Q25

CBSE 2019

Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.

3 Marks Area Problem + Existence ⭐ NCERT BASED

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✅ DETAILED SOLUTION

Step 1 Let length = x m. Then 2(x + breadth) = 80 ⟹ breadth = (40 – x)

Step 2 Area = x(40 – x) = 400 ⟹ 40x – x² = 400 ⟹ x² – 40x + 400 = 0

Step 3 D = (–40)² – 4(1)(400) = 1600 – 1600 = 0

D = 0 means the equation has real and equal roots. So yes, it IS possible!

Step 4 x = 40/2 = 20 m. Breadth = 40 – 20 = 20 m

∴ Yes, possible. Length = 20 m, Breadth = 20 m (it's a square)

Q26

CBSE 2019

A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

3 Marks Speed-Time Problem ⭐ REPEATED TYPE

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✅ DETAILED SOLUTION

Step 1 Let usual speed = x km/h. Time = 1500/x hours

Step 2 Increased speed = (x+100). New time = 1500/(x+100)

Step 3 1500/x – 1500/(x+100) = 1/2

1500 × 100 / [x(x+100)] = 1/2

x² + 100x = 300000

Step 4 x² + 100x – 300000 = 0 ⟹ x² + 600x – 500x – 300000 = 0

(x+600)(x–500) = 0 ⟹ x = 500 (reject –600)

∴ Usual speed = 500 km/h

Q27

CBSE 2021

Solve using quadratic formula: 2x² – 3x – 2 = 0

3 Marks Quadratic Formula

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✅ DETAILED SOLUTION

x = [–b ± √(b² – 4ac)] / 2a

Step 1 a = 2, b = –3, c = –2

Step 2 D = (–3)² – 4(2)(–2) = 9 + 16 = 25

Step 3 x = [3 ± √25] / 4 = (3 ± 5) / 4

Step 4 x = (3+5)/4 = 2 or x = (3–5)/4 = –1/2

∴ Roots are 2 and –1/2

Q28

CBSE 2023

The sum of ages of two friends is 20 years. Four years ago, the product of their ages was 48. Is this situation possible? If so, find their present ages.

3 Marks Age Problem + Existence ⭐ IMPORTANT

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✅ DETAILED SOLUTION

Step 1 Let one friend's age = x. Other = 20 – x

Step 2 4 years ago: (x–4)(16–x) = 48

16x – x² – 64 + 4x = 48

–x² + 20x – 112 = 0 ⟹ x² – 20x + 112 = 0

Step 3 D = (–20)² – 4(1)(112) = 400 – 448 = –48

D < 0 means no real roots exist. The given situation is NOT POSSIBLE.

∴ This situation is not possible (D = –48 < 0).

Q29

CBSE 2018

In a class test, the sum of Shefali's marks in Maths and English is 30. Had she got 2 marks more in Maths and 3 marks less in English, the product of their marks would have been 210. Find her marks in each subject.

3 Marks Number Word Problem

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✅ DETAILED SOLUTION

Step 1 Let Maths marks = x. English marks = 30 – x

Step 2 (x+2)(30–x–3) = 210 ⟹ (x+2)(27–x) = 210

27x – x² + 54 – 2x = 210

–x² + 25x – 156 = 0 ⟹ x² – 25x + 156 = 0

Step 3 x² – 12x – 13x + 156 = 0 ⟹ (x–12)(x–13) = 0

Case 1 x = 12: Maths = 12, English = 18

Case 2 x = 13: Maths = 13, English = 17

∴ Marks: (12, 18) or (13, 17)

Q30

CBSE 2018

A train travels a distance of 63 km and then 72 km at a speed 6 km/h more than original. Total time = 3 hours. Find original average speed.

3 Marks Speed-Time Problem ⭐ CHALLENGING

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✅ DETAILED SOLUTION

Step 1 Let original speed = x km/h

Step 2 Time for 63 km = 63/x ; Time for 72 km at (x+6) = 72/(x+6)

Step 3 63/x + 72/(x+6) = 3

[63(x+6) + 72x] / [x(x+6)] = 3

63x + 378 + 72x = 3x² + 18x

135x + 378 = 3x² + 18x

Step 4 3x² – 117x – 378 = 0 ⟹ x² – 39x – 126 = 0 (÷ 3)

x² – 42x + 3x – 126 = 0 ⟹ (x–42)(x+3) = 0

x = 42 (reject –3)

∴ Original average speed = 42 km/h

📐 KEY FORMULAS TO REMEMBER

Standard Form: ax² + bx + c = 0 (a ≠ 0)

Quadratic Formula: x = [–b ± √(b²–4ac)] / 2a

Discriminant: D = b² – 4ac
• D > 0 → Two distinct real roots
• D = 0 → Two equal real roots (repeated root = –b/2a)
• D < 0 → No real roots

Sum of roots: α + β = –b/a
Product of roots: αβ = c/a
Equation from roots: x² – (α+β)x + αβ = 0

Factorisation method: Split middle term such that
product of two parts = a × c and sum = b

💡 BOARD EXAM TIPS — QUADRATIC EQUATIONS

Always check D first: If asked "is the situation possible?", just compute D. If D < 0, say "not possible" — full marks!

Reject negative roots when the problem involves distance, speed, age, length — always state the reason.

For "find k" problems: Use D = 0 (equal roots) or D ≥ 0 (real roots) and solve for k. Don't forget k ≠ 0 check!

Speed-Time-Distance: Most common 3-mark word problem. Formula: Time = Distance/Speed. Set up equation carefully.

Expected marks: 6–8 marks from this chapter (MCQ + SA + Case Study).

Completing the Square is DELETED. Don't waste time on it. Use factorisation or quadratic formula only.