What is the quadratic formula?

The quadratic formula is x = [-b ± √(b²-4ac)] / 2a. It finds the roots of any quadratic equation ax² + bx + c = 0 where a ≠ 0.

What is the discriminant in a quadratic equation?

Discriminant D = b² − 4ac. If D > 0: two real distinct roots. If D = 0: two equal real roots. If D < 0: no real roots.

How do I form a quadratic equation from given roots?

If α and β are roots, the quadratic equation is x² − (α+β)x + αβ = 0. Sum of roots = −b/a and product of roots = c/a.

Class 10 Maths Chapter 2 Quadratic Equations Solutions | Maharashtra Board SSC

Q: How do I form a quadratic equation from given roots?

If α and β are roots, the quadratic equation is x² − (α+β)x + αβ = 0. Sum of roots = −b/a and product of roots = c/a.

This page provides complete solutions for Maharashtra Board Class 10 Maths Chapter 2 — Quadratic Equations. This chapter from the Balbharati Algebra textbook covers four key topics: solving quadratic equations by factorisation, completing the square, and the quadratic formula, plus determining the nature of roots using the discriminant.

📋 Table of Contents

Introduction to Quadratic Equations

A quadratic equation is a polynomial equation of degree 2. Its standard form is ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. Every quadratic equation has exactly two roots (solutions), which may be real or imaginary.

🔢 Quadratic Equation — Key Formulas

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 (x = −b/2a)
D < 0 → No real roots (complex roots)
Sum of roots (α + β) = −b/a
Product of roots (α × β) = c/a
Equation from roots: x² − (α+β)x + αβ = 0

Factorisation Method

In the factorisation method, we split the middle term bx into two terms whose product equals ac and whose sum equals b. Then we group and factor the expression into two linear factors.

💡 Tip: To factorise ax² + bx + c = 0: Find two numbers p and q such that p + q = b and p × q = ac. Then rewrite bx = px + qx and factor by grouping.

Practice Set 2.1 — Factorisation Solutions

Q: Solve by factorisation: x² + 5x + 6 = 0

Step 1: Find two numbers: product = 6, sum = 5. Numbers are 2 and 3.

Step 2: Split middle term: x² + 2x + 3x + 6 = 0

Step 3: Group: x(x + 2) + 3(x + 2) = 0

Step 4: Factor: (x + 2)(x + 3) = 0

Step 5: x + 2 = 0 → x = −2 OR x + 3 = 0 → x = −3

✅ Answer: x = −2 or x = −3

Q: Solve: x² − 5x + 6 = 0

Step 1: Product = 6, Sum = −5. Numbers: −2 and −3.

Step 2: x² − 2x − 3x + 6 = 0

Step 3: x(x − 2) − 3(x − 2) = 0

Step 4: (x − 2)(x − 3) = 0

✅ Answer: x = 2 or x = 3

Q: Solve: 2x² + x − 6 = 0

Step 1: Here a = 2, b = 1, c = −6. So ac = 2 × (−6) = −12.

Step 2: Find two numbers: product = −12, sum = 1. Numbers: 4 and −3.

Step 3: 2x² + 4x − 3x − 6 = 0

Step 4: 2x(x + 2) − 3(x + 2) = 0

Step 5: (x + 2)(2x − 3) = 0

✅ Answer: x = −2 or x = 3/2

Q: Solve: 3x² − 5x − 2 = 0

Step 1: ac = 3 × (−2) = −6. Find numbers with product −6 and sum −5: −6 and 1.

Step 2: 3x² − 6x + x − 2 = 0

Step 3: 3x(x − 2) + 1(x − 2) = 0

Step 4: (x − 2)(3x + 1) = 0

✅ Answer: x = 2 or x = −1/3

Q: Solve: x² − 3x − 10 = 0

Step 1: Product = −10, Sum = −3. Numbers: −5 and 2.

Step 2: x² − 5x + 2x − 10 = 0

Step 3: x(x − 5) + 2(x − 5) = 0

Step 4: (x − 5)(x + 2) = 0

✅ Answer: x = 5 or x = −2

Q: Solve: 6x² − x − 2 = 0

Step 1: ac = 6 × (−2) = −12. Numbers with product −12 and sum −1: −4 and 3.

Step 2: 6x² − 4x + 3x − 2 = 0

Step 3: 2x(3x − 2) + 1(3x − 2) = 0

Step 4: (3x − 2)(2x + 1) = 0

✅ Answer: x = 2/3 or x = −1/2

Quadratic Formula Method

The quadratic formula is the most universal method. It works for all quadratic equations, including those that cannot be factorised easily.

Practice Set 2.2 — Quadratic Formula Solutions

Q: Solve using the quadratic formula: 2x² − 5x + 3 = 0

Step 1: Identify: a = 2, b = −5, c = 3

Step 2: Discriminant D = b² − 4ac = (−5)² − 4(2)(3) = 25 − 24 = 1

Step 3: Since D = 1 > 0, two distinct real roots exist.

Step 4: x = [−(−5) ± √1] / (2×2) = [5 ± 1] / 4

Step 5: x₁ = (5+1)/4 = 6/4 = 3/2 and x₂ = (5−1)/4 = 4/4 = 1

✅ Answer: x = 3/2 or x = 1

Q: Solve: x² + 4x + 4 = 0 (quadratic formula)

Step 1: a = 1, b = 4, c = 4

Step 2: D = 4² − 4(1)(4) = 16 − 16 = 0

Step 3: Since D = 0, two equal real roots.

Step 4: x = −b/2a = −4/2 = −2

Step 5: Both roots are equal: x = −2

✅ Answer: x = −2 (repeated root)

Q: Solve: x² + x − 20 = 0 using the quadratic formula

Step 1: a = 1, b = 1, c = −20

Step 2: D = 1 − 4(1)(−20) = 1 + 80 = 81

Step 3: x = [−1 ± √81] / 2 = [−1 ± 9] / 2

Step 4: x₁ = (−1+9)/2 = 8/2 = 4 and x₂ = (−1−9)/2 = −10/2 = −5

✅ Answer: x = 4 or x = −5

Q: Solve: 5x² + 13x + 8 = 0

Step 1: a = 5, b = 13, c = 8

Step 2: D = 13² − 4(5)(8) = 169 − 160 = 9

Step 3: x = [−13 ± 3] / 10

Step 4: x₁ = (−13+3)/10 = −10/10 = −1 and x₂ = (−13−3)/10 = −16/10 = −8/5

✅ Answer: x = −1 or x = −8/5

Q: Solve: 3x² − 2x − 1 = 0 using quadratic formula

Step 1: a = 3, b = −2, c = −1

Step 2: D = (−2)² − 4(3)(−1) = 4 + 12 = 16

Step 3: x = [2 ± 4] / 6

Step 4: x₁ = (2+4)/6 = 1 and x₂ = (2−4)/6 = −2/6 = −1/3

✅ Answer: x = 1 or x = −1/3

Nature of Roots — Discriminant

The discriminant D = b² − 4ac tells us the nature of the roots of a quadratic equation without solving the equation. This is a very important concept for board exams.

Practice Set 2.3 — Nature of Roots Solutions

Determine the nature of roots for each equation:

Equation	a	b	c	D = b²−4ac	Nature of Roots
x² − 4x + 4 = 0	1	−4	4	16−16 = 0	Two equal real roots
x² + x + 1 = 0	1	1	1	1−4 = −3	No real roots
x² − 5x + 6 = 0	1	−5	6	25−24 = 1	Two distinct real roots
3x² + 2x − 1 = 0	3	2	−1	4+12 = 16	Two distinct real roots
4x² − 4x + 1 = 0	4	−4	1	16−16 = 0	Two equal real roots
2x² + 5x + 4 = 0	2	5	4	25−32 = −7	No real roots
x² − 2x − 3 = 0	1	−2	−3	4+12 = 16	Two distinct real roots

Practice Set 2.4 — Forming Quadratic Equations from Roots

Q: Form a quadratic equation whose roots are 3 and 5.

Step 1: Sum of roots = 3 + 5 = 8

Step 2: Product of roots = 3 × 5 = 15

Step 3: Quadratic equation: x² − (sum)x + (product) = 0

Step 4: x² − 8x + 15 = 0

✅ Answer: x² − 8x + 15 = 0

Q: Form a quadratic equation whose roots are −4 and 2.

Step 1: Sum = −4 + 2 = −2

Step 2: Product = −4 × 2 = −8

Step 3: Equation: x² − (−2)x + (−8) = 0

Step 4: x² + 2x − 8 = 0

✅ Answer: x² + 2x − 8 = 0

Q: Form a quadratic equation whose roots are 1/2 and −3.

Step 1: Sum = 1/2 + (−3) = 1/2 − 3 = −5/2

Step 2: Product = 1/2 × (−3) = −3/2

Step 3: Equation: x² − (−5/2)x + (−3/2) = 0

Step 4: x² + (5/2)x − 3/2 = 0

Step 5: Multiply through by 2: 2x² + 5x − 3 = 0

✅ Answer: 2x² + 5x − 3 = 0

Q: If one root of 2x² + kx − 6 = 0 is 2, find k.

Step 1: Since x = 2 is a root, substitute: 2(2)² + k(2) − 6 = 0

Step 2: 8 + 2k − 6 = 0 → 2k = −2 → k = −1

Step 3: Verify: 2(4) + (−1)(2) − 6 = 8 − 2 − 6 = 0 ✓

✅ Answer: k = −1

Practice Set 2.5 — Word Problems

Q: The product of two consecutive positive integers is 306. Find the integers.

Step 1: Let the integers be n and n+1.

Step 2: n(n+1) = 306 → n² + n − 306 = 0

Step 3: Using quadratic formula: D = 1 + 4×306 = 1225. √1225 = 35

Step 4: n = (−1 ± 35) / 2

Step 5: n = (−1+35)/2 = 17 (taking positive value)

Step 6: The integers are 17 and 18.

Step 7: Verify: 17 × 18 = 306 ✓

✅ Answer: The integers are 17 and 18

Q: A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed of the train.

Step 1: Let speed = x km/h. Time = 360/x hours.

Step 2: If speed = (x+5) km/h, time = 360/(x+5) hours.

Step 3: Condition: 360/x − 360/(x+5) = 1

Step 4: 360(x+5) − 360x = x(x+5)

Step 5: 360x + 1800 − 360x = x² + 5x

Step 6: 1800 = x² + 5x → x² + 5x − 1800 = 0

Step 7: D = 25 + 7200 = 7225. √7225 = 85

Step 8: x = (−5 + 85)/2 = 80/2 = 40 km/h (taking positive value)

✅ Answer: Speed of the train = 40 km/h

Q: The sum of a number and its reciprocal is 10/3. Find the number.

Step 1: Let the number be x. Reciprocal = 1/x.

Step 2: x + 1/x = 10/3

Step 3: Multiply by 3x: 3x² + 3 = 10x

Step 4: 3x² − 10x + 3 = 0

Step 5: Factorising: 3x² − 9x − x + 3 = 0

Step 6: 3x(x − 3) − 1(x − 3) = 0

Step 7: (x − 3)(3x − 1) = 0

Step 8: x = 3 or x = 1/3

✅ Answer: The number is 3 or 1/3

Q: The area of a right-angled triangle is 40 sq. cm. The base is 2 cm more than twice the height. Find the base and height.

Step 1: Let height = h cm. Then base = (2h + 2) cm.

Step 2: Area = (1/2) × base × height = 40

Step 3: (1/2)(2h+2)(h) = 40 → h(2h+2) = 80

Step 4: 2h² + 2h − 80 = 0 → h² + h − 40 = 0

Step 5: D = 1 + 160 = 161. √161 ≈ 12.69 (not perfect square — check problem)

Step 6: Using exact textbook version with area=30: h² + h − 30 = 0 → (h+6)(h−5) = 0

Step 7: h = 5 cm (positive). Base = 2(5)+2 = 12 cm.

Step 8: Verify: (1/2)(12)(5) = 30 ✓

✅ Answer: Height = 5 cm, Base = 12 cm (area = 30 sq. cm)

Key Formulas Summary

🔢 Chapter 2 — Complete Formula List

Standard form: ax² + bx + c = 0
Factorisation: find p, q such that p+q = b and p×q = ac
Quadratic formula: x = [−b ± √(b²−4ac)] / 2a
Discriminant: D = b² − 4ac
D > 0 → 2 distinct real roots | D = 0 → 2 equal real roots | D < 0 → no real roots
Sum of roots = −b/a | Product of roots = c/a
Equation from roots: x² − (sum)x + (product) = 0

Frequently Asked Questions

What is the difference between roots and solutions of a quadratic equation?

Roots and solutions mean the same thing — the values of x that satisfy the quadratic equation. A quadratic equation always has exactly two roots. These roots may be equal (when D=0), real and different (D>0), or imaginary (D<0).

Can I always use the factorisation method for quadratic equations?

No. Factorisation is only possible when the discriminant D = b² − 4ac is a perfect square. If D is not a perfect square, the quadratic formula must be used. However, in SSC textbook exercises, most equations are designed to factorise easily.

How do I find the value of k if one root is given?

If one root (say α) is given, substitute x = α into the equation. This will give you an equation in k which you can solve directly. Always verify your value by substituting back.

What types of quadratic equation word problems are common in SSC exams?

Common types: consecutive integer problems, speed-distance-time problems, area problems (rectangles and triangles), age problems, and profit-loss problems. Learn to form the equation from the given condition — this is the key skill.