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Quadratic Equations and Factorisation

🎓 Class 10 Mathematics CBSE Theory Ch 4 — Quadratic Equations ⏱ ~35 min

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4.1 Introduction

In Chapter 2, you studied different types of polynomials. A polynomial of degree 2 in the single variable \(x\) has the form \(a x^2 + b x + c\), with \(a \ne 0\). When this polynomial is set equal to zero, we obtain a quadratic equation^?.

Here is a classic situation that leads naturally to a quadratic equation: a rectangular hall's floor has to be designed. Its length is one metre more than twice its breadth. The area of the floor must be 300 m². Let the breadth be \(x\) metres. Then the length is \((2x + 1)\) metres, and the area is:

Fig 4.1 — Rectangular floor with breadth \(x\) m and length \((2x+1)\) m.

Area = length × breadth → \((2x + 1)\, x = 300\)

Expand: \(2x^2 + x = 300\), i.e. \(2x^2 + x - 300 = 0\).

This is a quadratic equation. Several everyday situations — from projectile motion to profit-maximisation — give rise to quadratic equations. Historically, the Babylonian mathematician (c. 400 BCE) gave an explicit formula for certain quadratics; the Indian mathematician Brahmagupta (598–665 CE) gave an explicit formula to solve quadratics of the form \(a x^2 + b x = c\). Later, Sridharacharya (c. 1025 CE) derived the general quadratic formula we use today.

4.2 Quadratic Equations

Standard Form

A quadratic equation in the variable \(x\) is an equation of the form \[ a x^2 + b x + c = 0,\qquad a \ne 0, \] where \(a, b, c\) are real numbers. This is called the standard form. When \(a = 0\) the equation is linear, not quadratic.

In fact, any equation of the form \(p(x) = 0\), where \(p(x)\) is a polynomial of degree 2 in \(x\), is a quadratic equation. If a quadratic equation is not yet in standard form, we rearrange it until it is.

Worked Example 1 — Framing Quadratic Equations

(i) John and Jivanti's marbles: John and Jivanti together have 45 marbles. Both lose 5 marbles each, and the product of the number of marbles they now have is 124. Form the equation.

Let John have \(x\) marbles. Jivanti has \((45 - x)\).

After losing 5 each: John has \((x - 5)\); Jivanti has \((45 - x - 5) = (40 - x)\).

Product condition: \((x - 5)(40 - x) = 124\) → \(40x - x^2 - 200 + 5x = 124\) → \(-x^2 + 45x - 200 = 124\) → \(\mathbf{x^2 - 45x + 324 = 0}\).

(ii) Toy-production puzzle: A cottage industry produces a certain number of toys in a day. The cost (in rupees) of production of each toy was found to be 55 minus the number of toys produced in the day. On a particular day, the total cost was Rs. 750. Form the equation.

Let number of toys = \(x\). Cost per toy = \((55 - x)\). Total cost = \(x(55 - x) = 750\).

Simplify: \(55x - x^2 = 750 \Rightarrow \mathbf{x^2 - 55x + 750 = 0}\).

Worked Example 2 — Is it Quadratic?

Check which of the following are quadratic equations:

(i) \((x - 2)^2 + 1 = 2x - 3\). LHS = \(x^2 - 4x + 4 + 1 = x^2 - 4x + 5\). RHS = \(2x - 3\). Bringing together: \(x^2 - 6x + 8 = 0\). Yes, it is quadratic.
(ii) \(x(x + 1) + 8 = (x + 2)(x - 2)\). LHS = \(x^2 + x + 8\). RHS = \(x^2 - 4\). So \(x + 12 = 0\) — linear. Not quadratic.
(iii) \(x(2x + 3) = x^2 + 1\). LHS = \(2x^2 + 3x\). Gives \(x^2 + 3x - 1 = 0\). Yes, quadratic.
(iv) \((x + 2)^3 = x^3 - 4\). LHS = \(x^3 + 6x^2 + 12x + 8\). Subtracting RHS: \(6x^2 + 12x + 12 = 0 \Rightarrow x^2 + 2x + 2 = 0\). Yes, quadratic.

Takeaway

Always bring every term to one side (so the other side is zero) before deciding whether the equation is truly of degree 2. A "hidden" \(x^3\) term may cancel — or an \(x^2\) term may cancel, leaving only a linear equation.

4.3 Solution of a Quadratic Equation by Factorisation

Consider the quadratic \(2x^2 - 3x + 1 = 0\). We can rewrite the LHS by splitting the middle term:

\(2x^2 - 3x + 1 = 2x^2 - 2x - x + 1 = 2x(x - 1) - 1(x - 1) = (2x - 1)(x - 1).\)

Since the product of two real numbers is zero only if at least one of them is zero:

\((2x - 1)(x - 1) = 0 \Rightarrow 2x - 1 = 0\) or \(x - 1 = 0 \Rightarrow x = \tfrac{1}{2}\) or \(x = 1\).

The values \(x = \tfrac{1}{2}\) and \(x = 1\) are the two roots^? of the quadratic equation. A real number \(\alpha\) is called a root of \(ax^2 + bx + c = 0\) if \(a\alpha^2 + b\alpha + c = 0\). Equivalently, \(\alpha\) is a zero of the polynomial \(ax^2 + bx + c\).

Key Fact

A quadratic equation can have at most two roots, because a quadratic polynomial has at most two zeroes.

Worked Example 3 — Factorisation by Splitting the Middle Term

Find the roots of \(6x^2 - x - 2 = 0\) by factorisation.

Product of first and last coefficients = \(6 \times (-2) = -12\). Find two numbers whose product is \(-12\) and sum is \(-1\) (the middle coefficient): \(-4\) and \(3\).

\(6x^2 - 4x + 3x - 2 = 2x(3x - 2) + 1(3x - 2) = (3x - 2)(2x + 1).\)

So \(3x - 2 = 0\) or \(2x + 1 = 0 \Rightarrow x = \tfrac{2}{3}\) or \(x = -\tfrac{1}{2}\).

Verification: \(6(\tfrac{2}{3})^2 - \tfrac{2}{3} - 2 = 6 \cdot \tfrac{4}{9} - \tfrac{2}{3} - 2 = \tfrac{8}{3} - \tfrac{2}{3} - 2 = 2 - 2 = 0\). ✓

Worked Example 4 — Perfect-Square Factor

Find the roots of \(2x^2 - x + \tfrac{1}{8} = 0\).

Multiply by 8 to clear the fraction: \(16x^2 - 8x + 1 = 0\).

This is a perfect square: \((4x - 1)^2 = 0 \Rightarrow 4x - 1 = 0\), i.e. \(x = \tfrac{1}{4}\) (repeated root).

Worked Example 5 — Word Problem

Find two consecutive positive integers whose product is 306.

Let the integers be \(x\) and \(x + 1\). Then \(x(x + 1) = 306 \Rightarrow x^2 + x - 306 = 0\).

Factor: find two numbers with product \(-306\) and sum \(1\): \(18\) and \(-17\). So \((x - 17)(x + 18) = 0 \Rightarrow x = 17\) (positive).

Answer: The integers are 17 and 18.

Worked Example 6 — Hypotenuse Problem

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Let base = \(x\) cm, altitude = \((x - 7)\) cm. By Pythagoras: \(x^2 + (x - 7)^2 = 13^2\).

Expand: \(x^2 + x^2 - 14x + 49 = 169 \Rightarrow 2x^2 - 14x - 120 = 0 \Rightarrow x^2 - 7x - 60 = 0\).

Factor: \((x - 12)(x + 5) = 0 \Rightarrow x = 12\) (reject \(-5\)).

Answer: Base = 12 cm, altitude = 5 cm.

Activity: Spot the Hidden Quadratic

Materials: paper, pen.

Take each of the following equations. Expand both sides and bring every term to the left.
Circle the highest-degree term. Decide: quadratic, linear, or neither?
\((a)\; (x + 1)^2 = 2(x - 3)\) \((b)\; x^2 - 2x = (-2)(3 - x)\) \((c)\; (x - 2)(x + 1) = (x - 1)(x + 3)\) \((d)\; (x + 2)^3 = 2x(x^2 - 1)\).

(a) \(x^2 + 2x + 1 = 2x - 6 \Rightarrow x^2 + 7 = 0\). Quadratic.
(b) \(x^2 - 2x = -6 + 2x \Rightarrow x^2 - 4x + 6 = 0\). Quadratic.
(c) \(x^2 - x - 2 = x^2 + 2x - 3 \Rightarrow -3x + 1 = 0\). Linear, not quadratic.
(d) \(x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x \Rightarrow -x^3 + 6x^2 + 14x + 8 = 0\). Cubic — not quadratic.

Interactive: Factorisation Helper

Enter coefficients \(a, b, c\) of \(ax^2 + bx + c = 0\). The tool searches for integer factors that split the middle term, if they exist.

Enter coefficients and click Factorise.

Exercise 4.1

Q1. Check whether the following are quadratic equations:
(i) \((x + 1)^2 = 2(x - 3)\) (ii) \(x^2 - 2x = (-2)(3 - x)\)
(iii) \((x - 2)(x + 1) = (x - 1)(x + 3)\) (iv) \((x - 3)(2x + 1) = x(x + 5)\)
(v) \((2x - 1)(x - 3) = (x + 5)(x - 1)\) (vi) \(x^2 + 3x + 1 = (x - 2)^2\)
(vii) \((x + 2)^3 = 2x(x^2 - 1)\) (viii) \(x^3 - 4x^2 - x + 1 = (x - 2)^3\)

(i) Quadratic: \(x^2 + 7 = 0\).
(ii) Quadratic: \(x^2 - 4x + 6 = 0\).
(iii) Not (reduces to linear \(-3x + 1 = 0\)).
(iv) Quadratic: \(x^2 - 10x - 3 = 0\) after expanding.
(v) Quadratic (after expansion yields degree-2 equation).
(vi) Quadratic: \(7x - 3 = 0\)... actually \(x^2 + 3x + 1 = x^2 - 4x + 4 \Rightarrow 7x - 3 = 0\) — not quadratic.
(vii) Not: cubic terms remain → \(x^3 - 6x^2 - 12x - 8 = 0\).
(viii) Quadratic: expand and simplify → \(2x^2 - 13x + 9 = 0\) (verify).

Q2. Represent the following situations as quadratic equations:
(i) Plot area 528 m²; length = 2(breadth) + 1.
(ii) Product of two consecutive integers = 306.
(iii) Rohan's mother is 26 years older. Product of their ages (in years) 3 years from now = 360.
(iv) A train travels 480 km at a uniform speed. Had it been 8 km/h slower, it would have taken 3 hours more.

(i) Let breadth = \(x\). \(x(2x + 1) = 528 \Rightarrow 2x^2 + x - 528 = 0\).
(ii) Let integers be \(x, x+1\). \(x(x+1) = 306 \Rightarrow x^2 + x - 306 = 0\).
(iii) Let Rohan's age = \(x\). Mother = \(x + 26\). In 3 years: \((x + 3)(x + 29) = 360 \Rightarrow x^2 + 32x - 273 = 0\).
(iv) Let speed = \(x\) km/h. \(\tfrac{480}{x - 8} - \tfrac{480}{x} = 3\). Multiply through: \(480x - 480(x - 8) = 3x(x - 8) \Rightarrow 3840 = 3x^2 - 24x \Rightarrow x^2 - 8x - 1280 = 0\).

Exercise 4.2

Q1. Find the roots of the following quadratic equations by factorisation:
(i) \(x^2 - 3x - 10 = 0\) (ii) \(2x^2 + x - 6 = 0\) (iii) \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
(iv) \(2x^2 - x + \tfrac{1}{8} = 0\) (v) \(100x^2 - 20x + 1 = 0\)

(i) \((x - 5)(x + 2) = 0 \Rightarrow x = 5, -2\).
(ii) Split \(x\) as \(4x - 3x\): \(2x^2 + 4x - 3x - 6 = 2x(x+2) - 3(x+2) = (2x-3)(x+2) = 0 \Rightarrow x = \tfrac{3}{2}, -2\).
(iii) Split \(7x = 2x + 5x\): \(\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = \sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = (\sqrt{2}x + 5)(x + \sqrt{2})\). Roots: \(x = -\tfrac{5}{\sqrt{2}}, -\sqrt{2}\).
(iv) (shown above) \(x = \tfrac{1}{4}\) (double).
(v) \((10x - 1)^2 = 0 \Rightarrow x = \tfrac{1}{10}\) (double).

Competency-Based Questions

Scenario: A community park committee is planning a rectangular flower garden. The length must be 3 m more than twice the breadth. The total area available for the garden is 54 m².

Q1. Form a quadratic equation in \(x\) (breadth) and solve by factorisation.

L3 Apply

\(x(2x + 3) = 54 \Rightarrow 2x^2 + 3x - 54 = 0\). Split \(3x = 12x - 9x\): \(2x^2 + 12x - 9x - 54 = 2x(x + 6) - 9(x + 6) = (2x - 9)(x + 6)\). So \(x = 4.5\) m (reject \(-6\)). Length = \(2(4.5) + 3 = 12\) m. Breadth 4.5 m, Length 12 m.

Q2. Analyse: one committee member reads the problem as "length is 3 m more than twice the square of breadth." Form that equation. Is it quadratic?

L4 Analyse

\(x(2x^2 + 3) = 54 \Rightarrow 2x^3 + 3x - 54 = 0\). This is cubic, not quadratic. Reading words carefully matters; a misplaced "square" changes the equation's degree and difficulty.

Q3. Evaluate: another committee member proposes a 6 m × 9 m garden "to save time". Does it satisfy both the 54 m² area and length=2·breadth+3 conditions?

L5 Evaluate

Area check: \(6 \times 9 = 54\) ✓. Length-relation check: \(2(6) + 3 = 15 \ne 9\). So the proposal satisfies area but violates the length-breadth relation. Reject unless the committee waives the relation.

Q4. Design a quadratic-equation problem from your daily life (e.g. fencing, painting, orchard). State the equation in standard form and solve it by factorisation.

L6 Create

Sample: "A farmer plants a square mango orchard and adds a 2-m wide gravel walk along two adjacent edges. Total area of orchard + walk = 144 m². Find the side of the orchard." Let side = \(x\). Then \((x + 2)^2 = 144 \Rightarrow x^2 + 4x - 140 = 0 \Rightarrow (x - 10)(x + 14) = 0 \Rightarrow x = 10\) m.

Assertion–Reason Questions

Assertion (A): \(x^2 + 2x + 2 = 0\) is a quadratic equation.
Reason (R): Any equation that, after simplification, has the standard form \(ax^2 + bx + c = 0\) with \(a \ne 0\) is quadratic.

(a) Both true, R explains A

(b) Both true, R doesn't explain A

(d) A false, R true

(a) — A has \(a = 1 \ne 0\). R captures the defining property exactly.

Assertion (A): A quadratic equation can have three real roots.
Reason (R): A polynomial of degree 2 has at most two zeroes.

(a) Both true, R explains A

(b) Both true, R doesn't explain A

(d) A false, R true

(d) — A is false (max 2 roots); R is true and in fact disproves A.

Assertion (A): The roots of \(2x^2 - x + \tfrac{1}{8} = 0\) are equal.
Reason (R): Multiplying by 8 gives \(16x^2 - 8x + 1 = (4x - 1)^2\), a perfect square.

(a) Both true, R explains A

(b) Both true, R doesn't explain A

(d) A false, R true

(a) — root is \(x = 1/4\), a repeated root. R correctly explains A.

Frequently Asked Questions

What is a quadratic equation?

A quadratic equation is an equation of the form ax^2 + bx + c = 0 where a, b, c are real numbers and a != 0. The highest power of the variable is 2. NCERT Class 10 Maths Chapter 4 starts with this definition.

How do you solve by factorisation?

Express the quadratic as a product of two linear factors: ax^2 + bx + c = a(x - r)(x - s). Then r and s are the roots. For example, x^2 - 5x + 6 = (x - 2)(x - 3), giving roots 2 and 3. NCERT Class 10 Chapter 4 teaches this.

What is splitting the middle term?

To factorise ax^2 + bx + c, find two numbers whose product is a*c and whose sum is b. Split bx as the sum of two terms using these numbers, then factor by grouping. NCERT Class 10 Maths Chapter 4 uses this technique.

What is a root of a quadratic?

A root is a value of the variable that makes the equation zero. A quadratic has up to two real roots. For ax^2 + bx + c = 0, the roots can be found by factorisation, completing the square, or the quadratic formula. NCERT Class 10 Chapter 4 defines this.

Can every quadratic be factorised easily?

No. Only quadratics with rational roots factor cleanly with integer coefficients. Others need completing the square or the quadratic formula. NCERT Class 10 Maths Chapter 4 covers all three methods.

Example: solve x^2 - 7x + 10 = 0.

Find two numbers with product 10 and sum -7: -2 and -5. So x^2 - 7x + 10 = (x - 2)(x - 5). Roots: x = 2 and x = 5. Verify: 4 - 14 + 10 = 0 and 25 - 35 + 10 = 0. NCERT Class 10 Chapter 4 example.

Frequently Asked Questions — Quadratic Equations

What is Quadratic Equations and Factorisation in NCERT Class 10 Mathematics?

Quadratic Equations and Factorisation is a key concept covered in NCERT Class 10 Mathematics, Chapter 4: Quadratic Equations. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Quadratic Equations and Factorisation step by step?

To solve problems on Quadratic Equations and Factorisation, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 10 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 4: Quadratic Equations?

The essential formulas of Chapter 4 (Quadratic Equations) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Quadratic Equations and Factorisation important for the Class 10 board exam?

Quadratic Equations and Factorisation is part of the NCERT Class 10 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Quadratic Equations and Factorisation?

Common mistakes in Quadratic Equations and Factorisation include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Quadratic Equations and Factorisation?

End-of-chapter NCERT exercises for Quadratic Equations and Factorisation cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 4, and solve at least one previous-year board paper to consolidate your understanding.

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