Chapter 4 of 14

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Rajasthan (RBSE)Class 10 Mathematics★ 5–7 marks · RBSE Class 10 Maths board (Unit — Algebra)

Quadratic Equations

RBSE Class 10 Maths Chapter 4 — the standard form ax² + bx + c = 0 and three ways to solve it (factorisation, completing the square, the quadratic formula), the discriminant test for the nature of the roots, and translating word problems into quadratics. Includes fully solved board-pattern problems with the discriminant reasoning shown.

⏱ 38 min readMedium difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Recognise and write a quadratic equation in standard form ax² + bx + c = 0
2Solve quadratic equations by factorisation (splitting the middle term)
3Solve by completing the square and derive the quadratic formula
4Apply the quadratic formula x = (−b ± √(b²−4ac))/2a
5Use the discriminant D = b² − 4ac to determine the nature of the roots and find unknown constants for equal roots
6Translate real-life situations into quadratic equations, solve them, and reject inadmissible roots

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Why this chapter matters

Quadratic equations are one of the highest-weight algebra chapters in Class 10 and a guaranteed scorer in the RBSE board — typically a direct solve, a discriminant/'equal roots' question, and a word problem. The quadratic formula and discriminant are reused throughout Class 11–12 and in physics.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Polynomials (Class 10)

Zeroes of a quadratic and the zero–coefficient relationships.

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Real Numbers (Class 10)

Surds and simplifying √ expressions in the formula.

Quadratic Equations — RBSE Class 10 (Mathematics)

A rectangular garden in Jodhpur is to have an area of 528 m², and its length is one more than twice its breadth. How wide is it? The moment you let breadth = x and write the area, you get $2 x^{2} + x - 528 = 0$ — a quadratic equation. This chapter is the toolkit for cracking exactly these "the answer is squared somewhere" problems.

1. Standard form

A quadratic equation in x is any equation that can be written as:

$a x^{2} + b x + c = 0, a \neq = 0$

where a, b, c are real numbers. The condition $a \neq = 0$ is essential — without the $x^{2}$ term it would only be linear. A root (or solution) is a value of x that satisfies the equation — and, from the last chapter, it is exactly a zero of the polynomial $a x^{2} + b x + c$ .

2. Method 1 — Solving by factorisation

Split the middle term so the product of the two parts equals $a \times c$ and their sum equals $b$ ; then group.

Example: $6 x^{2} - x - 2 = 0$ .

Need two numbers with product $a c = 6 \times (- 2) = - 12$ and sum $b = - 1$ . They are $- 4$ and $+ 3$ .
$6 x^{2} - 4 x + 3 x - 2 = 0 \Rightarrow 2 x (3 x - 2) + 1 (3 x - 2) = 0 \Rightarrow (3 x - 2) (2 x + 1) = 0$ .
So $x = \frac{2}{3}$ or $x = - \frac{1}{2}$ .

This is the fastest method when the factors are "nice". When they aren't, use the formula.

3. Method 2 — Completing the square

Turn the equation into "(something)² = number". This is the derivation behind the quadratic formula and is itself examinable.

For $x^{2} + 4 x - 5 = 0$ : $x^{2} + 4 x = 5 \Rightarrow x^{2} + 4 x + 4 = 5 + 4 \Rightarrow (x + 2)^{2} = 9 \Rightarrow x + 2 = \pm 3.$ So $x = 1$ or $x = - 5$ . (Add the square of half the coefficient of x to both sides.)

4. Method 3 — The quadratic formula

Completing the square on the general form gives the formula that always works:

$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Example: $2 x^{2} - 7 x + 3 = 0$ (a = 2, b = −7, c = 3). $x = \frac{7 \pm 49 - 24}{4} = \frac{7 \pm 25}{4} = \frac{7 \pm 5}{4} \Rightarrow x = 3 or x = \frac{1}{2} .$

5. The discriminant — nature of the roots

The quantity under the square root is the discriminant:

$D = b^{2} - 4 a c$

It tells you the nature of the roots before you solve:

Discriminant	Roots
$D > 0$	two distinct real roots
$D = 0$	two equal real roots (one repeated)
$D < 0$	no real roots (roots are imaginary)

This is one of the most-asked RBSE ideas — "find the value of k for which the equation has equal roots" simply means set D = 0 and solve for k.

6. Word problems — the real test

Most board marks come from forming the equation from a situation. The recipe:

Let the unknown be x and write the other quantities in terms of x.
Form an equation from the given condition (area, age, speed–time, etc.).
Bring it to standard form $a x^{2} + b x + c = 0$ .
Solve (factorise/formula).
Reject any root that is meaningless in context (negative length, negative age) and state the answer with units.

That rejection step is where careful students score and careless ones lose marks: a width cannot be negative, an age cannot be a fraction of a year if the problem says whole years, and so on.

7. Closing thought

Three methods, one formula, one discriminant — that is the whole chapter:

Factorise when it's clean; complete the square to understand why; reach for the quadratic formula when stuck.
The discriminant $b^{2} - 4 a c$ answers "how many real roots?" without solving.
For word problems, forming the equation and rejecting the impossible root are where the marks live.

Quadratics show up everywhere after this — coordinate geometry, trigonometry word sums, physics projectile motion, and all of Class 11–12. For the RBSE board, expect a direct solve, a discriminant/"equal roots" question, and a word problem. Drill all three.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Standard form

ax² + bx + c = 0, a ≠ 0

a, b, c real; a must be non-zero.

Quadratic formula

x = [−b ± √(b² − 4ac)] / 2a

Always works; derived by completing the square.

Discriminant

D = b² − 4ac

D>0 two distinct real, D=0 equal, D<0 no real roots.

Completing the square

Add (half the x-coefficient)² to both sides

Turns LHS into a perfect square (x + k)².

Sum & product of roots

α + β = −b/a, αβ = c/a

Carried over from Polynomials; handy for checks.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Forgetting the ± in the quadratic formula

✓ A quadratic has up to TWO roots. Always write both x = (−b + √D)/2a and x = (−b − √D)/2a.

WATCH OUT

✗ Splitting the middle term to sum b but not product ac

✓ The two numbers must have product = a×c AND sum = b. Both conditions, or the grouping fails.

WATCH OUT

✗ Saying D = 0 gives no real roots

✓ D = 0 gives two EQUAL real roots (one repeated value). It is D < 0 that gives no real roots.

WATCH OUT

✗ Keeping a negative or absurd root in a word problem

✓ Always reject roots that make no physical sense (negative length, age, number of items) and state which root you keep and why.

WATCH OUT

✗ Mis-substituting signs of b and c into the formula

✓ Write a, b, c with their signs first. For 2x² − 7x + 3, b = −7 so −b = +7. A sign slip ruins the whole answer.

WATCH OUT

✗ Treating an equation as quadratic when a = 0

✓ If the x² coefficient is 0, it is linear, not quadratic. Check a ≠ 0 before applying these methods.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex 4.1

Exercise 4.1

Identifying quadratics and forming equations from situations

Questions

Ex 4.2

Exercise 4.2

Solving by factorisation; word problems

Questions

Ex 4.3

Exercise 4.3

Quadratic formula, completing the square, and word problems

Questions

Ex 4.4

Exercise 4.4

Discriminant — nature of roots and finding unknown constants

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Standard form

Is (x − 2)² = x² − 4 a quadratic equation? Justify.

▶ Show solution

Step 1 — Expand LHS: x² − 4x + 4 = x² − 4.

Step 2 — Simplify: −4x + 4 = −4 ⇒ −4x + 8 = 0 (the x² terms cancel).

Step 3 — No x² term remains, so it is linear, not quadratic.

✦ Answer: No — it reduces to a linear equation.

Q2EASY· Factorisation

Solve x² − 5x + 6 = 0 by factorisation.

▶ Show solution

Step 1 — Two numbers with product 6, sum −5: −2 and −3.

Step 2 — (x − 2)(x − 3) = 0.

✦ Answer: x = 2 or x = 3.

Q3EASY· Discriminant

Find the discriminant of 2x² − 4x + 3 = 0 and state the nature of its roots.

▶ Show solution

Step 1 — D = b² − 4ac = (−4)² − 4(2)(3) = 16 − 24 = −8.

Step 2 — D < 0 ⇒ no real roots.

✦ Answer: D = −8; no real roots.

Q4MEDIUM· Factorisation

Solve 6x² − x − 2 = 0 by splitting the middle term.

▶ Show solution

Step 1 — Product ac = 6×(−2) = −12, sum = −1 → numbers −4 and 3.

Step 2 — 6x² − 4x + 3x − 2 = 0 ⇒ 2x(3x − 2) + 1(3x − 2) = 0 ⇒ (3x − 2)(2x + 1) = 0.

Step 3 — x = 2/3 or x = −1/2.

✦ Answer: x = 2/3 or x = −1/2.

Q5MEDIUM· Quadratic formula

Solve 2x² − 7x + 3 = 0 using the quadratic formula.

▶ Show solution

Step 1 — a = 2, b = −7, c = 3; D = 49 − 24 = 25.

Step 2 — x = [7 ± √25]/4 = (7 ± 5)/4.

Step 3 — x = 12/4 = 3 or x = 2/4 = 1/2.

✦ Answer: x = 3 or x = 1/2.

Q6MEDIUM· Completing the square

Solve x² + 4x − 5 = 0 by completing the square.

▶ Show solution

Step 1 — x² + 4x = 5.

Step 2 — Add (4/2)² = 4 to both sides: x² + 4x + 4 = 9 ⇒ (x + 2)² = 9.

Step 3 — x + 2 = ±3 ⇒ x = 1 or x = −5.

✦ Answer: x = 1 or x = −5.

Q7HARD· Equal roots

Find the value(s) of k for which the equation kx² − 2kx + 6 = 0 has equal roots (k ≠ 0).

▶ Show solution

Step 1 — Equal roots ⇒ D = 0. Here a = k, b = −2k, c = 6.

Step 2 — D = (−2k)² − 4(k)(6) = 4k² − 24k = 0.

Step 3 — 4k(k − 6) = 0 ⇒ k = 0 or k = 6. Since k ≠ 0, k = 6.

✦ Answer: k = 6.

Q8HARD· Word problem (area)

The length of a rectangular plot is one metre more than twice its breadth, and its area is 528 m². Find its dimensions.

▶ Show solution

Step 1 — Let breadth = x m; length = (2x + 1) m.

Step 2 — Area: x(2x + 1) = 528 ⇒ 2x² + x − 528 = 0.

Step 3 — D = 1 + 4(2)(528) = 1 + 4224 = 4225; √D = 65.

Step 4 — x = (−1 ± 65)/4 ⇒ x = 16 or x = −16.5. Reject the negative breadth.

Step 5 — Breadth = 16 m, length = 2(16) + 1 = 33 m.

✦ Answer: 16 m × 33 m.

Q9HARD· Word problem (speed)

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 for the journey. Find the speed of the train.

▶ Show solution

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

Step 2 — At (x + 5) km/h: time = 360/(x + 5), which is 1 hour less.

Step 3 — 360/x − 360/(x + 5) = 1 ⇒ 360(x + 5 − x) = x(x + 5) ⇒ 1800 = x² + 5x.

Step 4 — x² + 5x − 1800 = 0; D = 25 + 7200 = 7225; √D = 85.

Step 5 — x = (−5 ± 85)/2 ⇒ x = 40 or x = −45. Reject negative speed.

✦ Answer: Speed of the train = 40 km/h.

Q10HARD· Word problem (numbers)

The product of two consecutive positive integers is 306. Form a quadratic equation and find the integers.

▶ Show solution

Step 1 — Let the smaller integer be x; the next is (x + 1).

Step 2 — Product: x(x + 1) = 306 ⇒ x² + x − 306 = 0.

Step 3 — D = 1² − 4(1)(−306) = 1 + 1224 = 1225; √D = 35.

Step 4 — x = (−1 ± 35)/2 ⇒ x = 17 or x = −18. Reject the negative (integers must be positive).

Step 5 — So x = 17 and the next integer is 18.

✦ Answer: x² + x − 306 = 0; the integers are 17 and 18.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Standard form ax² + bx + c = 0 with a ≠ 0; a root is a value satisfying it.
•Factorisation: split the middle term so product = ac and sum = b, then group.
•Completing the square: add (half the x-coefficient)² to both sides to form (x + k)².
•Quadratic formula x = [−b ± √(b² − 4ac)]/2a — keep the ±.
•Discriminant D = b² − 4ac: D>0 two distinct real, D=0 equal, D<0 no real roots.
•'Equal roots' means D = 0 — solve for the unknown constant.
•Word problems: let x = unknown, form the equation, solve, reject inadmissible roots, give units.
•Sum of roots = −b/a, product = c/a — useful for quick checks.

Rajasthan (RBSE) marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 5–7 marks

Question type	Marks each	Typical count	What it tests
MCQ / very short	1	1	Standard form, discriminant value, nature of roots
Short answer	2	1–2	Solve by factorisation / formula / completing the square
Short answer	3	1	Equal-roots (find k); a simple word problem
Long answer	4	1	Speed–time / area / age word problem

Prep strategy

Be fluent in all three methods, but default to the quadratic formula when factors aren't obvious
Memorise D = b² − 4ac and the three nature-of-root cases
Practise 'find k for equal roots' — it is just D = 0
For word problems, write the 'let x = …' line, form the equation, and ALWAYS state the rejected root

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Projectile motion

The height of a launched object is a quadratic in time; setting it to zero finds when it lands — pure quadratic-equation work.

Area and fencing problems

Designing a plot or garden of a fixed area with a length–breadth relation leads straight to a quadratic, as in the 528 m² plot.

Speed, distance and time

'If the speed were 5 km/h more, the trip is an hour shorter' problems become quadratics — common in travel and logistics.

Business break-even

Revenue and cost models are often quadratic; their roots give the break-even quantities.

Optimisation

Finding the maximum area for a given perimeter, or minimum material for a box, uses the vertex of a quadratic.

Engineering design

Parabolic trajectories, arches and antenna shapes are governed by quadratic relationships.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Write a, b, c with signs before substituting into the formula or discriminant.
Always present both roots (the ± gives two), then interpret.
For nature-of-roots, compute D and quote the matching case explicitly.
In word problems, never skip the 'reject the negative/absurd root' line — it carries a mark.
Show the equation-forming step clearly; most word-problem marks are for setting it up correctly.
Verify a found root by substituting back if time permits — a quick safety check.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Relationship between roots and coefficients (Vieta) used to build new equations whose roots are transformed (α², 1/α, etc.).
Nature of roots with rational vs irrational discriminant; when roots are rational.
Equations reducible to quadratic form (biquadratic, reciprocal equations).
The role of the discriminant in conic sections and in classifying curves later on.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

RBSE Class 10 Board (BSER Ajmer)Very high — solve + discriminant + word problem almost every year

NTSE / state scholarshipMedium — quadratic and discriminant MCQs

JEE FoundationVery high — quadratics and the discriminant are core Class 11 algebra

Maths Olympiad (NMTC)Medium — root–coefficient manipulation problems

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Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Yes. RBSE prescribes the NCERT Mathematics textbook for Class 10, so the chapter content and exercises are identical. RBSE (BSER Ajmer) sets the exam pattern and marking.

Try factorisation first — it is fastest when the factors are simple. If you cannot factorise quickly, use the quadratic formula, which always works. Completing the square is mainly asked when the question names it explicitly.

D = b² − 4ac tells you the nature of the roots without solving: D > 0 → two different real roots, D = 0 → two equal real roots, D < 0 → no real roots (they are imaginary).

A quadratic gives two algebraic roots, but a physical quantity (length, speed, age, number of items) cannot be negative or otherwise meaningless. You discard the inadmissible root and keep only the sensible one.

No. Many quadratics have irrational or non-obvious factors. Those are best solved with the quadratic formula or by completing the square; factorisation is only convenient when the factors are 'nice' integers/fractions.

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