Quadratic Equations — Class 10 Notes

CBSE 2027 — dates, rules and official papers →

Maths · Ch 4Class 10 · CBSE 2026–27High weightage

Quadratic Equations

What the board actually asks: solving by factorisation & the quadratic formula, the discriminant (nature of roots), and word problems — these carry almost every mark here.

NCERT VERBATIM sourced, word-for-wordAUTHORED LazyTopper teaching scaffolding

The Big Idea AUTHORED

One question, a few tools.

A quadratic equation is any equation you can write as $a x^{2} + b x + c = 0$ , where $a \neq = 0$ . The whole chapter answers one question: what values of $x$ make this true? Those values are the roots. You'll learn to find them (factorisation, the quadratic formula), decide whether they exist (the discriminant), relate them to the coefficients, and turn real situations into equations you can solve.

Formulae

Standard form

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

Tip: Move everything to one side first.

Quadratic Formula

Quadratic formula

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

Tip: Compute D = b² − 4ac first, then substitute D.

Discriminant

D = b^{2} - 4 a c

Tip: Write the value of D as its own step.

Nature of roots

D > 0

: distinct ·

D = 0

: equal ·

D < 0

: none real

Tip: State D > 0 / = 0 / < 0 in words before concluding.

Roots & coefficients

α + β = - \frac{b}{a} \cdot α β = \frac{c}{a} \cdot x^{2} - (α + β) x + α β = 0

Tip: Sum −b/a, product c/a, for checking.

Solving by Factorisation (Splitting the Middle Term)

Find two numbers whose product is

a c

and whose sum is

b

; split the middle term

b x

with them; factorise by grouping; set each factor equal to 0.

Condition for Real and Equal Roots

D = 0

, the equation has two equal real roots, each equal to

- \frac{b}{2 a}

Shortcuts & memory tricks →

Definitions the board rewards NCERT VERBATIM

State these in NCERT’s exact words

Quadratic equation

NCERT verbatim — memorise this exact wording· Ch 4,

“A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers, a ≠ 0.”

What it means: any equation that simplifies to degree 2. If the

x^{2}

term survives simplification, it is quadratic — always simplify FIRST before deciding (NCERT's Example 2 warning).

Root of a quadratic equation

NCERT verbatim — memorise this exact wording· Ch 4,

“In general, a real number α is called a root of the quadratic equation ax² + bx + c = 0, a ≠ 0 if aα² + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation.”

What it means: a value of

x

that makes the equation true — plug it in and the LHS becomes

0

. Same thing as a zero of the polynomial

a x^{2} + b x + c

Discriminant

NCERT verbatim — memorise this exact wording· Ch 4,

“Since b² – 4ac determines whether the quadratic equation ax² + bx + c = 0 has real roots or not, b² – 4ac is called the discriminant of this quadratic equation.”

What it means:

D = b^{2} - 4 a c

. When a question asks about the NATURE of roots (or “find

k

”), compute

D

first — its sign answers the question without solving for

x

Nature of roots

NCERT verbatim — memorise this exact wording· Ch 4,

“So, a quadratic equation ax² + bx + c = 0 has (i) two distinct real roots, if b² – 4ac > 0, (ii) two equal real roots, if b² – 4ac = 0, (iii) no real roots, if b² – 4ac < 0.”

What it means:

D > 0

: cuts the x-axis twice ·

D = 0

: just touches it ·

D < 0

: never meets it. In answers, always say “no real roots”, not “no roots”.

Quadratic formula

NCERT verbatim — memorise this exact wording· Ch 4,

“Quadratic formula: The roots of a quadratic equation ax² + bx + c = 0 are given by (−b ± √(b² − 4ac)) / 2a, provided b² – 4ac ≥ 0.”

What it means:

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

— the fallback that always works when

b^{2} - 4 a c \geq 0

, e.g. when the middle term won't split.

Key terms — NCERT verbatim (state word-for-word)

Standard form: That is, ax² + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation. (Ch 4, )
Zeroes = roots: Note that the zeroes of the quadratic polynomial ax² + bx + c and the roots of the quadratic equation ax² + bx + c = 0 are the same. (Ch 4, )
At most two roots: You have observed, in Chapter 2, that a quadratic polynomial can have at most two zeroes. So, any quadratic equation can have atmost two roots. (Ch 4, )
Roots by factorisation: If we can factorise ax² + bx + c, a ≠ 0, into a product of two linear factors, then the roots of the quadratic equation ax² + bx + c = 0 can be found by equating each factor to zero. (Ch 4, )

Must-know concepts AUTHORED FRAMING

…and where they’re tested

Solving by factorisation

Split the middle term so the product of the two parts equals

a \cdot c

, factor by grouping into two linear factors, then equate each factor to zero — the roots fall out.

2–3 mark “find the roots”

The quadratic formula

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

, valid when

b^{2} - 4 a c \geq 0

. The fallback when factorisation is hard — write

a

b

c

with their signs before substituting.

2–3 marks

Nature of roots via the discriminant

D = b^{2} - 4 a c

D > 0

→ two distinct real roots ·

D = 0

→ two equal real roots ·

D < 0

→ no real roots. The graph shows it best:

Figurey = ax² + bx + c — where the curve meets the x-axis = the real roots

D > 0

two distinct roots (cuts twice)

D = 0

two equal roots (just touches)

D < 0

no real roots (never meets)

Left: D > 0 cuts twice (two distinct roots) · Middle: D = 0 just touches (two equal roots) · Right: D < 0 never meets (no real roots).

“find k” for equal roots — up to 5 marks (CBSE 2024, 30/5/1 Q32a)

Roots & coefficients (sum & product)

For roots

α, β

α + β = - \frac{b}{a}

and

α β = \frac{c}{a}

. Build an equation from given roots:

x^{2} - (α + β) x + α β = 0

. ⚑ Syllabus note: catalogued under Polynomials (quadratic zeroes–coefficient relationship) — applied here.

Verify roots · form-the-equation

Word problems → equation

Translate the situation (area, ages, speed, consecutive numbers) into

a x^{2} + b x + c = 0

, solve, and REJECT the root the situation forbids — a negative length or age, or an excluded value like

k \neq = - 1

Long applied problem — a fixture of Sections C/D

Worked examples — real solved problems NCERT VERBATIM

Solved exactly as the source does them

Every problem is a real sourced item, quoted verbatim with its page; the step-by-step decoding and mark-logic are LazyTopper-authored coaching.

Ch 4, Example 4,

“Find the roots of the quadratic equation 6x² – x – 2 = 0.”

Split the middle term (NCERT's split).

6 x^{2} - x - 2 = 6 x^{2} + 3 x - 4 x - 2

— the two parts multiply to

6 x^{2} \times (- 2) \dots

i.e. product

- 12 x^{2}

, sum

- x

.Concept: split the middle term

1 mark

Group and factor.

= 3 x (2 x + 1) - 2 (2 x + 1) = (3 x - 2) (2 x + 1)

.Concept: factor by grouping

1 mark

Equate each factor to zero.The roots of

6 x^{2} - x - 2 = 0

are the values of

x

for which

(3 x - 2) (2 x + 1) = 0

: so

3 x - 2 = 0

2 x + 1 = 0

.Concept: zero-product rule

½ mark

State BOTH roots.

x = \frac{2}{3}

x = - \frac{1}{2}

. NCERT then verifies both satisfy the equation.Concept: verify both roots

½ mark

Mark-scheme logic (authored): 3-mark shape: 1 for the split, 1 for the factorisation, 1 for BOTH roots. Writing only one root is the classic lost mark — NCERT itself verifies both.

Ch 4, Example 7,

“Find the discriminant of the quadratic equation 2x² – 4x + 3 = 0, and hence find the nature of its roots.”

Ch 4, Example 9,

“Find the discriminant of the equation 3x² – 2x + 1/3 = 0 and hence find the nature of its roots. Find them, if they are real.”

Ch 4, Example 8,

“A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed 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?”

Ch 4,

“Represent the following situations in the form of quadratic equations : … (ii) The product of two consecutive positive integers is 306. We need to find the integers.”

“Find the value of 'k' for which the quadratic equation (k + 1)x² – 6(k + 1)x + 3(k + 9) = 0, k ≠ – 1 has real and equal roots.”

Problem-solving strategies AUTHORED

“real / equal / no roots” or “find k”?Go straight to

D = b^{2} - 4 a c

. Don't solve for

x

“Find the roots”?Factorise first; if the middle term won't split, use the formula.

“Roots given / form the equation”?Sum & product →

x^{2} - (α + β) x + α β = 0

Word problem?Name

x

→ build the equation → standard form → solve → reject the impossible root.

Coefficient holds the unknown (like

(k + 1) x^{2}

)?Check

a \neq = 0

— an excluded value in the question is a mark waiting to be lost.

Common mistakes that cost marks

Where students lose easy marks

Forgetting to reject the invalid root — negative length / age / count, or an excluded value like

k \neq = - 1

Sign slips in

D

— when

b

is negative,

(- 4)^{2} = + 16

, not

- 16

Sign error in form-the-equation — the middle term is

- (α + β)

, not

+

Writing “no roots” when

D < 0

— say “no real roots”.

Formulae

Standard form

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

Tip: Move everything to one side first.

Quadratic Formula

Quadratic formula

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

Tip: Compute D = b² − 4ac first, then substitute D.

Discriminant

D = b^{2} - 4 a c

Tip: Write the value of D as its own step.

Nature of roots

D > 0

: distinct ·

D = 0

: equal ·

D < 0

: none real

Tip: State D > 0 / = 0 / < 0 in words before concluding.

Roots & coefficients

α + β = - \frac{b}{a} \cdot α β = \frac{c}{a} \cdot x^{2} - (α + β) x + α β = 0

Tip: Sum −b/a, product c/a, for checking.

Solving by Factorisation (Splitting the Middle Term)

Find two numbers whose product is

a c

and whose sum is

b

; split the middle term

b x

with them; factorise by grouping; set each factor equal to 0.

Condition for Real and Equal Roots

D = 0

, the equation has two equal real roots, each equal to

- \frac{b}{2 a}

Shortcuts & Tricks

Set D = 0, solve for k; reject any k that makes a = 0 (e.g. in (k−1)x² + …, reject k = 1).

Safe in written answers

"Real roots": D ≥ 0. "Two distinct real roots": D > 0.

MCQ / check only: don't write this as working

If a and c have opposite signs then −4ac > 0, so D > 0: two distinct real roots.

MCQ / check only: don't write this as working

For "sum/product of roots", use −b/a and c/a; do not solve.

MCQ / check only: don't write this as working

Substitute each root back into the equation.

Safe in written answers

Memory Tricks

Bhai ka square minus chaar Aloo-Chacha.

D = b² − 4ac. Bhai = b (squared), chaar = 4, Aloo = a, Chacha = c.

Plus = Pakke do alag · Zero = Zindagi bhar ek jaisa (do barabar) · Minus = Mayoos (real root nahi).

D>0 / D=0 / D<0. P→two distinct real roots; Z→two equal real roots; M→no real roots.

Minus Bhai, plus-minus root-D, poore upar wale ko do Aloo se bhaag.

x = (−b ± √D)/(2a). minus b, ± √D, over 2a.

Mindmap

- - Split the middle term
  - Factor by grouping
  - Zero-product → two roots
- - x = (−b ± √(b²−4ac)) / 2a
  - Valid when b²−4ac ≥ 0
  - Fallback when it won't factor
- - D = b² − 4ac
  - D > 0 → two distinct real
  - D = 0 → two equal real
  - D < 0 → no real roots
  - “find k” lives here
- - α + β = −b/a
  - αβ = c/a
  - x² − (α+β)x + αβ = 0
  - ⚑ catalogued under Polynomials
- - Name the unknown → build the equation
  - Reduce to standard form
  - REJECT the impossible root
- - Forgetting to reject a root
  - Sign slips in D
  - Say “no REAL roots”
  - Check a ≠ 0

The quadratic formula — where it comes from

NCERT's history box (p.39) credits Sridharacharya (C.E. 1025) with the quadratic formula, derived “by the method of completing the square”. In the 2026-27 syllabus, completing the square is NOT an assessed solution method — you solve by factorisation or the formula. Use this derivation to understand where the formula and the $b^{2} - 4 a c \geq 0$ condition come from, not as an exam technique.

Start from the standard form.

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

Divide through by

a

— allowed precisely because

a \neq = 0

x^{2} + \frac{b}{a} x + \frac{c}{a} = 0

Move the constant across.

x^{2} + \frac{b}{a} x = - \frac{c}{a}

Complete the square: add

(\frac{b}{2 a})^{2}

to both sides.

x^{2} + \frac{b}{a} x + (\frac{b}{2 a})^{2} = (\frac{b}{2 a})^{2} - \frac{c}{a}

The left side is now a perfect square.

(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}

Take square roots.

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

Real only when

b^{2} - 4 a c \geq 0

— this is exactly where the discriminant condition comes from.

Isolate

x

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

3 competency questions from the CBSE board pattern, with the full step-marked solution for each — the marks are shown per step, as the official scheme awards them.

3 marksSection C

The product of two consecutive positive odd integers is 483. Find the two integers.

Step-marked solution

[1 mark] Let the smaller odd integer be x; the next is x + 2. x(x + 2) = 483 ⇒ x² + 2x − 483 = 0.
[1 mark] Split the middle term: x² + 23x − 21x − 483 = 0 ⇒ (x + 23)(x − 21) = 0 ⇒ x = 21 or x = −23.
[1 mark] Reject x = −23 (integers are positive). The integers are 21 and 23 (21 × 23 = 483 ✓).

5 marksSection D

One positive number is 3 more than twice another positive number. If their product is 90, find the two numbers.

Step-marked solution

[1 mark] Let the smaller number be x; the larger number is (2x + 3). Given product: x(2x + 3) = 90.
[1 mark] Standard form: 2x² + 3x − 90 = 0.
[1 mark] Apply the quadratic formula: x = (−3 ± √(9 + 720)) / 4 = (−3 ± 27) / 4.
[1 mark] Roots x = 6 or x = −7.5. Reject x = −7.5 because the numbers are stated to be positive.
[1 mark] So x = 6 and the larger number is 2(6) + 3 = 15.

5 marksSection D

The difference of two natural numbers is 3 and the difference of their reciprocals is 3/28. Find the two numbers.

Step-marked solution

[1 mark] Let the larger number be x; the smaller is (x − 3). Given 1/(x − 3) − 1/x = 3/28.
[1 mark] Combine the left side: [x − (x − 3)] / [x(x − 3)] = 3/28 ⇒ 3/(x² − 3x) = 3/28, so x² − 3x = 28. Standard form: x² − 3x − 28 = 0.
[1 mark] Factorise: (x − 7)(x + 4) = 0.
[1 mark] Roots x = 7 or x = −4. Reject x = −4 because the numbers are natural (positive).
[1 mark] So x = 7 and the smaller number is 7 − 3 = 4.

Practise Quadratic Equations competency questions

Read NCERT Chapter 4 on LazyTopper (jemh104)

Official NCERT chapter PDF (jemh104) — ncert.nic.in

Sources: NCERT 2026-27 (tap to view)

Source ledger PROVENANCE

Source edition: NCERT Reprint 2026-27, Class 10 Mathematics, Ch 4 (jemh104.pdf, printed pages 38–48). One row per sourced item — every verbatim, example and NCERT figure traces to its page.

Item	Type	Source
Quadratic equation	verbatim	NCERT 2026-27 Ch4
Root of a quadratic equation	verbatim	NCERT 2026-27 Ch4
Discriminant	verbatim	NCERT 2026-27 Ch4
Nature of roots	verbatim	NCERT 2026-27 Ch4
Quadratic formula	verbatim	NCERT 2026-27 Ch4 (Summary 4)
Standard form	verbatim	NCERT 2026-27 Ch4
Zeroes = roots	verbatim	NCERT 2026-27 Ch4
At most two roots	verbatim	NCERT 2026-27 Ch4
Roots by factorisation	verbatim	NCERT 2026-27 Ch4 (Summary 3)
Factorisation · NCERT Example 4	example	NCERT 2026-27 Ch4 Example 4
Discriminant → nature · NCERT Example 7	example	NCERT 2026-27 Ch4 Example 7
Equal roots — find them · NCERT Example 9	example	NCERT 2026-27 Ch4 Example 9
Word problem · NCERT Example 8	example	NCERT 2026-27 Ch4 Example 8 –46
Form the equation · NCERT Exercise 4.1 Q2(ii)	example	NCERT 2026-27 Ch4 Exercise 4.1 Q2(ii) –42
“Find k” for equal roots · CBSE 2024	example	CBSE 2024 board paper 30/5/1 Q32(a)

Quadratic Equations · spec-rendered note · NCERT Reprint 2026-27, Class 10 Mathematics, Ch 4 (jemh104.pdf, printed pages 38–48)

Practise writing answers the way examiners mark them → Guide