Quadratic Equations: Class 10 Maths · Chapter 4

Build them from situations, solve them, and let the discriminant say what to expect

Mathematics · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

55 min
26 practice questions
13 MCQs

What you will learn in Quadratic Equations

Recognise a quadratic equation, write it in the standard form ax² + bx + c = 0, and spot equations that only look quadratic
Translate situations about areas, numbers, ages, speeds and prices into quadratic equations
Solve quadratic equations by splitting the middle term and the zero-product property, and check roots by substitution
Use the quadratic formula to find real roots, including roots that involve surds
Use the discriminant b² − 4ac to decide whether the roots are distinct, equal or not real, and to find coefficients that give equal roots
Decide whether a real-life situation is possible, and reject roots that do not fit it

Original companion text aligned to NCERT Mathematics Class 10, Chapter 4, pp. 38–48. Not affiliated with or endorsed by NCERT.

4.1 Introduction

In Chapter 2 (Polynomials) you met quadratic polynomials such as $3 x^{2} - 5 x - 50.$ This chapter turns them into questions. Put a quadratic polynomial on one side of an equals sign and 0 on the other, and you have a quadratic equation (द्विघात समीकरण). Solving it means finding the values of $x$ that make the polynomial zero.

Such equations appear whenever an unknown ends up multiplied by itself. Think of the area of a rectangle with unknown sides, or a speed that sits in the denominator of a time. Think of a price that depends on how many items are sold, or the height of a ball thrown upwards. Here is one.

For the school's Annual Day, the art teacher wants a rectangular stage with an area of $50 m^{2}$ . To suit the hall, its length must be 5 m less than three times its breadth. How wide should the stage be?

If the breadth is $x$ metres, three times it less 5 gives the length, $(3 x - 5)$ metres. Area = length × breadth, so:

\begin{aligned} x (3 x - 5) & = 50 \\ 3 x^{2} - 5 x & = 50 \\ 3 x^{2} - 5 x - 50 & = 0 \end{aligned}

So the breadth is a number $x$ that makes $3 x^{2} - 5 x - 50$ zero. Guessing is slow (Fig. 4.1): $x = 4$ gives an area of only $28 m^{2}$ , while $x = 6$ overshoots to $78 m^{2}$ . In this chapter you will learn two dependable methods: factorisation and the quadratic formula. You will also meet a quick test that tells you in advance whether an equation has any real solution at all.

4/4 $x = 4$ is too small and $x = 6$ too big. We need a method.

Fig. 4.1An area condition with an unknown side leads to a quadratic equation.NCERT Fig. 4.1 · p. 38

A top view of a rectangular stage drawn with plank lines, labelled area 50 square metres. Step 1 marks both sides with question marks. Step 2 replaces them: the breadth is x and the length is 3x minus 5. Step 3 adds the area equation x times (3x minus 5) equals 50 below the stage, and its standard form 3x squared minus 5x minus 50 equals 0. Step 4 adds a guess-and-check panel: x equal to 4 gives an area of 28, too small, and x equal to 6 gives 78, too big, so guessing is slow.

Timeline

Quadratic equations through history

c. 1800 BCE
Babylonian clay tablets
Scribes worked out the sides of a field from its area and the sum of two of its sides: a quadratic problem in disguise.
c. 300 BCE
Euclid's Elements
Book II finds such lengths with geometry, by cutting up and rearranging squares and rectangles.
628 CE
Brahmagupta
His Brahmasphutasiddhanta states, in verse, a rule for solving equations of the type $a x^{2} + b x = c .$
c. 820 CE
Al-Khwarizmi
His book on al-jabr sorted equations of degree up to 2 into six standard types and solved each; the word algebra comes from its title.
c. 900–1025 CE
Sridharacharya's rule
He solved quadratics by completing the square. Bhaskara II later quoted his rule; estimates of Sridhara's dates vary.
1145
Abraham bar Hiyya
His treatise on measuring land, known in Latin as Liber embadorum, showed Europe how to solve every type of quadratic.

Quick check

1 of 2

A rectangular notice board is 6 cm longer than it is wide, and its area is

216 c m^{2} .

If the width is

w

cm, which equation must

w

satisfy?

w^{2} + 6 = 216

4 w + 12 = 216

w^{2} + 6 w - 216 = 0

w^{2} - 6 w - 216 = 0

Show answer

Answer (C) $w^{2} + 6 w - 216 = 0$

Area

= w (w + 6) = 216

gives

w^{2} + 6 w - 216 = 0.

Option B adds up the four sides (the perimeter), option A adds 6 instead of multiplying, and option D would make the length

w - 6.

4.2 Quadratic Equations

Quadratic equations come in many shapes, but each can be tidied into one. Move every term to the left-hand side, collect like terms and arrange them by falling powers of $x$ . The result is the standard form (मानक रूप).

Standard form

\begin{array}{c} a x^{2} + b x + c = 0 \\ (a \neq 0) \end{array}

(4.1)

where

$x$: the variable (the unknown)
$a, b, c$: real numbers: $a$ multiplies $x^{2}$ , $b$ multiplies $x$ , and $c$ is the constant term

Definition

Quadratic equation

An equation in one variable $x$ that can be rearranged into $a x^{2} + b x + c = 0$ with real coefficients $a,$ $b,$ $c$ and $a \neq 0.$ After rearranging, its left side is a polynomial of degree 2.

Example $5 x^{2} - 2 x - 7 = 0,$ with $a = 5,$ $b = - 2$ and $c = - 7.$

Why insist on $a \neq 0$ ? With $a = 0$ the $x^{2}$ term vanishes and $b x + c = 0$ is only a linear equation. The other two coefficients may well be zero, and the terms may start out in any order:

$9 - 4 x^{2} = 0$ becomes $- 4 x^{2} + 9 = 0$ : here $a = - 4,$ $b = 0$ and $c = 9.$
$x^{2} = 6 x$ becomes $x^{2} - 6 x = 0$ : here $a = 1,$ $b = - 6$ and $c = 0.$
$3 x + 2 - x^{2} = 0$ becomes $- x^{2} + 3 x + 2 = 0.$ Multiplying by $- 1$ gives $x^{2} - 3 x - 2 = 0,$ another standard form of the same equation (often easier, since $a > 0$ ).

In real problems an equation is rarely handed to you. You build it from a situation: name the unknown, express every other quantity in terms of it, and use the condition you are given. The stone path in Fig. 4.2 is a typical case.

3/3The lawn left is $(20 - 2 x)$ m by $(14 - 2 x)$ m, and it covers $160 m^{2}$ .

Fig. 4.2A path of width

x

trims

2 x

from each dimension of the lawn.NCERT Example 1 · p. 40

A rectangle labelled 20 m along the top and 14 m down the side represents a school lawn, shaded green. In step 2 a stone path appears along the inside of all four edges, and four double arrows mark its width x on each side. In step 3 the inner lawn is outlined with a dashed border and labelled 20 minus 2x across and 14 minus 2x down, with its area of 160 square metres written in the middle.

Example 1NCERT Example 1 · p. 39

From words to an equation

Write each situation as a quadratic equation in standard form. (You will solve them in Practice 4.2.)

A school lawn is 20 m long and 14 m wide. A path of uniform width runs along the inside of all four edges, and the lawn left in the middle covers $160 m^{2}$ . We want the width of the path.
Some friends share the ₹480 cost of a cake equally. Had two more friends joined, each would have paid ₹20 less. We want the number of friends.

Try it yourself first. Then reveal the solution one step at a time.

Situation (ii) did not look quadratic: it had fractions, and clearing them produced the $x^{2}$ term. The reverse also happens. An equation can look quadratic, or even cubic, and turn into something else once it is simplified (Fig. 4.3).

4/4A is linear ( $3 x - 1 = 0$ ); B is quadratic ( $x^{2} - 3 x - 1 = 0$ ).

Fig. 4.3Decide only after simplifying: the highest power that survives decides the type.NCERT p. 40

Two cards. Card A shows x(x + 5) = (x + 1) squared and card B shows x cubed + 4x squared = (x + 1) cubed. In step 2 each card shows both sides expanded: x squared + 5x = x squared + 2x + 1, and x cubed + 4x squared = x cubed + 3x squared + 3x + 1. In step 3 a note says the x squared terms cancel in A and the x cubed terms cancel in B, leaving 3x − 1 = 0 and x squared − 3x − 1 = 0. In step 4 the verdicts appear: A has degree 1 and is linear, so its card gets a dashed border; B has degree 2 and is quadratic, so its card gets a thick border.

Example 2NCERT Example 2 · p. 40

Quadratic or not?

Which of these are quadratic equations?

$(x + 3)^{2} - 4 = 3 x + 7$
$(x - 1) (x + 4) = x^{2} + 2 x - 6$
$2 x^{2} - 3 = (x - 1)^{2} + x$
$(x - 1)^{3} = x^{3} + 2 x$

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Written as

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

with

a > 0,

the equation

7 - 2 x^{2} = 5 x

has

(a, b, c)

equal to

(2, 5, - 7)

(- 2, 5, 7)

(2, - 5, 7)

(7, 5, - 2)

Show answer

Answer (A) $(2, 5, - 7)$

Add

2 x^{2}

to both sides and subtract 7:

0 = 2 x^{2} + 5 x - 7.

a = 2,

b = 5,

c = - 7.

Option C forgets that each term changes sign when it crosses the equals sign.

Practice · 3 questions

Practice 4.1: questions with answers

1
Short answerEasyNCERT Exercise 4.1, Q1
Simplify each equation and decide whether it is quadratic.
1. $(x - 4)^{2} = 3 (x + 2)$
  Show solution
  $x^{2} - 8 x + 16 = 3 x + 6,$ so $x^{2} - 11 x + 10 = 0.$ Quadratic.
2. $(x + 3) (x - 2) = (x + 1) (x + 5)$
  Show solution
  $x^{2} + x - 6 = x^{2} + 6 x + 5,$ so $- 5 x - 11 = 0.$ The $x^{2}$ terms cancel: not quadratic.
3. $(x + 1)^{3} = x^{2} (x + 3)$
  Show solution
  $x^{3} + 3 x^{2} + 3 x + 1 = x^{3} + 3 x^{2},$ so $3 x + 1 = 0.$ The cubes and the squares cancel: not quadratic (linear).
4. $x^{3} - 2 x^{2} + 5 = (x - 1)^{3}$
  Show solution
  $x^{3} - 2 x^{2} + 5 = x^{3} - 3 x^{2} + 3 x - 1,$ so $x^{2} - 3 x + 6 = 0.$ Quadratic.
2
Quick answerMediumNCERT Exercise 4.1, Q1
Which of these are quadratic equations? Choose all that apply.
$2 x (x - 1) = x^{2} + 4$ $(x - 2)^{2} = x^{2} + 1$ $x (x + 3) (x - 1) = x^{3} + 5$ $(2 x + 1)^{2} = 4 x^{2} - 3$
Select all that apply
Show answer
Answer
(A) $2 x (x - 1) = x^{2} + 4$
(C) $x (x + 3) (x - 1) = x^{3} + 5$
A gives $x^{2} - 2 x - 4 = 0$ (quadratic). B gives $- 4 x + 3 = 0$ (linear). C: the cubes cancel and $2 x^{2} - 3 x - 5 = 0$ is left (quadratic). D gives $4 x + 4 = 0$ (linear).
3
Short answerMediumNCERT Exercise 4.1, Q2
Write each situation as a quadratic equation in standard form. Say what your variable stands for.
1. A plot shaped like a right-angled triangle has a base 6 m longer than its height, and its area is $108 m^{2}$ . We want the height.
  Show solution
  Let the height be $x$ m; the base is $(x + 6)$ m.
  Area $= \frac{1}{2} x (x + 6) = 108,$ so $x^{2} + 6 x = 216$ and $x^{2} + 6 x - 216 = 0.$
2. Two consecutive positive even numbers multiply to give 224. We want the numbers.
  Show solution
  Let the smaller number be $x$ ; the next even number is $x + 2.$
  $x (x + 2) = 224$ gives $x^{2} + 2 x - 224 = 0.$
3. Ishaan's father is 25 years older than Ishaan. Five years ago, their ages (in years) multiplied to 150. We want Ishaan's present age.
  Show solution
  Let Ishaan be $x$ years old now; his father is $x + 25.$ Five years ago they were $x - 5$ and $x + 20.$
  $(x - 5) (x + 20) = 150$ gives $x^{2} + 15 x - 100 = 150,$ so $x^{2} + 15 x - 250 = 0.$
4. Riya cycles 48 km to her grandparents' village at a steady speed. Had she ridden 4 km/h faster, the trip would have taken 1 hour less. We want her speed.
  Show solution
  Let her speed be $x$ km/h. The trip takes $\frac{48}{x}$ hours, or $\frac{48}{x + 4}$ hours at the faster speed.
  $\frac{48}{x} - \frac{48}{x + 4} = 1$
  Multiply by $x (x + 4)$ : $48 (x + 4) - 48 x = x (x + 4),$ so $192 = x^{2} + 4 x .$
  Standard form: $x^{2} + 4 x - 192 = 0.$

4.3 Solution of a Quadratic Equation by Factorisation

What does it mean to solve $3 x^{2} - 5 x - 2 = 0 ?$ We want every value of $x$ that turns the left side into 0. Testing a value is easy: $x = 1$ gives $3 - 5 - 2 = - 4,$ so 1 fails, while $x = 2$ gives $12 - 10 - 2 = 0,$ so 2 works. A value that works is called a root (मूल) or a solution of the equation; we also say it satisfies the equation.

Definition

Root of a quadratic equation

The real number $α$ is a root of $a x^{2} + b x + c = 0$ when $a α^{2} + b α + c = 0$ : putting $α$ in place of $x$ turns the left side into zero.

Example2 is a root of $3 x^{2} - 5 x - 2 = 0,$ since $12 - 10 - 2 = 0.$

How many roots can there be? At most two. The reason comes from Chapter 2. The roots of the equation are just the zeroes of the polynomial $a x^{2} + b x + c$ on its left side. They are the $x$ -values where its graph, a parabola, meets the $x$ -axis, and a parabola cannot meet that axis more than twice.

PropertyZero-product property

If the product of two real numbers is zero, at least one of them is zero: $p q = 0$ means $p = 0$ or $q = 0.$

This property is why factorising solves equations. Once $a x^{2} + b x + c$ is written as a product of two linear factors, the equation breaks into two small linear equations, one per factor. To find the factors we use splitting the middle term from Class IX. Look for two numbers whose product is $a c$ and whose sum is $b$ . Use them to split $b x$ into two terms, then factorise in pairs. Algebra tiles show why this works (Fig. 4.4).

3/4Its sides are $2 x + 1$ and $x + 3,$ so $2 x^{2} + 7 x + 3 = (2 x + 1) (x + 3) .$

Fig. 4.4Factorising means rebuilding the expression as a rectangle: its sides are the factors.NCERT p. 42

Twelve tiles: two large blue squares standing for x squared, seven orange strips standing for x and three small green squares standing for 1, with a key at the bottom right. Step 1 shows the loose pieces and asks whether they can make one rectangle. Step 2 slides them together: the two big squares and one upright strip form the top row, and three rows of two flat strips and a small square fill the bottom, with no gaps. Step 3 marks the sides of the rectangle as 2x + 1 across and x + 3 down, so 2x squared + 7x + 3 = (2x + 1)(x + 3). Step 4 sets the product to zero, giving x = −1/2 or x = −3.

Table 4.1Searching for the split of

3 x^{2} - 11 x + 10

: the product must be

a c = 30

and the sum

b = - 11.

Pair with product 30	Sum
$- 1$ and $- 30$	$- 31$
$- 2$ and $- 15$	$- 17$
$- 3$ and $- 10$	$- 13$
$- 5$ and $- 6$	$- 11$ ✓

Example 3NCERT Example 3 · p. 42

Factorise, then solve

Solve $3 x^{2} - 11 x + 10 = 0$ by factorising.

Try it yourself first. Then reveal the solution one step at a time.

Example 4NCERT Example 4 · p. 43

When the product $a c$ is negative

Solve $10 x^{2} + 3 x - 4 = 0$ by factorisation.

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 43

A repeated root with surds

Solve $5 x^{2} - 2 \sqrt{15} x + 3 = 0.$

Try it yourself first. Then reveal the solution one step at a time.

Spotting a perfect square saves time. Here $5 x^{2} = (\sqrt{5} x)^{2},$ $3 = (\sqrt{3})^{2}$ and $2 \sqrt{15} x = 2 (\sqrt{5} x) (\sqrt{3}),$ so the left side is $(\sqrt{5} x - \sqrt{3})^{2}$ at sight.

Example 6NCERT Example 6 · p. 44

Back to the stage

Find the breadth and length of the Annual Day stage from Section 4.1.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

The roots of

x^{2} - x - 12 = 0

are

- 4

and

3

4

and

- 3

4

and

3

12

and

- 1

Show answer

Answer (B) $4$ and $- 3$

x^{2} - x - 12 = (x - 4) (x + 3),

x = 4

x = - 3.

Option A has the signs the wrong way round: for

x = 3,

9 - 3 - 12 = - 6 \neq 0.

Practice · 8 questions

Practice 4.2: questions with answers

1
NumericalEasyNCERT Exercise 4.2, Q1
Solve by factorisation.
1. $x^{2} + 2 x - 35 = 0$
  Show solution
  Numbers with product $- 35$ and sum $2$ : $7$ and $- 5.$
  $(x + 7) (x - 5) = 0,$ so $x = 5$ or $x = - 7.$
2. $3 x^{2} - x - 10 = 0$
  Show solution
  $a c = - 30$ and $b = - 1$ : the numbers are $- 6$ and $5.$
  $\begin{aligned} 3 x^{2} - 6 x + 5 x - 10 \\ = 3 x (x - 2) + 5 (x - 2) \\ = (3 x + 5) (x - 2) \end{aligned}$
  So $x = 2$ or $x = - \frac{5}{3} .$
3. $\sqrt{3} x^{2} + 10 x + 7 \sqrt{3} = 0$
  Show solution
  $a c = \sqrt{3} \cdot 7 \sqrt{3} = 21$ and $b = 10$ : the numbers are $7$ and $3.$
  $\begin{aligned} \sqrt{3} x^{2} + 7 x + 3 x + 7 \sqrt{3} \\ = x (\sqrt{3} x + 7) \\ + \sqrt{3} (\sqrt{3} x + 7) \\ = (\sqrt{3} x + 7) (x + \sqrt{3}) \end{aligned}$
  So $x = - \sqrt{3}$ or $x = - \frac{7}{\sqrt{3}} = - \frac{7 \sqrt{3}}{3} .$
4. $9 x^{2} - 3 x + \frac{1}{4} = 0$
  Show solution
  Multiply by 4: $36 x^{2} - 12 x + 1 = 0.$
  This is $(6 x)^{2} - 2 (6 x) (1) + 1^{2} = (6 x - 1)^{2} .$ So $x = \frac{1}{6}$ twice: a repeated root.
Short answerMediumNCERT Exercise 4.2, Q2
Solve the two equations you formed in Example 1. How wide is the path, and how many friends were there?
Show solution
(i) $x^{2} - 17 x + 30 = (x - 2) (x - 15) = 0,$ so $x = 2$ or $x = 15.$ A 15 m path cannot fit inside a lawn only 14 m wide, so the path is 2 m wide. Check: the inner lawn is $16 \times 10 = 160 m^{2} .$
(ii) $x^{2} + 2 x - 48 = (x + 8) (x - 6) = 0,$ and a count cannot be $- 8,$ so there were 6 friends. Check: ₹80 each for 6 friends, ₹60 each for 8, which is ₹20 less.
NumericalEasyNCERT Exercise 4.2, Q3
Two numbers add up to 31 and multiply to 228. What are they?
Show solution
$x (31 - x) = 228$ gives $x^{2} - 31 x + 228 = 0.$
$228 = 12 \times 19$ and $12 + 19 = 31,$ so $(x - 12) (x - 19) = 0.$ The numbers are 12 and 19.
NumericalMediumNCERT Exercise 4.2, Q4
The squares of two consecutive odd positive integers add up to 290. Find the integers.
Show solution
Let them be $x$ and $x + 2.$ Then $x^{2} + (x + 2)^{2} = 290,$ so $2 x^{2} + 4 x - 286 = 0,$ that is $x^{2} + 2 x - 143 = 0.$
$(x + 13) (x - 11) = 0,$ and $x$ must be positive, so $x = 11.$ The integers are 11 and 13: $121 + 169 = 290.$
Long answerMediumNCERT Exercise 4.2, Q5
A rectangular park has a straight path along one diagonal. The longer side of the park is 30 m more than the shorter side, and the path is 60 m longer than the shorter side. Find the sides of the park and the length of the path.
Show solution
Let the shorter side be $x$ m; the longer side is $x + 30$ and the path is $x + 60.$ By Pythagoras, $x^{2} + (x + 30)^{2} = (x + 60)^{2} .$
Expanding: $2 x^{2} + 60 x + 900 = x^{2} + 120 x + 3600,$ so $x^{2} - 60 x - 2700 = 0.$
Product $- 2700,$ sum $- 60$ : the numbers are $- 90$ and $30,$ so $(x - 90) (x + 30) = 0$ and $x = 90.$
The park is 90 m by 120 m and the path is 150 m long. Check: $8100 + 14400 = 22500 = 150^{2} .$
Long answerMediumNCERT Exercise 4.2, Q6
A stall at a Diwali mela sells handmade diyas. One evening the price of each diya, in rupees, was 4 less than three times the number of diyas sold, and the stall took ₹160 in all. How many diyas were sold, and at what price?
Show solution
Let $x$ diyas be sold, at $(3 x - 4)$ rupees each. Then $x (3 x - 4) = 160,$ so $3 x^{2} - 4 x - 160 = 0.$
Here $a c = - 480$ and $b = - 4$ : the numbers are $- 24$ and $20.$
$\begin{aligned} 3 x^{2} - 24 x + 20 x - 160 \\ = 3 x (x - 8) + 20 (x - 8) \\ = (3 x + 20) (x - 8) \end{aligned}$
So $x = 8$ (the root $- \frac{20}{3}$ is rejected). 8 diyas at ₹20 each: $8 \times 20 = 160.$
Quick answerEasy
Which of these is a root of $2 x^{2} - x - 6 = 0 ?$
$\frac{3}{2}$ $- 2$ $2$ $3$
Show answer
Answer (C) $2$
$2 (4) - 2 - 6 = 0.$ Factorising, $(2 x + 3) (x - 2) = 0,$ so the other root is $- \frac{3}{2},$ not $\frac{3}{2} .$
Quick answerMedium
If $x = 3$ is a root of $x^{2} + k x - 12 = 0,$ find $k .$
Show answer
Answer 1
$9 + 3 k - 12 = 0$ gives $k = 1.$ Then $x^{2} + x - 12 = (x + 4) (x - 3),$ so the other root is $- 4.$

4.4 Nature of Roots

Factorising is quick when it works, but many quadratics have no tidy factors. To split the middle term of $x^{2} + 4 x - 3$ you would need two whole numbers with product $- 3$ and sum 4, and there are none. The roots exist all the same; we just need a method that does not depend on luck.

The quadratic formula

The quadratic formula (द्विघाती सूत्र) works for every quadratic equation in standard form. It builds the roots directly from the coefficients:

Quadratic formula

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

(4.2)

Read off $a,$ $b$ and $c$ with their signs and substitute. The $\pm$ sign gives two roots: take $+$ for one root and $-$ for the other. The formula needs $b^{2} - 4 a c \geq 0,$ because a negative number has no real square root; what happens otherwise is the subject of What the discriminant tells us.

TheoremWhy the formula works (optional: not in the current syllabus)

If $a \neq 0$ and $b^{2} - 4 a c \geq 0,$ the two numbers given by (4.2) are exactly the roots of $a x^{2} + b x + c = 0.$

Divide by $a$ and move the constant across: $x^{2} + \frac{b}{a} x = - \frac{c}{a} .$ Add ${(\frac{b}{2 a})}^{2}$ to both sides, which makes the left side a perfect square. This is completing the square, the method behind Sridharacharya's rule:

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

When $b^{2} - 4 a c \geq 0,$ take square roots of both sides: $x + \frac{b}{2 a} = \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} .$ Subtracting $\frac{b}{2 a}$ gives (4.2). Every step can be reversed, so these are the roots and there are no others.

ExampleNCERT p. 44

Roots that factorisation cannot find

Solve $x^{2} + 4 x - 3 = 0.$

Try it yourself first. Then reveal the solution one step at a time.

What the discriminant tells us

Everything interesting in (4.2) happens under the square root. So give that number a name: we call $D = b^{2} - 4 a c$ the discriminant (विविक्तकर). The two roots are then $- \frac{b}{2 a} + \frac{\sqrt{D}}{2 a}$ and $- \frac{b}{2 a} - \frac{\sqrt{D}}{2 a} .$ They sit symmetrically, one on each side of $- \frac{b}{2 a}$ and at the same distance from it. As $D$ shrinks, they slide towards each other (Fig. 4.5).

1/4 $D = 16 > 0$ : two distinct real roots, 2 and 6, each 2 from the centre line.

Fig. 4.5The roots are

- \frac{b}{2 a} \pm \frac{\sqrt{D}}{2 a}

: equally spaced about the dashed line, closer together as

D

falls.NCERT p. 45

Axes with a dashed vertical line at x = 4, labelled x = −b/2a = 4. An upward parabola sits on this line. In step 1, y = x squared − 8x + 12 crosses the x-axis at 2 and 6, which are rings, and arrows of equal length run from the dashed line to each root; D = 16. In step 2 the parabola rises to y = x squared − 8x + 15, crossing at 3 and 5 with shorter arrows; D = 4. In step 3 it rises to y = x squared − 8x + 16 and just touches the axis at 4, a repeated root; D = 0. In step 4 it rises to y = x squared − 8x + 18 and no longer meets the axis; D = −8, no real roots.

Definition

Discriminant

For the quadratic equation $a x^{2} + b x + c = 0,$ the number $D = b^{2} - 4 a c .$ Its sign tells you how many real roots there are before you solve.

ExampleFor $2 x^{2} - 7 x + 6 = 0,$ $D = 49 - 48 = 1 > 0.$

Discriminant

D = b^{2} - 4 a c

(4.3)

Since $\sqrt{D}$ decides everything, the sign of $D$ gives the nature of the roots (मूलों की प्रकृति):

If $D > 0,$ $\sqrt{D}$ is a positive real number, so the two roots are different: two distinct real roots.
If $D = 0,$ both roots are $- \frac{b}{2 a} \pm 0,$ the same number: two equal real roots (a repeated root).
If $D < 0,$ $\sqrt{D}$ does not exist among the real numbers (every real square is at least 0): no real roots.

Table 4.2The discriminant and the nature of the roots of

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

Discriminant	Nature of the roots	Graph of $y = a x^{2} + b x + c$
$D > 0$	two distinct real roots	crosses the $x$ -axis twice
$D = 0$	two equal real roots, each $- \frac{b}{2 a}$	touches the $x$ -axis once
$D < 0$	no real roots	never meets the $x$ -axis

InteractiveDiscriminant explorer

An explorer with two tabs. In the a, b, c tab you set the three coefficients with sliders or plus and minus buttons, or pick a preset such as two roots, equal roots or no real roots. The panel shows the equation, the working D = b squared − 4ac with your numbers substituted, a bar that shows whether D is negative, zero or positive, the nature of the roots, and the roots from the quadratic formula. A graph shows the parabola, the dashed line x = −b/2a and rings where it meets the x-axis. In the Find k tab you pick an equation with an unknown coefficient k, slide k, and see D written in terms of k; the panel tells you when D = 0 gives equal roots, and warns when k makes the x squared term vanish.

Fig. 4.6Change

a,

b

and

c

(or a single unknown

k

) and watch

D,

the roots and the graph change together.

Example 7NCERT Example 7 · p. 45

Discriminant first

Without solving, decide whether $3 x^{2} + 5 x + 4 = 0$ has real roots. Use the discriminant.

Try it yourself first. Then reveal the solution one step at a time.

The discriminant also settles practical questions: can this situation happen at all? Build the equation for the situation. If $D < 0,$ no real number fits, so the situation is impossible.

3/4 $D = 4 > 0$ : possible. The bed is 8 m by 6 m.

Fig. 4.7With 28 m of fence the diagonal can be anything from about 9.9 m to just under 14 m: 10 m works, 9 m does not.NCERT Fig. 4.2 · p. 45

A green rectangle with a dashed fence, sides labelled x m and (14 − x) m, and a note that the 28 m fence means two adjacent sides add up to 14 m. Step 2 adds a dotted stone path along the diagonal, a right-angle mark at the opposite corner, and the equations x squared + (14 − x) squared = 100, x squared − 14x + 48 = 0 and D = 4 > 0. Step 3 replaces the side labels with 8 m and 6 m and shows the roots x = 8 or x = 6. Step 4 fades the bed and draws a dashed 7 m square whose diagonal is about 9.9 m, the shortest possible; D = −136 < 0, so a 9 m path is impossible.

Example 8NCERT Example 8 · p. 45

Is the flower bed possible?

A gardener has exactly 28 m of fencing to go round a rectangular flower bed. She also wants a straight stone path, exactly 10 m long, from one corner of the bed to the opposite corner. Is this possible? If so, find the sides of the bed. What if the path must be 9 m long?

Try it yourself first. Then reveal the solution one step at a time.

Example 9NCERT Example 9 · p. 46

Equal roots

What kind of roots does $5 x^{2} - 4 x + \frac{4}{5} = 0$ have? Solve it if they are real.

Try it yourself first. Then reveal the solution one step at a time.

Example

Board-exam style: find $k$ for equal roots

Find the values of $k$ for which $x^{2} - 2 k x + 7 k - 12 = 0$ has equal roots, and find the root in each case.

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 4

The equation

4 x^{2} - 12 x + 9 = 0

has

two distinct real rootstwo equal real rootsno real roots

Show answer

Answer (B) two equal real roots

D = 144 - 144 = 0,

so the roots are equal:

(2 x - 3)^{2} = 0

gives

x = \frac{3}{2}

twice. A non-zero constant term does not by itself mean two different roots.

Practice · 6 questions

Practice 4.3: questions with answers

1
NumericalMediumNCERT Exercise 4.3, Q1
Use the discriminant to describe the roots of each equation, then solve the ones that have real roots.
1. $3 x^{2} - 2 x + 4 = 0$
  Show solution
  $D = (- 2)^{2} - 4 (3) (4) = 4 - 48 = - 44 < 0$ : no real roots.
2. $5 x^{2} - 2 \sqrt{5} x + 1 = 0$
  Show solution
  $D = (2 \sqrt{5})^{2} - 4 (5) (1) = 20 - 20 = 0$ : two equal real roots.
  Each root is $- \frac{b}{2 a} = \frac{2 \sqrt{5}}{10} = \frac{\sqrt{5}}{5} .$
3. $3 x^{2} - 8 x + 2 = 0$
  Show solution
  $D = 64 - 24 = 40 > 0$ : two distinct real roots.
  $x = \frac{8 \pm \sqrt{40}}{6} = \frac{8 \pm 2 \sqrt{10}}{6} = \frac{4 \pm \sqrt{10}}{3},$ about $2.39$ and $0.28.$
2
NumericalMediumNCERT Exercise 4.3, Q2
For which value(s) of $k$ are the two roots of each equation equal?
1. $5 x^{2} + k x + 5 = 0$
  Show solution
  $D = k^{2} - 4 (5) (5) = k^{2} - 100 = 0,$ so $k = 10$ or $k = - 10.$
2. $k x^{2} + 4 k x + 12 = 0$
  Show solution
  $D = 16 k^{2} - 48 k = 16 k (k - 3) .$ So $D = 0$ when $k = 0$ or $k = 3.$
  But $k = 0$ removes the $x^{2}$ term (the equation becomes $12 = 0$ ), so $k = 3.$ Check: $3 x^{2} + 12 x + 12 = 3 (x + 2)^{2} .$
Short answerMediumNCERT Exercise 4.3, Q3
A community vegetable garden is planned as a rectangle three times as long as it is wide, covering $432 m^{2}$ . Is that possible? Give its dimensions if it is.
Show solution
Let the breadth be $x$ m. Then $3 x \cdot x = 432,$ so $3 x^{2} - 432 = 0.$
$D = 0^{2} - 4 (3) (- 432) = 5184 > 0,$ so it is possible.
$x^{2} = 144$ gives $x = 12$ (a breadth cannot be $- 12$ ). The garden is 12 m by 36 m.
4
Short answerMediumNCERT Exercise 4.3, Q4
The present ages of two cousins add up to 25 years. For each condition, decide whether it is possible; if it is, find their ages.
1. Five years ago, their ages multiplied to 100.
  Show solution
  $(x - 5) (20 - x) = 100$ gives $- x^{2} + 25 x - 100 = 100,$ so $x^{2} - 25 x + 200 = 0.$
  $D = 625 - 800 = - 175 < 0$ : not possible.
2. Five years ago, their ages multiplied to 50.
  Show solution
  Now $x^{2} - 25 x + 150 = 0,$ with $D = 625 - 600 = 25 > 0.$
  $x = \frac{25 \pm 5}{2} = 15$ or $10.$ The cousins are 15 and 10 (five years ago 10 and 5, product 50).
5
Short answerHardNCERT Exercise 4.3, Q5
A rectangular badminton practice area is to be marked out with a perimeter of 36 m.
1. Could it enclose $81 m^{2}$ ? Give the sides if so.
  Show solution
  Sides $x$ and $18 - x$ : $x (18 - x) = 81$ gives $x^{2} - 18 x + 81 = 0.$
  $D = 324 - 324 = 0$ : possible, with equal roots $x = 9.$ The area is a 9 m square.
2. Could it enclose $85 m^{2}$ ?
  Show solution
  $x^{2} - 18 x + 85 = 0$ has $D = 324 - 340 = - 16 < 0$ : not possible. With a 36 m perimeter, the square's $81 m^{2}$ is the largest area there is.
Quick answerMedium
The equation $x^{2} + 4 x + k = 0$ has real roots exactly when
$k < 4$ $k \leq 4$ $k \geq 4$ $k > 4$
Show answer
Answer (B) $k \leq 4$
Real roots, distinct or equal, need $D = 16 - 4 k \geq 0,$ that is $k \leq 4.$ Option A leaves out $k = 4,$ where the two roots are equal and still real.

4.5 Summary

Recap

Quadratic Equations: key points and summary

Any equation that simplifies to $a x^{2} + b x + c = 0,$ with real coefficients and $a \neq 0,$ is quadratic; that is its standard form (Section 4.2).
Simplify both sides before deciding whether an equation is quadratic: $x^{2}$ or $x^{3}$ terms may cancel (Example 2).
A root makes the equation true. The roots of $a x^{2} + b x + c = 0$ are the zeroes of $a x^{2} + b x + c,$ and there are at most two (Section 4.3).
To factorise, split $b x$ using two numbers with product $a c$ and sum $b,$ then set each factor equal to zero (zero-product property).
In word problems, check every root against the situation and reject the impossible ones (Example 6).
Quadratic formula: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a},$ valid whenever $b^{2} - 4 a c \geq 0$ ((4.2)).
The discriminant $D = b^{2} - 4 a c$ gives the nature of the roots: $D > 0$ two distinct real roots, $D = 0$ two equal real roots $- \frac{b}{2 a},$ $D < 0$ no real roots (Table 4.2).
To test whether a situation is possible, form its equation and look at the sign of $D$ (Example 8).

5/5Substitute each root, and reject any that cannot fit the situation.

Fig. 4.8One route through any quadratic equation. (“

D

a square” assumes whole-number coefficients: then the roots are rational and the factors are easy to find.)NCERT p. 47

A flowchart built up one box at a time. Step 1 shows the first box: write the equation in standard form. Step 2 adds an arrow to a second box: find D = b squared − 4ac. Step 3 splits into two boxes: D < 0, no real roots, stop (dashed border); and D ≥ 0, real roots exist. Step 4 adds two boxes below the second one: if D is a perfect square, factorise; for any D ≥ 0, use formula (4.2). Step 5 adds a final box: substitute to check, and reject roots that do not fit the problem.

Flash cards · 6

Key terms

1 / 6

Board-exam practice

These questions follow the formats of the CBSE Class 10 board paper: multiple choice, assertion–reason, short and long answers and a case study. Try each one before opening the solution.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 4.1, Q1	Practice 4.1 · Q1
NCERT Exercise 4.1, Q1	Practice 4.1 · Q2
NCERT Exercise 4.1, Q2	Practice 4.1 · Q3
NCERT Exercise 4.2, Q1	Practice 4.2 · Q1
NCERT Exercise 4.2, Q2	Practice 4.2 · Q2
NCERT Exercise 4.2, Q3	Practice 4.2 · Q3
NCERT Exercise 4.2, Q4	Practice 4.2 · Q4
NCERT Exercise 4.2, Q5	Practice 4.2 · Q5
NCERT Exercise 4.2, Q6	Practice 4.2 · Q6
NCERT Exercise 4.3, Q1	Practice 4.3 · Q1
NCERT Exercise 4.3, Q2	Practice 4.3 · Q2
NCERT Exercise 4.3, Q3	Practice 4.3 · Q3
NCERT Exercise 4.3, Q4	Practice 4.3 · Q4
NCERT Exercise 4.3, Q5	Practice 4.3 · Q5

Quick check: MCQs on Quadratic Equations

4.1 Introduction2 questions
4.2 Quadratic Equations3 questions
4.3 Solution of a Quadratic Equation by Factorisation4 questions
What the discriminant tells us4 questions

Glossary · 9

Key terms (English–Hindi)

discriminantविविक्तकरvivictakar: The number $D = b^{2} - 4 a c .$ Its sign tells whether the roots are distinct, equal or not real.First seen in the chapter →

equal rootsबराबर मूलbaraabar mool: Two roots with the same value, $- \frac{b}{2 a};$ they occur when $D = 0.$ First seen in the chapter →

nature of rootsमूलों की प्रकृतिmoolon ki prakriti: Whether a quadratic equation has two distinct real roots, two equal real roots or no real roots.First seen in the chapter →

quadratic equationद्विघात समीकरणdvighaat samikaran: An equation that can be written as $a x^{2} + b x + c = 0,$ with $a,$ $b,$ $c$ real and $a \neq 0.$ First seen in the chapter →
quadratic formulaद्विघाती सूत्रdvighaati sootra: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a},$ the roots of $a x^{2} + b x + c = 0$ when $b^{2} - 4 a c \geq 0.$ First seen in the chapter →

rootमूलmool: A number that makes the equation true when put in place of $x .$ A quadratic equation has at most two roots.First seen in the chapter →

splitting the middle termमध्य पद का विभाजनmadhya pad ka vibhaajan: Writing $b x$ as two terms whose coefficients have product $a c$ and sum $b,$ so the expression factorises in pairs.First seen in the chapter →
standard formमानक रूपmaanak roop: A quadratic equation written as $a x^{2} + b x + c = 0$ : every term on the left, in falling powers of $x .$ First seen in the chapter →

zero of a polynomialशून्यकshoonyak: A value of $x$ at which the polynomial equals 0; the zeroes of $a x^{2} + b x + c$ are the roots of $a x^{2} + b x + c = 0.$ First seen in the chapter →

Quadratic Equations: Class 10 Maths · Chapter 4

What you will learn in Quadratic Equations

4.1 Introduction

4.2 Quadratic Equations

Practice 4.1: questions with answers

4.3 Solution of a Quadratic Equation by Factorisation

Practice 4.2: questions with answers

4.4 Nature of Roots

The quadratic formula

What the discriminant tells us

Practice 4.3: questions with answers

4.5 Summary

Quadratic Equations: key points and summary

Board-exam practice

Practice questions with answers (NCERT “Board-exam practice”)

Find your NCERT exercise question

Quick check: MCQs on Quadratic Equations

Key terms (English–Hindi)

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