Quadratic Equations: Class 10 Maths Chapter 4 Hinglish mein

Situations se equations banao, unhe solve karo, aur discriminant se pehle hi jaano ki kya milega

Mathematics · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

55 min
26 practice questions
13 MCQs

Is chapter mein aap kya seekhenge

Quadratic equation ko pehchano, use standard form ax² + bx + c = 0 mein likho, aur un equations ko pakdo jo sirf quadratic dikhti hain
Areas, numbers, ages, speeds aur prices wali situations ko quadratic equations mein badlo
Middle term split karke aur zero-product property se quadratic equations solve karo, aur substitution se roots check karo
Quadratic formula se real roots nikaalo, surds wale roots bhi
Discriminant b² − 4ac se tay karo ki roots distinct hain, equal hain ya real nahi hain, aur equal roots dene wale coefficients nikaalo
Tay karo ki koi real-life situation possible hai ya nahi, aur jo roots usmein fit nahi hote unhe reject karo

NCERT Mathematics Class 10, Chapter 4, pp. 38–48 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

4.1 Introduction

Chapter 2 (Polynomials) mein tum quadratic polynomials se mil chuke ho, jaise $3 x^{2} - 5 x - 50.$ Yeh chapter unhe questions mein badalta hai. Equals sign ke ek taraf quadratic polynomial rakho aur doosri taraf 0, to tumhare paas ek quadratic equation (द्विघात समीकरण) hai. Ise solve karne ka matlab hai $x$ ki woh values dhoondhna jo polynomial ko zero bana dein.

Aisi equations tab aati hain jab koi unknown khud se hi multiply ho jaata hai. Unknown sides wale rectangle ka area socho, ya ek speed jo time ke denominator mein baithi ho. Ek price socho jo is par depend kare ki kitne items bikte hain, ya upar phenki gayi ball ki height. Yeh raha ek example.

School ke Annual Day ke liye art teacher ko $50 m^{2}$ area wala ek rectangular stage chahiye. Hall mein fit hone ke liye uski length, uski breadth ke teen guna se 5 m kam honi chahiye. Stage kitna chauda hona chahiye?

Agar breadth $x$ metres hai, to uske teen guna mein se 5 ghatane par length milti hai, $(3 x - 5)$ metres. Area = length × breadth, isliye:

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

To breadth ek aisa number $x$ hai jo $3 x^{2} - 5 x - 50$ ko zero bana deta hai. Guess karna slow hai (Fig. 4.1): $x = 4$ se area sirf $28 m^{2}$ aata hai, jabki $x = 6$ se area zyada hokar $78 m^{2}$ ho jaata hai. Is chapter mein tum do bharosemand methods seekhoge: factorisation aur quadratic formula. Tum ek quick test bhi dekhoge jo pehle hi bata deta hai ki equation ka koi real solution hai bhi ya nahi.

4/4 $x = 4$ bahut chhota hai aur $x = 6$ bahut bada. Hume ek method chahiye.

Fig. 4.1Unknown side wali area condition se quadratic equation banti hai.NCERT Fig. 4.1 · p. 38

Plank lines ke saath bana ek rectangular stage ka top view, jis par area 50 square metres likha hai. Step 1 mein dono sides par question marks hain. Step 2 mein unki jagah aata hai: breadth x hai aur length 3x minus 5. Step 3 stage ke neeche area equation x times (3x minus 5) equals 50 jodta hai, aur uska standard form 3x squared minus 5x minus 50 equals 0. Step 4 ek guess-and-check panel jodta hai: x equal to 4 se area 28 aata hai, bahut chhota, aur x equal to 6 se 78, bahut bada, isliye guess karna slow hai.

Timeline

History mein quadratic equations

c. 1800 BCE
Babylon ki clay tablets
Scribes ne kisi field ke area aur uski do sides ke sum se uski sides nikaali: yeh ek chhupa hua quadratic problem tha.
c. 300 BCE
Euclid ki Elements
Book II aisi lengths geometry se nikaalti hai, squares aur rectangles ko kaat kar aur dobara jama kar.
628 CE
Brahmagupta
Unki Brahmasphutasiddhanta mein verse ke roop mein ek rule diya gaya hai jo is type ki equations solve karta hai: $a x^{2} + b x = c .$
c. 820 CE
Al-Khwarizmi
al-jabr par unki book ne degree 2 tak ki equations ko chhe standard types mein baanta aur har type ko solve kiya; algebra shabd isi book ke title se aaya hai.
c. 900–1025 CE
Sridharacharya ka rule
Unhone quadratics ko completing the square se solve kiya. Baad mein Bhaskara II ne unka rule quote kiya; Sridhara ki dates ke estimates alag-alag hain.
1145
Abraham bar Hiyya
Zameen naapne par likhe unke treatise ne, jo Latin mein Liber embadorum kehlata hai, Europe ko har type ka quadratic solve karna sikhaya.

Quick check

2 mein se 1

Ek rectangular notice board ki length uski width se 6 cm zyada hai, aur uska area hai

216 c m^{2} .

Agar width

w

cm hai, to

w

ko kaun si equation satisfy karni chahiye?

w^{2} + 6 = 216

4 w + 12 = 216

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

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

Answer dekho

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

Area

= w (w + 6) = 216

, isse milta hai

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

Option B chaaron sides ko jodta hai (perimeter), option A 6 ko multiply karne ki jagah jodta hai, aur option D mein length ho jaati

w - 6.

4.2 Quadratic Equations

Quadratic equations kai shapes mein aati hain, lekin har ek ko saja kar ek hi shape mein laaya ja sakta hai. Har term ko left-hand side par le jao, like terms ko jodo aur unhe $x$ ki ghatti powers ke order mein lagao. Jo milta hai woh hai standard form (मानक रूप).

Standard form

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

(4.1)

jahan

$x$: variable (yaani unknown)
$a, b, c$: real numbers: $a$ , $x^{2}$ ko multiply karta hai, $b$ , $x$ ko multiply karta hai, aur $c$ constant term hai

Definition

Quadratic equation

Ek variable $x$ mein aisi equation jise $a x^{2} + b x + c = 0$ mein rearrange kiya ja sake, jahan coefficients $a,$ $b,$ $c$ real hon aur $a \neq 0.$ Rearrange karne ke baad iski left side degree 2 ka polynomial hoti hai.

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

$a \neq 0$ par itna zor kyun? $a = 0$ hone par $x^{2}$ term gayab ho jaata hai aur $b x + c = 0$ sirf ek linear equation reh jaati hai. Baaki do coefficients zero ho sakte hain, aur shuru mein terms kisi bhi order mein ho sakte hain:

$9 - 4 x^{2} = 0$ ban jaata hai $- 4 x^{2} + 9 = 0$ : yahan $a = - 4,$ $b = 0$ aur $c = 9.$
$x^{2} = 6 x$ ban jaata hai $x^{2} - 6 x = 0$ : yahan $a = 1,$ $b = - 6$ aur $c = 0.$
$3 x + 2 - x^{2} = 0$ ban jaata hai $- x^{2} + 3 x + 2 = 0.$ $- 1$ se multiply karne par milta hai $x^{2} - 3 x - 2 = 0,$ jo usi equation ka ek aur standard form hai (aksar zyada aasaan, kyunki $a > 0$ ).

Real problems mein equation tumhe ready-made kam hi milti hai. Tum use situation se banate ho: unknown ko naam do, baaki har quantity ko uske terms mein likho, aur di gayi condition use karo. Fig. 4.2 ka stone path ek typical case hai.

3/3Bacha hua lawn $(20 - 2 x)$ m by $(14 - 2 x)$ m hai, aur yeh $160 m^{2}$ cover karta hai.

Fig. 4.2

x

width ka path lawn ki har dimension se

2 x

kam kar deta hai.NCERT Example 1 · p. 40

Ek rectangle jiske upar 20 m aur side mein 14 m likha hai, green shade wala school lawn dikhata hai. Step 2 mein chaaron edges ke andar ki taraf ek stone path aata hai, aur chaar double arrows har side par uski width x dikhate hain. Step 3 mein andar wale lawn ko dashed border se outline kiya gaya hai, us par across 20 minus 2x aur neeche ki taraf 14 minus 2x likha hai, aur beech mein uska area 160 square metres likha hai.

Example 1NCERT Example 1 · p. 39

Words se equation tak

Har situation ko standard form mein quadratic equation ki tarah likho. (Inhe tum Practice 4.2 mein solve karoge.)

Ek school lawn 20 m lamba aur 14 m chauda hai. Chaaron edges ke andar ki taraf uniform width ka ek path hai, aur beech mein bacha lawn $160 m^{2}$ cover karta hai. Hume path ki width chahiye.
Kuch dost ek cake ka ₹480 ka kharcha barabar share karte hain. Agar do aur dost jud jaate, to har ek ko ₹20 kam dene padte. Hume doston ka number chahiye.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Situation (ii) quadratic dikhti nahi thi: usmein fractions the, aur unhe hataane par $x^{2}$ term bana. Ulta bhi hota hai. Koi equation quadratic, ya cubic bhi, dikh sakti hai aur simplify karne par kuch aur ban jaati hai (Fig. 4.3).

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

Fig. 4.3Simplify karne ke baad hi tay karo: jo highest power bachti hai, wahi type tay karti hai.NCERT p. 40

Do cards. Card A par x(x + 5) = (x + 1) squared hai aur card B par x cubed + 4x squared = (x + 1) cubed. Step 2 mein har card par dono sides expand hokar dikhti hain: x squared + 5x = x squared + 2x + 1, aur x cubed + 4x squared = x cubed + 3x squared + 3x + 1. Step 3 mein ek note batata hai ki A mein x squared terms cancel hote hain aur B mein x cubed terms, aur bachta hai 3x − 1 = 0 aur x squared − 3x − 1 = 0. Step 4 mein faisle aate hain: A ki degree 1 hai aur woh linear hai, isliye uske card ko dashed border milta hai; B ki degree 2 hai aur woh quadratic hai, isliye uske card ko thick border milta hai.

Example 2NCERT Example 2 · p. 40

Quadratic hai ya nahi?

Inmein se kaun si quadratic equations hain?

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

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Jab equation

7 - 2 x^{2} = 5 x

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

ki tarah likhte hain, jahan

a > 0,

(a, b, c)

barabar hai

(2, 5, - 7)

(- 2, 5, 7)

(2, - 5, 7)

(7, 5, - 2)

Answer dekho

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

Dono sides mein

2 x^{2}

jodo aur 7 ghatao:

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

a = 2,

b = 5,

c = - 7.

Option C bhool jaata hai ki equals sign paar karne par har term ka sign badal jaata hai.

Practice · 3 questions

Practice 4.1: answers ke saath

1
Short answerEasyNCERT Exercise 4.1, Q1
Har equation ko simplify karo aur tay karo ki woh quadratic hai ya nahi.
1. $(x - 4)^{2} = 3 (x + 2)$
  Solution dikhao
  $x^{2} - 8 x + 16 = 3 x + 6,$ isliye $x^{2} - 11 x + 10 = 0.$ Quadratic.
2. $(x + 3) (x - 2) = (x + 1) (x + 5)$
  Solution dikhao
  $x^{2} + x - 6 = x^{2} + 6 x + 5,$ isliye $- 5 x - 11 = 0.$ $x^{2}$ terms cancel ho jaate hain: quadratic nahi hai.
3. $(x + 1)^{3} = x^{2} (x + 3)$
  Solution dikhao
  $x^{3} + 3 x^{2} + 3 x + 1 = x^{3} + 3 x^{2},$ isliye $3 x + 1 = 0.$ Cubes aur squares dono cancel ho jaate hain: quadratic nahi hai (linear hai).
4. $x^{3} - 2 x^{2} + 5 = (x - 1)^{3}$
  Solution dikhao
  $x^{3} - 2 x^{2} + 5 = x^{3} - 3 x^{2} + 3 x - 1,$ isliye $x^{2} - 3 x + 6 = 0.$ Quadratic.
2
Quick answerMediumNCERT Exercise 4.1, Q1
Inmein se kaun si quadratic equations hain? Jitne sahi hain, sab chuno.
$2 x (x - 1) = x^{2} + 4$ $(x - 2)^{2} = x^{2} + 1$ $x (x + 3) (x - 1) = x^{3} + 5$ $(2 x + 1)^{2} = 4 x^{2} - 3$
Saare sahi options chuno
Answer dekho
Answer
(A) $2 x (x - 1) = x^{2} + 4$
(C) $x (x + 3) (x - 1) = x^{3} + 5$
A se milta hai $x^{2} - 2 x - 4 = 0$ (quadratic). B se milta hai $- 4 x + 3 = 0$ (linear). C: cubes cancel ho jaate hain aur $2 x^{2} - 3 x - 5 = 0$ bachta hai (quadratic). D se milta hai $4 x + 4 = 0$ (linear).
3
Short answerMediumNCERT Exercise 4.1, Q2
Har situation ko standard form mein quadratic equation ki tarah likho. Batao ki tumhara variable kya dikhata hai.
1. Right-angled triangle jaise shape wale ek plot ka base uski height se 6 m zyada hai, aur uska area $108 m^{2}$ hai. Hume height chahiye.
  Solution dikhao
  Maano height $x$ m hai; to base $(x + 6)$ m hai.
  Area $= \frac{1}{2} x (x + 6) = 108,$ isliye $x^{2} + 6 x = 216$ aur $x^{2} + 6 x - 216 = 0.$
2. Do consecutive positive even numbers ka product 224 hai. Hume woh numbers chahiye.
  Solution dikhao
  Maano chhota number $x$ hai; agla even number hai $x + 2.$
  $x (x + 2) = 224$ se milta hai $x^{2} + 2 x - 224 = 0.$
3. Ishaan ke papa Ishaan se 25 saal bade hain. Paanch saal pehle unki ages (saalon mein) ka product 150 tha. Hume Ishaan ki abhi ki age chahiye.
  Solution dikhao
  Maano Ishaan abhi $x$ saal ka hai; uske papa ki age hai $x + 25.$ Paanch saal pehle unki ages thi $x - 5$ aur $x + 20.$
  $(x - 5) (x + 20) = 150$ se milta hai $x^{2} + 15 x - 100 = 150,$ isliye $x^{2} + 15 x - 250 = 0.$
4. Riya steady speed se cycle chala kar 48 km door apne nana-nani ke gaon jaati hai. Agar woh 4 km/h tez chalaati, to trip mein 1 ghanta kam lagta. Hume uski speed chahiye.
  Solution dikhao
  Maano uski speed $x$ km/h hai. Trip mein $\frac{48}{x}$ ghante lagte hain, ya tez speed par $\frac{48}{x + 4}$ ghante.
  $\frac{48}{x} - \frac{48}{x + 4} = 1$
  $x (x + 4)$ se multiply karo: $48 (x + 4) - 48 x = x (x + 4),$ isliye $192 = x^{2} + 4 x .$
  Standard form: $x^{2} + 4 x - 192 = 0.$

4.3 Factorisation se Quadratic Equation ka Solution

Is equation ko solve karne ka kya matlab hai: $3 x^{2} - 5 x - 2 = 0 ?$ Hume $x$ ki har woh value chahiye jo left side ko 0 bana de. Kisi value ko test karna aasaan hai: $x = 1$ rakhne par aata hai $3 - 5 - 2 = - 4,$ isliye 1 fail hai, jabki $x = 2$ rakhne par aata hai $12 - 10 - 2 = 0,$ isliye 2 kaam karta hai. Jo value kaam karti hai use equation ka root (मूल) ya solution kehte hain; hum yeh bhi kehte hain ki woh equation ko satisfy karti hai.

Definition

Quadratic equation ka root

Real number $α$ , $a x^{2} + b x + c = 0$ ka root hai jab $a α^{2} + b α + c = 0$ : yaani $x$ ki jagah $α$ rakhne par left side zero ho jaati hai.

Example2 is equation ka root hai: $3 x^{2} - 5 x - 2 = 0,$ kyunki $12 - 10 - 2 = 0.$

Kitne roots ho sakte hain? Zyada se zyada do. Iski wajah Chapter 2 se aati hai. Equation ke roots bas left side wale polynomial $a x^{2} + b x + c$ ke zeroes hain. Yeh woh $x$ -values hain jahan iska graph, yaani ek parabola, $x$ -axis se milta hai, aur parabola us axis se do baar se zyada nahi mil sakta.

PropertyZero-product property

Agar do real numbers ka product zero hai, to unmein se kam se kam ek zero hai: $p q = 0$ ka matlab hai $p = 0$ ya $q = 0.$

Isi property ki wajah se factorise karke equations solve hoti hain. Jab $a x^{2} + b x + c$ ko do linear factors ke product ki tarah likh dete hain, to equation do chhoti linear equations mein toot jaati hai, har factor ke liye ek. Factors dhoondhne ke liye hum Class IX wala splitting the middle term use karte hain. Aise do numbers dhoondho jinka product $a c$ ho aur sum $b$ ho. Unse $b x$ ko do terms mein todo, phir pairs mein factorise karo. Algebra tiles dikhate hain ki yeh kyun kaam karta hai (Fig. 4.4).

3/4Iski sides hain $2 x + 1$ aur $x + 3,$ isliye $2 x^{2} + 7 x + 3 = (2 x + 1) (x + 3) .$

Fig. 4.4Factorise karne ka matlab hai expression ko ek rectangle ki tarah dobara banana: uski sides hi factors hain.NCERT p. 42

Barah tiles: x squared ke liye do bade blue squares, x ke liye saat orange strips aur 1 ke liye teen chhote green squares, aur neeche right mein ek key. Step 1 khule pieces dikhata hai aur poochta hai ki kya inse ek rectangle ban sakta hai. Step 2 unhe khiska kar jodta hai: do bade squares aur ek khadi strip upar wali row banate hain, aur do leti strips aur ek chhote square wali teen rows neeche ka hissa bharti hain, bina kisi gap ke. Step 3 rectangle ki sides par across 2x + 1 aur neeche ki taraf x + 3 likhta hai, isliye 2x squared + 7x + 3 = (2x + 1)(x + 3). Step 4 product ko zero ke barabar rakhta hai, jisse x = −1/2 ya x = −3 milta hai.

Table 4.1

3 x^{2} - 11 x + 10

ka split dhoondhna: product

a c = 30

hona chahiye aur sum

b = - 11.

Product 30 wala pair	Sum
$- 1$ aur $- 30$	$- 31$
$- 2$ aur $- 15$	$- 17$
$- 3$ aur $- 10$	$- 13$
$- 5$ aur $- 6$	$- 11$ ✓

Example 3NCERT Example 3 · p. 42

Pehle factorise karo, phir solve karo

$3 x^{2} - 11 x + 10 = 0$ ko factorise karke solve karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 4NCERT Example 4 · p. 43

Jab product $a c$ negative ho

$10 x^{2} + 3 x - 4 = 0$ ko factorisation se solve karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 5NCERT Example 5 · p. 43

Surds ke saath repeated root

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

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Perfect square pehchaan lo to time bachta hai. Yahan $5 x^{2} = (\sqrt{5} x)^{2},$ $3 = (\sqrt{3})^{2}$ aur $2 \sqrt{15} x = 2 (\sqrt{5} x) (\sqrt{3}),$ isliye left side dekhte hi pata chal jaata hai ki yeh $(\sqrt{5} x - \sqrt{3})^{2}$ hai.

Example 6NCERT Example 6 · p. 44

Wapas stage par

Section 4.1 wale Annual Day stage ki breadth aur length nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

x^{2} - x - 12 = 0

ke roots hain

- 4

aur

3

4

aur

- 3

4

aur

3

12

aur

- 1

Answer dekho

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

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

isliye

x = 4

x = - 3.

Option A mein signs ulte hain: jab

x = 3,

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

Practice · 8 questions

Practice 4.2: answers ke saath

1
NumericalEasyNCERT Exercise 4.2, Q1
Factorisation se solve karo.
1. $x^{2} + 2 x - 35 = 0$
  Solution dikhao
  Product $- 35$ aur sum $2$ wale numbers: $7$ aur $- 5.$
  $(x + 7) (x - 5) = 0,$ isliye $x = 5$ ya $x = - 7.$
2. $3 x^{2} - x - 10 = 0$
  Solution dikhao
  $a c = - 30$ aur $b = - 1$ : numbers hain $- 6$ aur $5.$
  $\begin{aligned} 3 x^{2} - 6 x + 5 x - 10 \\ = 3 x (x - 2) + 5 (x - 2) \\ = (3 x + 5) (x - 2) \end{aligned}$
  To $x = 2$ ya $x = - \frac{5}{3} .$
3. $\sqrt{3} x^{2} + 10 x + 7 \sqrt{3} = 0$
  Solution dikhao
  $a c = \sqrt{3} \cdot 7 \sqrt{3} = 21$ aur $b = 10$ : numbers hain $7$ aur $3.$
  $\begin{aligned} \sqrt{3} x^{2} + 7 x + 3 x + 7 \sqrt{3} \\ = x (\sqrt{3} x + 7) \\ + \sqrt{3} (\sqrt{3} x + 7) \\ = (\sqrt{3} x + 7) (x + \sqrt{3}) \end{aligned}$
  To $x = - \sqrt{3}$ ya $x = - \frac{7}{\sqrt{3}} = - \frac{7 \sqrt{3}}{3} .$
4. $9 x^{2} - 3 x + \frac{1}{4} = 0$
  Solution dikhao
  4 se multiply karo: $36 x^{2} - 12 x + 1 = 0.$
  Yeh hai $(6 x)^{2} - 2 (6 x) (1) + 1^{2} = (6 x - 1)^{2} .$ To $x = \frac{1}{6}$ do baar: ek repeated root.
Short answerMediumNCERT Exercise 4.2, Q2
Example 1 mein banayi do equations solve karo. Path kitna chauda hai, aur kitne dost the?
Solution dikhao
(i) $x^{2} - 17 x + 30 = (x - 2) (x - 15) = 0,$ isliye $x = 2$ ya $x = 15.$ Sirf 14 m chaude lawn ke andar 15 m ka path fit nahi ho sakta, isliye path 2 m chauda hai. Check: andar wala lawn hai $16 \times 10 = 160 m^{2} .$
(ii) $x^{2} + 2 x - 48 = (x + 8) (x - 6) = 0,$ aur ginti ho nahi sakti $- 8,$ isliye 6 dost the. Check: 6 doston ke liye har ek ka ₹80, 8 ke liye har ek ka ₹60, jo ₹20 kam hai.
NumericalEasyNCERT Exercise 4.2, Q3
Do numbers ka sum 31 hai aur product 228. Woh kaun se hain?
Solution dikhao
$x (31 - x) = 228$ se milta hai $x^{2} - 31 x + 228 = 0.$
$228 = 12 \times 19$ aur $12 + 19 = 31,$ isliye $(x - 12) (x - 19) = 0.$ Numbers hain 12 aur 19.
NumericalMediumNCERT Exercise 4.2, Q4
Do consecutive odd positive integers ke squares ka sum 290 hai. Integers nikaalo.
Solution dikhao
Maano woh hain $x$ aur $x + 2.$ To $x^{2} + (x + 2)^{2} = 290,$ isliye $2 x^{2} + 4 x - 286 = 0,$ yaani $x^{2} + 2 x - 143 = 0.$
$(x + 13) (x - 11) = 0,$ aur $x$ positive hona chahiye, isliye $x = 11.$ Integers hain 11 aur 13: $121 + 169 = 290.$
Long answerMediumNCERT Exercise 4.2, Q5
Ek rectangular park mein ek diagonal ke saath seedha path bana hai. Park ki lambi side chhoti side se 30 m zyada hai, aur path chhoti side se 60 m lamba hai. Park ki sides aur path ki length nikaalo.
Solution dikhao
Maano chhoti side $x$ m hai; lambi side $x + 30$ hai aur path $x + 60.$ Pythagoras se, $x^{2} + (x + 30)^{2} = (x + 60)^{2} .$
Expand karne par: $2 x^{2} + 60 x + 900 = x^{2} + 120 x + 3600,$ isliye $x^{2} - 60 x - 2700 = 0.$
Product $- 2700,$ sum $- 60$ : numbers hain $- 90$ aur $30,$ isliye $(x - 90) (x + 30) = 0$ aur $x = 90.$
Park 90 m by 120 m ka hai aur path 150 m lamba hai. Check: $8100 + 14400 = 22500 = 150^{2} .$
Long answerMediumNCERT Exercise 4.2, Q6
Diwali mela mein ek stall handmade diyas bechta hai. Ek shaam har diya ki price, rupees mein, beche gaye diyas ke number ke teen guna se 4 kam thi, aur stall ki kul kamaai ₹160 hui. Kitne diyas bike, aur kis price par?
Solution dikhao
Maano $x$ diyas bike, har ek $(3 x - 4)$ rupees ka. To $x (3 x - 4) = 160,$ isliye $3 x^{2} - 4 x - 160 = 0.$
Yahan $a c = - 480$ aur $b = - 4$ : numbers hain $- 24$ aur $20.$
$\begin{aligned} 3 x^{2} - 24 x + 20 x - 160 \\ = 3 x (x - 8) + 20 (x - 8) \\ = (3 x + 20) (x - 8) \end{aligned}$
To $x = 8$ (root $- \frac{20}{3}$ reject hota hai). ₹20 wale 8 diyas: $8 \times 20 = 160.$
Quick answerEasy
Inmein se kaun sa is equation ka root hai: $2 x^{2} - x - 6 = 0 ?$
$\frac{3}{2}$ $- 2$ $2$ $3$
Answer dekho
Answer (C) $2$
$2 (4) - 2 - 6 = 0.$ Factorise karne par $(2 x + 3) (x - 2) = 0,$ isliye doosra root hai $- \frac{3}{2},$ na ki $\frac{3}{2} .$
Quick answerMedium
Agar $x = 3$ is equation ka root hai: $x^{2} + k x - 12 = 0,$ to nikaalo $k .$
Answer dekho
Answer 1
$9 + 3 k - 12 = 0$ se milta hai $k = 1.$ Phir $x^{2} + x - 12 = (x + 4) (x - 3),$ isliye doosra root hai $- 4.$

4.4 Roots ki Nature

Jab factorise ho jaaye to yeh jaldi ho jaata hai, lekin kai quadratics ke saaf-suthre factors nahi hote. $x^{2} + 4 x - 3$ ka middle term split karne ke liye tumhe do whole numbers chahiye jinka product $- 3$ ho aur sum 4, aur aise koi numbers hain hi nahi. Roots phir bhi exist karte hain; bas hume aisa method chahiye jo luck par depend na kare.

Quadratic formula

Quadratic formula (द्विघाती सूत्र) standard form ki har quadratic equation ke liye kaam karta hai. Yeh coefficients se seedhe roots bana deta hai:

Quadratic formula

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

(4.2)

$a,$ $b$ aur $c$ ko unke signs ke saath padho aur substitute karo. $\pm$ sign do roots deta hai: ek root ke liye $+$ lo aur doosre ke liye $-$ . Formula ke liye $b^{2} - 4 a c \geq 0,$ zaroori hai, kyunki negative number ka koi real square root nahi hota; warna kya hota hai, yeh Discriminant hume kya batata hai ka topic hai.

TheoremFormula kyun kaam karta hai (optional: current syllabus mein nahi)

Agar $a \neq 0$ aur $b^{2} - 4 a c \geq 0,$ to (4.2) se milne wale do numbers hi is equation ke roots hain: $a x^{2} + b x + c = 0.$

$a$ se divide karo aur constant ko doosri taraf le jao: $x^{2} + \frac{b}{a} x = - \frac{c}{a} .$ Dono sides mein ${(\frac{b}{2 a})}^{2}$ jodo, jisse left side perfect square ban jaati hai. Ise completing the square kehte hain, yahi Sridharacharya ke rule ke peeche ka method hai:

\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}

Jab $b^{2} - 4 a c \geq 0,$ ho, to dono sides ka square root lo: $x + \frac{b}{2 a} = \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a} .$ $\frac{b}{2 a}$ ghatane par (4.2) milta hai. Har step ulta bhi kiya ja sakta hai, isliye yahi roots hain aur koi doosre nahi.

ExampleNCERT p. 44

Woh roots jo factorisation nahi dhoondh sakta

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

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Discriminant hume kya batata hai

(4.2) mein saari interesting cheezein square root ke andar hoti hain. To us number ko ek naam do: hum $D = b^{2} - 4 a c$ ko discriminant (विविक्तकर) kehte hain. Tab do roots hain $- \frac{b}{2 a} + \frac{\sqrt{D}}{2 a}$ aur $- \frac{b}{2 a} - \frac{\sqrt{D}}{2 a} .$ Yeh symmetrically baithe hain, $- \frac{b}{2 a}$ ke dono taraf ek-ek, aur usse barabar doori par. Jaise-jaise $D$ chhota hota hai, yeh ek doosre ki taraf khisakte hain (Fig. 4.5).

1/4 $D = 16 > 0$ : do distinct real roots, 2 aur 6, dono centre line se 2 door.

Fig. 4.5Roots hain

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

: dashed line ke dono taraf barabar doori par, aur

D

ghatne par paas-paas aate hain.NCERT p. 45

Axes ke saath x = 4 par ek dashed vertical line, jis par x = −b/2a = 4 likha hai. Is line par ek upward parabola hai. Step 1 mein y = x squared − 8x + 12, x-axis ko 2 aur 6 par cross karta hai, jo rings hain, aur dashed line se har root tak barabar lambai ke arrows jaate hain; D = 16. Step 2 mein parabola upar uthkar y = x squared − 8x + 15 ban jaata hai, 3 aur 5 par cross karta hai aur arrows chhote ho jaate hain; D = 4. Step 3 mein yeh upar uthkar y = x squared − 8x + 16 ban jaata hai aur axis ko bas 4 par chhoota hai, ek repeated root; D = 0. Step 4 mein yeh upar uthkar y = x squared − 8x + 18 ban jaata hai aur ab axis se nahi milta; D = −8, koi real roots nahi.

Definition

Discriminant

Is quadratic equation ke liye: $a x^{2} + b x + c = 0,$ yeh number hai $D = b^{2} - 4 a c .$ Iska sign solve karne se pehle hi bata deta hai ki kitne real roots hain.

ExampleIs equation ke liye: $2 x^{2} - 7 x + 6 = 0,$ $D = 49 - 48 = 1 > 0.$

Discriminant

D = b^{2} - 4 a c

(4.3)

Kyunki sab kuch $\sqrt{D}$ tay karta hai, $D$ ka sign nature of the roots (मूलों की प्रकृति) batata hai:

Agar $D > 0,$ to $\sqrt{D}$ ek positive real number hai, isliye dono roots alag hain: do distinct real roots.
Agar $D = 0,$ to dono roots hain $- \frac{b}{2 a} \pm 0,$ yaani same number: do equal real roots (ek repeated root).
Agar $D < 0,$ to real numbers mein $\sqrt{D}$ exist nahi karta (har real square kam se kam 0 hota hai): koi real roots nahi.

Table 4.2

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

ka discriminant aur roots ki nature

Discriminant	Roots ki nature	$y = a x^{2} + b x + c$ ka graph
$D > 0$	do distinct real roots	$x$ -axis ko do baar cross karta hai
$D = 0$	do equal real roots, har ek $- \frac{b}{2 a}$	$x$ -axis ko ek baar chhoota hai
$D < 0$	koi real roots nahi	$x$ -axis se kabhi nahi milta

InteractiveDiscriminant explorer

Do tabs wala ek explorer. a, b, c tab mein tum sliders ya plus aur minus buttons se teeno coefficients set karte ho, ya koi preset chunte ho jaise two roots, equal roots ya no real roots. Panel dikhata hai: equation, tumhare numbers substitute karke D = b squared − 4ac ki working, ek bar jo batata hai ki D negative, zero ya positive hai, roots ki nature, aur quadratic formula se roots. Ek graph parabola, dashed line x = −b/2a aur x-axis se milne wali jagahon par rings dikhata hai. Find k tab mein tum unknown coefficient k wali equation chunte ho, k ko slide karte ho, aur k ke terms mein likha D dekhte ho; panel batata hai ki kab D = 0 se equal roots milte hain, aur warn karta hai jab k se x squared term gayab ho jaata hai.

Fig. 4.6

a,

b

aur

c

(ya ek unknown

k

) badlo aur dekho ki

D,

roots aur graph saath-saath kaise badalte hain.

Example 7NCERT Example 7 · p. 45

Pehle discriminant

Bina solve kiye tay karo ki $3 x^{2} + 5 x + 4 = 0$ ke real roots hain ya nahi. Discriminant use karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Discriminant practical questions bhi suljhata hai: kya yeh situation ho bhi sakti hai? Situation ki equation banao. Agar $D < 0,$ to koi real number fit nahi hota, isliye situation impossible hai.

3/4 $D = 4 > 0$ : possible hai. Bed 8 m by 6 m ka hai.

Fig. 4.728 m fence ke saath diagonal lagbhag 9.9 m se lekar 14 m se thoda kam tak kuch bhi ho sakta hai: 10 m chalega, 9 m nahi.NCERT Fig. 4.2 · p. 45

Dashed fence wala ek green rectangle, jiski sides par x m aur (14 − x) m likha hai, aur ek note ki 28 m fence ka matlab hai do adjacent sides ka sum 14 m hai. Step 2 diagonal ke saath ek dotted stone path, opposite corner par right-angle mark, aur equations x squared + (14 − x) squared = 100, x squared − 14x + 48 = 0 aur D = 4 > 0 jodta hai. Step 3 side labels ki jagah 8 m aur 6 m likhta hai aur roots x = 8 ya x = 6 dikhata hai. Step 4 bed ko fade karta hai aur ek dashed 7 m square banata hai jiska diagonal lagbhag 9.9 m hai, jo sabse chhota possible hai; D = −136 < 0, isliye 9 m path impossible hai.

Example 8NCERT Example 8 · p. 45

Kya flower bed possible hai?

Ek gardener ke paas rectangular flower bed ke chaaron taraf lagane ke liye bilkul 28 m fencing hai. Use bed ke ek corner se opposite corner tak ek seedha stone path bhi chahiye, bilkul 10 m lamba. Kya yeh possible hai? Agar haan, to bed ki sides nikaalo. Agar path 9 m lamba hona ho to?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 9NCERT Example 9 · p. 46

Equal roots

$5 x^{2} - 4 x + \frac{4}{5} = 0$ ke roots kis tarah ke hain? Agar real hain to ise solve karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example

Board-exam style: equal roots ke liye $k$ nikaalo

$k$ ki woh values nikaalo jinke liye $x^{2} - 2 k x + 7 k - 12 = 0$ ke equal roots hon, aur har case mein root bhi nikaalo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Equation

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

do distinct real roots haindo equal real roots hainkoi real roots nahi hain

Answer dekho

Answer (B) do equal real roots hain

D = 144 - 144 = 0,

isliye roots equal hain:

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

x = \frac{3}{2}

do baar milta hai. Sirf non-zero constant term hone se do alag roots nahi ho jaate.

Practice · 6 questions

Practice 4.3: answers ke saath

1
NumericalMediumNCERT Exercise 4.3, Q1
Har equation ke roots ko discriminant se describe karo, phir jinke real roots hain unhe solve karo.
1. $3 x^{2} - 2 x + 4 = 0$
  Solution dikhao
  $D = (- 2)^{2} - 4 (3) (4) = 4 - 48 = - 44 < 0$ : koi real roots nahi.
2. $5 x^{2} - 2 \sqrt{5} x + 1 = 0$
  Solution dikhao
  $D = (2 \sqrt{5})^{2} - 4 (5) (1) = 20 - 20 = 0$ : do equal real roots.
  Har root hai $- \frac{b}{2 a} = \frac{2 \sqrt{5}}{10} = \frac{\sqrt{5}}{5} .$
3. $3 x^{2} - 8 x + 2 = 0$
  Solution dikhao
  $D = 64 - 24 = 40 > 0$ : do distinct real roots.
  $x = \frac{8 \pm \sqrt{40}}{6} = \frac{8 \pm 2 \sqrt{10}}{6} = \frac{4 \pm \sqrt{10}}{3},$ lagbhag $2.39$ aur $0.28.$
2
NumericalMediumNCERT Exercise 4.3, Q2
$k$ ki kis value (ya values) ke liye har equation ke dono roots equal hain?
1. $5 x^{2} + k x + 5 = 0$
  Solution dikhao
  $D = k^{2} - 4 (5) (5) = k^{2} - 100 = 0,$ isliye $k = 10$ ya $k = - 10.$
2. $k x^{2} + 4 k x + 12 = 0$
  Solution dikhao
  $D = 16 k^{2} - 48 k = 16 k (k - 3) .$ To $D = 0$ jab $k = 0$ ya $k = 3.$
  Lekin $k = 0$ se $x^{2}$ term hat jaata hai (equation ban jaati hai $12 = 0$ ), isliye $k = 3.$ Check: $3 x^{2} + 12 x + 12 = 3 (x + 2)^{2} .$
Short answerMediumNCERT Exercise 4.3, Q3
Ek community vegetable garden ka plan hai: ek rectangle jo apni width se teen guna lamba ho aur $432 m^{2}$ cover kare. Kya yeh possible hai? Agar hai, to uski dimensions batao.
Solution dikhao
Maano breadth $x$ m hai. To $3 x \cdot x = 432,$ isliye $3 x^{2} - 432 = 0.$
$D = 0^{2} - 4 (3) (- 432) = 5184 > 0,$ isliye yeh possible hai.
$x^{2} = 144$ se milta hai $x = 12$ (breadth $- 12$ nahi ho sakti). Garden 12 m by 36 m ka hai.
4
Short answerMediumNCERT Exercise 4.3, Q4
Do cousins ki abhi ki ages ka sum 25 saal hai. Har condition ke liye tay karo ki woh possible hai ya nahi; agar hai, to unki ages nikaalo.
1. Paanch saal pehle unki ages ka product 100 tha.
  Solution dikhao
  $(x - 5) (20 - x) = 100$ se milta hai $- x^{2} + 25 x - 100 = 100,$ isliye $x^{2} - 25 x + 200 = 0.$
  $D = 625 - 800 = - 175 < 0$ : possible nahi hai.
2. Paanch saal pehle unki ages ka product 50 tha.
  Solution dikhao
  Ab $x^{2} - 25 x + 150 = 0,$ aur $D = 625 - 600 = 25 > 0.$
  $x = \frac{25 \pm 5}{2} = 15$ ya $10.$ Cousins 15 aur 10 saal ke hain (paanch saal pehle 10 aur 5, product 50).
5
Short answerHardNCERT Exercise 4.3, Q5
Ek rectangular badminton practice area mark karna hai jiska perimeter 36 m ho.
1. Kya yeh $81 m^{2}$ gher sakta hai? Agar haan, to sides batao.
  Solution dikhao
  Sides $x$ aur $18 - x$ : $x (18 - x) = 81$ se milta hai $x^{2} - 18 x + 81 = 0.$
  $D = 324 - 324 = 0$ : possible hai, aur equal roots hain $x = 9.$ Area ek 9 m square hai.
2. Kya yeh $85 m^{2}$ gher sakta hai?
  Solution dikhao
  $x^{2} - 18 x + 85 = 0$ ka $D = 324 - 340 = - 16 < 0$ : possible nahi hai. 36 m perimeter ke saath square ka $81 m^{2}$ hi sabse bada area hai.
Quick answerMedium
Equation $x^{2} + 4 x + k = 0$ ke real roots theek tab hote hain jab
$k < 4$ $k \leq 4$ $k \geq 4$ $k > 4$
Answer dekho
Answer (B) $k \leq 4$
Real roots ke liye, chahe distinct hon ya equal, chahiye $D = 16 - 4 k \geq 0,$ yaani $k \leq 4.$ Option A chhod deta hai $k = 4,$ jahan dono roots equal hain aur phir bhi real hain.

4.5 Summary

Recap

Summary aur key points

Koi bhi equation jo simplify hokar ban jaaye $a x^{2} + b x + c = 0,$ jahan coefficients real hon aur $a \neq 0,$ quadratic hai; yahi uska standard form hai (Section 4.2).
Koi equation quadratic hai ya nahi, yeh tay karne se pehle dono sides simplify karo: $x^{2}$ ya $x^{3}$ terms cancel ho sakte hain (Example 2).
Root equation ko sahi bana deta hai. $a x^{2} + b x + c = 0$ ke roots is polynomial ke zeroes hain: $a x^{2} + b x + c,$ aur yeh zyada se zyada do hote hain (Section 4.3).
Factorise karne ke liye, aise do numbers se $b x$ ko todo jinka product $a c$ ho aur sum $b,$ phir har factor ko zero ke barabar rakho (zero-product property).
Word problems mein har root ko situation se check karo aur impossible roots ko reject karo (Example 6).
Quadratic formula: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a},$ jo tab valid hai jab $b^{2} - 4 a c \geq 0$ ((4.2)).
Discriminant $D = b^{2} - 4 a c$ roots ki nature batata hai: $D > 0$ do distinct real roots, $D = 0$ do equal real roots $- \frac{b}{2 a},$ $D < 0$ koi real roots nahi (Table 4.2).
Koi situation possible hai ya nahi, yeh test karne ke liye uski equation banao aur $D$ ka sign dekho (Example 8).

5/5Har root ko substitute karo, aur jo situation mein fit nahi ho sakta use reject karo.

Fig. 4.8Kisi bhi quadratic equation ke liye ek raasta. (“

D

ek square” ka matlab hai ki coefficients whole numbers hain: tab roots rational hote hain aur factors aasaani se mil jaate hain.)NCERT p. 47

Ek-ek box karke banta flowchart. Step 1 pehla box dikhata hai: equation ko standard form mein likho. Step 2 ek arrow ke saath doosra box jodta hai: D = b squared − 4ac nikaalo. Step 3 do boxes mein bant jaata hai: D < 0, koi real roots nahi, ruk jao (dashed border); aur D ≥ 0, real roots hain. Step 4 doosre box ke neeche do boxes jodta hai: agar D perfect square hai, to factorise karo; kisi bhi D ≥ 0 ke liye formula (4.2) use karo. Step 5 aakhri box jodta hai: substitute karke check karo, aur jo roots problem mein fit nahi hote unhe reject karo.

Flash cards · 6

Key terms

1 / 6

Board-exam practice

Yeh questions CBSE Class 10 board paper ke formats follow karte hain: multiple choice, assertion–reason, short aur long answers aur ek case study. Solution kholne se pehle har ek ko khud try karo.

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
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

4.1 Introduction2 questions
4.2 Quadratic Equations3 questions
4.3 Factorisation se Quadratic Equation ka Solution4 questions
Discriminant hume kya batata hai4 questions

Glossary · 9

Key terms (English + Hindi)

discriminantविविक्तकरvivictakar: Number $D = b^{2} - 4 a c .$ Iska sign batata hai ki roots distinct hain, equal hain ya real nahi hain.Chapter mein pehli baar yahan aaya →

equal rootsबराबर मूलbaraabar mool: Same value wale do roots, $- \frac{b}{2 a};$ yeh tab aate hain jab $D = 0.$ Chapter mein pehli baar yahan aaya →

nature of rootsमूलों की प्रकृतिmoolon ki prakriti: Quadratic equation ke do distinct real roots hain, do equal real roots hain, ya koi real roots nahi hain.Chapter mein pehli baar yahan aaya →

quadratic equationद्विघात समीकरणdvighaat samikaran: Aisi equation jise is tarah likha ja sake: $a x^{2} + b x + c = 0,$ jahan $a,$ $b,$ $c$ real hon aur $a \neq 0.$ Chapter mein pehli baar yahan aaya →
quadratic formulaद्विघाती सूत्रdvighaati sootra: $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a},$ jo $a x^{2} + b x + c = 0$ ke roots deta hai jab $b^{2} - 4 a c \geq 0.$ Chapter mein pehli baar yahan aaya →

rootमूलmool: Woh number jo equation ko sahi bana de jab use iski jagah rakhte hain: $x .$ Quadratic equation ke zyada se zyada do roots hote hain.Chapter mein pehli baar yahan aaya →

splitting the middle termमध्य पद का विभाजनmadhya pad ka vibhaajan: $b x$ ko aise do terms mein likhna jinke coefficients ka product $a c$ ho aur sum $b,$ taaki expression pairs mein factorise ho jaaye.Chapter mein pehli baar yahan aaya →
standard formमानक रूपmaanak roop: Quadratic equation ko $a x^{2} + b x + c = 0$ ki tarah likhna: har term left mein, ghatti powers ke order mein, jahan variable hai $x .$ Chapter mein pehli baar yahan aaya →

zero of a polynomialशून्यकshoonyak: $x$ ki woh value jis par polynomial 0 ho jaata hai; $a x^{2} + b x + c$ ke zeroes hi is equation ke roots hain: $a x^{2} + b x + c = 0.$ Chapter mein pehli baar yahan aaya →

Quadratic Equations: Class 10 Maths Chapter 4 Hinglish mein

Is chapter mein aap kya seekhenge

4.1 Introduction

4.2 Quadratic Equations

Practice 4.1: answers ke saath

4.3 Factorisation se Quadratic Equation ka Solution

Practice 4.2: answers ke saath

4.4 Roots ki Nature

Quadratic formula

Discriminant hume kya batata hai

Practice 4.3: answers ke saath

4.5 Summary

Summary aur key points

Board-exam practice

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

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye