Class 10 Maths · Chapter 4
Short answer:
Is chapter me total 3 exercises hain — Exercise 4.1 (11 questions: quadratic equation check karna aur real-life situations ko quadratic equation ke form me likhna), Exercise 4.2 (6 questions: factorisation method se roots nikalna, jisme John-Jivanti marbles aur cottage industry toys jaise word problems bhi hain), aur Exercise 4.3 (5 questions: discriminant se Nature of Roots — real distinct, real equal, ya no real roots — plus mango grove, friends' ages aur rectangular park jaise application problems). Rationalised syllabus ke hisaab se sirf factorisation method use hota hai roots nikalne ke liye — quadratic formula ka poora derivation ab is chapter me nahi padhaya jaata; discriminant sirf Nature of Roots samajhne ke liye use hota hai.
Quadratic equation woh equation hai jisme variable ka highest power 2 ho — ax²+bx+c=0 (a≠0). Class 9 me hum sirf linear equations (degree 1) solve karte the; ab hum degree 2 wali equations solve karna seekhenge, jaise x²-5x+6=0. Real life me quadratic equations kaafi jagah aati hain — area problems (rectangular plot, park), speed-distance-time problems, ages ke problems, aur production-cost problems. Is chapter me hum ye equations factorisation se solve karenge — matlab equation ko do linear factors ke product ke form me todenge, aur phir har factor ko zero maan kar roots nikalenge. Saath hi hum ye bhi seekhenge ki bina equation solve kiye, sirf discriminant (b²-4ac) dekh kar kaise pata karein ki roots real hain ya nahi, aur real hue to distinct hain ya equal.
Chapter 4 Summary — 5 Minute Revision
1. Standard Form
Koi bhi equation quadratic equation kehlati hai agar use ax² + bx + c = 0 ke form me likha ja sake, jahan a, b, c real numbers hain aur a ≠ 0. Agar equation ko simplify karne ke baad x² (ya higher power) present rehta hai, tabhi woh quadratic hai — agar x² wale terms cancel ho jaayein aur sirf x reh jaaye, to woh linear ban jaati hai, quadratic nahi.
Check karne ka tareeka: Diye gaye expression ko fully expand (simplify) karo, RHS ko LHS me le aao, aur dekho highest power kya bacha.
2. Representing Real-Life Situations as Quadratic Equations
Word problems me sabse important step hai variable define karna (jaise "breadth = x m rakho") aur phir diye gaye conditions ko equation me convert karna. Common patterns:
- Area problems: length × breadth = area — agar length aur breadth dono variable me expressed hain (ek dusre ke terms me), to product quadratic ban jaata hai.
- Consecutive integers: agar ek integer x hai to agla x+1 — inka product quadratic deta hai.
- Ages: present age x, kuch saal baad/pehle ki age (x ± n) — product ya sum conditions se quadratic banta hai.
- Speed-distance-time: time = distance/speed — jab speed change hone se time change ho, to equation me x variable denominator me aata hai jo simplify karne par quadratic ban jaata hai.
3. Solution by Factorisation
Agar quadratic polynomial ax²+bx+c ko do linear factors (px+q)(rx+s) ke product me likha ja sake, to equation ke roots x = -q/p aur x = -s/r hote hain (kyunki product zero tabhi hoga jab koi ek factor zero ho — zero product rule).
Middle term splitting ka tareeka: ax²+bx+c me middle term "bx" ko do parts me todo — aise do numbers dhoondo jinka product = a×c ho aur sum = b ho. Fir un do naye terms ko alag-alag group karke common factor nikalo.
| Step | Kya karna hai |
|---|---|
| 1 | Equation ko ax²+bx+c=0 form me lao (a>0 rakhna easy hota hai) |
| 2 | a×c nikalo, aur do numbers dhoondo jinka product = a×c aur sum = b ho |
| 3 | bx ko un do numbers ke terms me split karo |
| 4 | Pehle do terms aur last do terms me se common factor nikalo (grouping) |
| 5 | Common binomial factor nikal kar (factor1)(factor2)=0 likho |
| 6 | Har factor ko zero rakh kar dono roots nikalo |
↔ Table ko side me swipe karein
Jab dono factors same nikalte hain (jaise (2x-1)²=0), tab equation ke do roots equal hote hain — isko "repeated root" bhi kehte hain.
4. Nature of Roots — Discriminant
Bina equation solve kiye, sirf coefficients a, b, c se hum pata laga sakte hain ki roots kaisi honge. Iske liye discriminant D = b² - 4ac nikalte hain.
| Discriminant | Nature of Roots |
|---|---|
| D > 0 | Do real aur distinct (alag-alag) roots |
| D = 0 | Do real aur equal roots (repeated root) |
| D < 0 | Koi real root nahi (roots exist nahi karte real numbers me) |
↔ Table ko side me swipe karein
Ye concept application problems me bahut kaam aata hai — jaise "kya aisa rectangle possible hai?" type questions me hum equation banate hain aur uska discriminant check karte hain. Agar D<0 aata hai to answer hota hai "nahi, ye situation possible nahi hai" — bina roots nikale hi hum conclude kar sakte hain.
Poore Class 10 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q9)
Q1. Check whether the following are quadratic equations:
(i) (x+1)² = 2(x-3)
(ii) x²-2x = (-2)(3-x)
(iii) (x-2)(x+1) = (x-1)(x+3)
(iv) (x-3)(2x+1) = x(x+5)
(v) (2x-1)(x-3) = (x+5)(x-1)
(vi) x²+3x+1 = (x-2)²
(vii) (x+2)³ = 2x(x²-1)
(viii) x³-4x²-x+1 = (x-2)³
Har expression ko expand karke simplify karte hain:
(i) (x+1)² = 2(x-3) ⇒ x²+2x+1 = 2x-6 ⇒ x²+7 = 0 ⇒ Quadratic hai (x² term bacha)
(ii) x²-2x = -2(3-x) ⇒ x²-2x = -6+2x ⇒ x²-4x+6 = 0 ⇒ Quadratic hai
(iii) (x-2)(x+1)=(x-1)(x+3) ⇒ x²-x-2 = x²+2x-3 ⇒ -3x+1 = 0 ⇒ x² cancel ho gaya ⇒ Quadratic NAHI hai (linear ban gayi)
(iv) (x-3)(2x+1)=x(x+5) ⇒ 2x²-5x-3 = x²+5x ⇒ x²-10x-3 = 0 ⇒ Quadratic hai
(v) (2x-1)(x-3)=(x+5)(x-1) ⇒ 2x²-7x+3 = x²+4x-5 ⇒ x²-11x+8 = 0 ⇒ Quadratic hai
(vi) x²+3x+1=(x-2)² ⇒ x²+3x+1 = x²-4x+4 ⇒ 7x-3 = 0 ⇒ x² cancel ⇒ Quadratic NAHI hai
(vii) (x+2)³=2x(x²-1) ⇒ x³+6x²+12x+8 = 2x³-2x ⇒ -x³+6x²+14x+8 = 0 ⇒ x³ term bacha ⇒ Quadratic NAHI hai (degree 3 hai)
(viii) x³-4x²-x+1=(x-2)³ ⇒ RHS = x³-6x²+12x-8 ⇒ x³-4x²-x+1-x³+6x²-12x+8=0 ⇒ 2x²-13x+9 = 0 ⇒ x³ cancel ho gaya ⇒ Quadratic hai
Q2. Represent the following situations in the form of quadratic equations (aur values bhi nikalo):
(i) Area of a rectangular plot is 528 m². Length is one more than twice its breadth. Find length and breadth.
(ii) Product of two consecutive positive integers is 306. Find the integers.
(iii) Rohan's mother is 26 years older than him. Product of their ages 3 years from now will be 360. Find Rohan's present age.
(iv) A train travels 480 km at a uniform speed. If speed had been 8 km/h less, it would have taken 3 hours more to cover the same distance. Find the speed.
(i) Breadth = x m, to length = (2x+1) m.
Area: x(2x+1) = 528 ⇒ 2x²+x-528 = 0
Middle term split: product = 2×(-528) = -1056, sum = 1 ⇒ numbers 33 aur -32
2x²+33x-32x-528 = 0 ⇒ x(2x+33) - 16(2x+33) = 0 ⇒ (2x+33)(x-16) = 0
x = 16 (x = -33/2 reject, breadth negative nahi ho sakti)
Breadth = 16 m, Length = 2(16)+1 = 33 m (check: 16×33 = 528 ✓)
(ii) Chota integer = x, agla = x+1.
x(x+1) = 306 ⇒ x²+x-306 = 0
product=-306, sum=1 ⇒ numbers 18, -17
x²+18x-17x-306=0 ⇒ x(x+18)-17(x+18)=0 ⇒ (x+18)(x-17)=0
x = 17 (x=-18 reject, positive integers chahiye)
Integers = 17 aur 18
(iii) Rohan ki present age = x, mother ki age = x+26.
3 saal baad: (x+3)(x+26+3) = 360 ⇒ (x+3)(x+29) = 360
x²+32x+87 = 360 ⇒ x²+32x-273 = 0
product=-273, sum=32 ⇒ numbers 39, -7
x²+39x-7x-273=0 ⇒ x(x+39)-7(x+39)=0 ⇒ (x+39)(x-7)=0
x = 7 (x=-39 reject)
Rohan ki present age = 7 saal
(iv) Speed = x km/h. Time liya = 480/x hours. Speed (x-8) par time = 480/(x-8), jo 3 hours zyada hai.
480/(x-8) - 480/x = 3
480x - 480(x-8) = 3x(x-8) ⇒ 480×8 = 3x²-24x ⇒ 3840 = 3x²-24x
Dono side 3 se divide: x²-8x-1280 = 0
product=-1280, sum=-8 ⇒ numbers -40, 32
x²-40x+32x-1280=0 ⇒ x(x-40)+32(x-40)=0 ⇒ (x-40)(x+32)=0
x = 40 (x=-32 reject, speed negative nahi ho sakti)
Train ki speed = 40 km/h
Q3. Find the roots of the following quadratic equations by factorisation:
(i) x²-3x-10=0
(ii) 2x²+x-6=0
(iii) √2 x²+7x+5√2=0
(iv) 2x²-x+1/8=0
(v) 100x²-20x+1=0
(i)
x²-3x-10=0 ⇒ numbers jinka product=-10, sum=-3 ⇒ -5, 2
(x-5)(x+2)=0 ⇒ x = 5, -2
(ii)
2x²+x-6=0 ⇒ product=2×(-6)=-12, sum=1 ⇒ 4, -3
2x²+4x-3x-6=0 ⇒ 2x(x+2)-3(x+2)=0 ⇒ (2x-3)(x+2)=0
x = 3/2, -2
(iii)
√2x²+7x+5√2=0 ⇒ product=√2×5√2=10, sum=7 ⇒ 5, 2
√2x²+5x+2x+5√2=0 ⇒ x(√2x+5) + √2(√2x+5)=0 [kyunki √2(√2x+5)=2x+5√2]
(√2x+5)(x+√2)=0 ⇒ x = -5/√2 = -5√2/2, x = -√2
(iv) Dono side 8 se multiply karo:
2x²-x+1/8=0 ⇒ 16x²-8x+1=0 ⇒ product=16, sum=-8 ⇒ -4, -4
16x²-4x-4x+1=0 ⇒ 4x(4x-1)-1(4x-1)=0 ⇒ (4x-1)²=0
x = 1/4, 1/4 (equal roots)
(v)
100x²-20x+1=0 ⇒ product=100, sum=-20 ⇒ -10, -10
100x²-10x-10x+1=0 ⇒ 10x(10x-1)-1(10x-1)=0 ⇒ (10x-1)²=0
x = 1/10, 1/10 (equal roots)
Q4. (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find how many marbles they had at the beginning.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. Find the number of toys produced that day.
(i) John ke marbles = x, to Jivanti ke = 45-x. 5-5 khone ke baad: John = x-5, Jivanti = 40-x.
(x-5)(40-x) = 124 ⇒ 40x-x²-200+5x = 124 ⇒ -x²+45x-324=0 ⇒ x²-45x+324=0
product=324, sum=-45 ⇒ -36, -9
x²-36x-9x+324=0 ⇒ x(x-36)-9(x-36)=0 ⇒ (x-36)(x-9)=0
x = 36 ya x = 9
John = 36, Jivanti = 9 (ya John = 9, Jivanti = 36) — dono answers valid hain kyunki question me specify nahi kiya kisne kitne rakhe.
(ii) Toys produced = x, cost per toy = (55-x) rupees.
Total cost: x(55-x) = 750 ⇒ 55x-x²=750 ⇒ x²-55x+750=0
product=750, sum=-55 ⇒ -30, -25
x²-30x-25x+750=0 ⇒ x(x-30)-25(x-30)=0 ⇒ (x-30)(x-25)=0
x = 30 ya x = 25 — dono answers valid hain (dono positive aur cost per toy bhi positive rehti hai)
Q5. Find the nature of the roots of the following quadratic equations. If real roots exist, find them:
(i) 2x²-3x+5=0
(ii) 3x²-4√3x+4=0
(iii) 2x²-6x+3=0
(i) a=2, b=-3, c=5.
D = b²-4ac = (-3)²-4(2)(5) = 9-40 = -31
D < 0 ⇒ koi real root nahi hai.
(ii) a=3, b=-4√3, c=4.
D = (-4√3)²-4(3)(4) = 48-48 = 0
D = 0 ⇒ real aur equal roots hain. Chunki D=0, expression ek perfect square hai:
3x²-4√3x+4 = (√3x-2)² [check: (√3x)²=3x², 2×√3x×2=4√3x, 2²=4 ✓]
(√3x-2)²=0 ⇒ x = 2/√3 = 2√3/3 (dono roots)
(iii) a=2, b=-6, c=3.
D = (-6)²-4(2)(3) = 36-24 = 12
D > 0 ⇒ real aur distinct roots hain. Yahan D=12 ek perfect square nahi hai, isliye factorisation se roots nahi nikalte — quadratic formula use karenge:
x = (-b ± √D) / 2a = (6 ± √12) / 4 = (6 ± 2√3) / 4
x = (3 + √3)/2 ya x = (3 - √3)/2
Q6. (i) Find the values of k for each of the following quadratic equations, so that they have two equal roots:
(a) 2x²+kx+3=0
(b) kx(x-2)+6=0
(a) a=2, b=k, c=3. Equal roots ke liye D=0.
D = k²-4(2)(3) = k²-24 = 0 ⇒ k²=24 ⇒ k = ±2√6
(b) kx(x-2)+6=0 ⇒ kx²-2kx+6=0. Quadratic hone ke liye k≠0. a=k, b=-2k, c=6.
D = (-2k)²-4(k)(6) = 4k²-24k = 0 ⇒ 4k(k-6)=0 ⇒ k=0 ya k=6
k=0 reject karenge (tab equation quadratic hi nahi rahegi). k = 6
Q7. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Breadth = x m, length = 2x m.
Area: 2x × x = 800 ⇒ 2x² = 800 ⇒ x² = 400 ⇒ x²-400=0
(x-20)(x+20)=0 ⇒ x = 20 (x=-20 reject, breadth negative nahi ho sakti)
D yahan bhi check karein to positive hi milega (D = 0²-4(2)(-800) > 0, real distinct roots) — is se pehle hi confirm ho jaata hai ki possible hai.
Haan, possible hai. Breadth = 20 m, Length = 40 m.
Q8. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Ek friend ki age = x, dusre ki = 20-x. 4 saal pehle: (x-4) aur (16-x).
(x-4)(16-x) = 48 ⇒ 16x-x²-64+4x = 48 ⇒ -x²+20x-112=0 ⇒ x²-20x+112=0
D = (-20)²-4(1)(112) = 400-448 = -48
D < 0 ⇒ koi real root nahi. Ye situation possible NAHI hai — factorisation se koi real solution nahi milta, discriminant khud negative aakar bata deta hai ki aisi ages exist nahi karti.
Q9. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Perimeter 80 m ⇒ l+b = 40 ⇒ b = 40-l.
Area: l(40-l) = 400 ⇒ 40l-l²=400 ⇒ l²-40l+400=0
D = (-40)²-4(1)(400) = 1600-1600 = 0 ⇒ equal roots — perfect square hai:
l²-40l+400 = (l-20)² = 0 ⇒ l = 20
Haan, possible hai. Length = 20 m, Breadth = 40-20 = 20 m — yani park actually ek square nikalta hai.
Important Equations — Ek Nazar Me
| Concept | Formula / Rule |
|---|---|
| Standard form | ax² + bx + c = 0, jahan a ≠ 0 |
| Discriminant | D = b² - 4ac (nature of roots batata hai) |
| Zero product rule | Agar (px+q)(rx+s) = 0, to px+q=0 ya rx+s=0 |
| Quadratic formula | x = (-b ± √D) / 2a (D≥0 hone par, khaas kar jab factorisation aasaan na ho ya roots irrational hon) |
| Nature of roots (D>0) | Do real aur distinct roots |
| Nature of roots (D=0) | Do real aur equal (repeated) roots |
| Nature of roots (D<0) | Koi real root nahi |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Middle term ko galat split karna. Numbers dhoondte waqt sirf sum b check karke le lete hain, product a×c check karna bhool jaate hain — dono conditions (product AND sum) simultaneously satisfy honi chahiye.
- a=0 wale case ko quadratic maan lena. Agar simplify karne ke baad x² wala term cancel ho jaaye (jaise coefficient of x² zero nikle), to equation quadratic nahi rehti — linear ban jaati hai. Highest degree term ka coefficient zero nahi hona chahiye.
- Word problems me negative/invalid root reject na karna. Age, length, breadth, speed, number of items — ye sab quantities negative nahi ho sakti. Dono roots nikalne ke baad context check karke invalid root discard karna zaroori hai, warna final answer galat aa jaata hai.
- Sign errors word-problem equation set up karte waqt. Jaise '8 km/h kam speed' likhte waqt (x+8) likh dena instead of (x-8), ya '3 hours zyada' ko time subtract kar dena instead of add — is se poori equation hi galat ban jaati hai.
- Discriminant ka sign confuse karna. D>0 ko 'no roots' aur D<0 ko 'real roots' samajh lena — ulta yaad rakhna common mistake hai. Yaad rakho: D negative = roots negative territory me chale gaye (real nahi bache), D positive = do alag real roots.
- b² nikalte waqt sign bhool jaana. Jab b negative ho (jaise -6), to b² hamesha positive hi aata hai — (-6)²=36, na ki -36. Isi tarah -4ac me bhi agar c negative ho to -4ac positive ho jaata hai — dono jagah sign ki galti bahut aam hai.
Board-Style Important Questions
- 1 mark: State whether the equation x² + 5x - 3 = 2x² is a quadratic equation, giving reason.
- 2 marks: Find the roots of the quadratic equation 6x²-x-2=0 by factorisation.
- 2 marks: Find the value of k for which the quadratic equation kx²+2x+1=0 has two equal real roots.
- 3 marks: The product of two consecutive positive integers is 240. Represent this situation as a quadratic equation and find the integers by factorisation.
- 3 marks: Is it possible to design a rectangular garden whose perimeter is 60 m and area is 200 m²? If so, find its length and breadth by factorisation; if not, justify using the discriminant.
Quick Quiz — Score Check Karein
Q1. Class 10 Maths Ch4 (Quadratic Equations): Standard form ax²+bx+c=0 me condition kya honi chahiye?
Q2. (x-2)(x+1) = (x-1)(x+3) ko simplify karne par kya milta hai?
Q3. x²-3x-10=0 ko factorisation se solve karo. Roots kya hain?
Q4. Middle term splitting method me ax²+bx+c ke liye do numbers dhoondte waqt kya condition hoti hai?
Q5. Discriminant D = b²-4ac agar D < 0 ho, to roots ki nature kya hogi?
Q6. 3x²-4√3x+4=0 ka discriminant D nikalo. Roots ki nature kya hai?
Q7. Ek rectangular plot ka area 528 m² hai aur length, breadth ke double se ek zyada hai (length = 2x+1, breadth = x). Breadth kitni hai?
Q8. 2x²+kx+3=0 ke do equal roots ho, iske liye k ki value kya hogi?
Q9. Do dosto ki ages ka sum 20 saal hai. 4 saal pehle unki ages ka product 48 tha. Ye situation possible hai ya nahi?
Q10. Rationalised syllabus ke hisaab se, Class 10 Chapter 4 me roots nikalne ke liye konsa method use hota hai (quadratic formula ka derivation nahi padhaya jaata)?
Aksar Poochhe Jaane Wale Sawaal
Quadratic equation kya hoti hai aur ye linear equation se kaise different hai?
Quadratic equation woh equation hai jisme variable ka highest power exactly 2 hota hai (ax²+bx+c=0, a≠0). Linear equation me highest power 1 hota hai. Quadratic equation ke generally 2 roots hote hain, jabki linear ka sirf 1 root hota hai.
Kya har quadratic equation ko factorisation se solve kiya ja sakta hai?
Jab equation ke roots rational numbers hon (yani discriminant ek perfect square ho), tab factorisation aasaani se ho jaata hai. Agar discriminant perfect square nahi hai (jaise D=12), to roots irrational hote hain aur unhe simple integer-based middle-term splitting se factor karna practical nahi rehta — waha quadratic formula [x = (-b ± √D)/2a] use karke exact irrational roots nikale ja sakte hain.
Discriminant kya batata hai aur kyu important hai?
Discriminant D=b²-4ac bina equation solve kiye ye bata deta hai ki roots real hain ya nahi, aur real hue to distinct hain ya equal. Ye bahut kaam aata hai jab hume sirf ye janna ho ki koi situation (jaise ek rectangle) possible hai ya nahi, bina actual roots nikale.
Word problem me equation set up karne ka best tareeka kya hai?
Sabse pehle unknown quantity ko x maan kar define karo (jaise 'breadth = x m'). Fir dusri quantities ko x ke terms me likho. Aakhir me diya gaya condition (area, product, time difference, etc.) equation ke form me likho. Hamesha units aur real-world constraints (negative na ho) dhyan me rakho.
Agar dono roots equal aa jaayein to iska matlab kya hai?
Iska matlab hai discriminant zero hai, aur quadratic expression ek perfect square hai — jaise (x-3)²=0. Geometrically, ye tab hota hai jab parabola x-axis ko sirf ek point par touch karta hai, cross nahi karta.
Nature of Roots wale questions me negative discriminant ka matlab situation impossible hai?
Application problems (jaise ages ya rectangle design) me agar discriminant negative aa jaaye, to iska matlab hai koi real solution exist nahi karta — matlab woh di gayi situation real world me possible nahi hai. Ye conclude karne ke liye poori equation solve karne ki zaroorat nahi, sirf D ka sign dekhna kaafi hai.
Class 10 Maths — Saare Chapters
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