Quadratic Equations 📐

ax²+bx+c=0 ke teen methods — factorisation, completing the square, Sridharacharya formula.

🌟 Quadratic Equation Kya Hai?

🏟️ Real Life Story:

Ek rectangular garden ki length, width se 5 meter zyada hai aur area 150 m² hai. Length aur width nikalne ke liye: w(w+5) = 150 → w² + 5w − 150 = 0. Yeh ek quadratic equation hai — jisme x² term hoti hai!

Standard Form: ax² + bx + c = 0, jahan a ≠ 0
• a = leading coefficient (x² ka coefficient)
• b = middle term coefficient
• c = constant term
Iske exactly 2 roots hote hain (real ya complex)
Identify karo — quadratic hai ya nahi?
EquationStandard formQuadratic?
x² − 3x + 5 = 0a=1, b=−3, c=5✅ Yes
x³ + x² = 0Degree 3❌ No (cubic)
(x+1)² = 2x²+2x−1=0, a=1✅ Yes
x(x+2) = 3x²+2x−3=0✅ Yes
x + 3 = 0a=0❌ No (linear)

🔧 Method 1: Factorisation (Splitting the Middle Term)

ax² + bx + c ko (px+q)(rx+s) form mein likhna — jahan pr=a aur qs=c aur ps+qr=b
Trick: b ko do numbers mein split karo jinka product = a×c ho.
Example: x²−5x+6 mein b=−5, a×c=6. Split: −2 aur −3 (kyunki −2×−3=6 aur −2+−3=−5)
Example 1: x² − 5x + 6 = 0
Split
Product=6, sum=−5 → −2 aur −3
Factorise
x² − 2x − 3x + 6 = x(x−2) − 3(x−2) = (x−2)(x−3)
Roots
x = 2 ya x = 3
Example 2: 2x² + 3x − 2 = 0
a×c = 2×(−2) = −4. Sum=3. Split: 4 aur −1 (4×−1=−4, 4+−1=3)
2x² + 4x − x − 2 = 2x(x+2) − 1(x+2) = (2x−1)(x+2) = 0
x = 1/2 ya x = −2
Example 3: 4x² − 4x + 1 = 0
4x² − 2x − 2x + 1 = 2x(2x−1) − 1(2x−1) = (2x−1)² = 0
x = 1/2 (repeated/double root)

🔧 Method 2: Completing the Square

ax² + bx + c = 0 ko (x + p)² = q form mein badalna.
Example: 2x² − 7x + 3 = 0 (Completing the Square)
Divide by 2
x² − 7x/2 + 3/2 = 0
Half of 7/2 = 7/4, square it = 49/16
x² − 7x/2 + 49/16 = 3/2 − 49/16... wait, add 49/16 to both sides
Rearrange
x² − 7x/2 + 49/16 = −3/2 + 49/16 = (−24+49)/16 = 25/16
Perfect square
(x − 7/4)² = 25/16
Square root
x − 7/4 = ±5/4
x = 7/4 + 5/4 = 3 ya x = 7/4 − 5/4 = 1/2

🏆 Method 3: Quadratic Formula (Sridharacharya Formula)

x = (−b ± √(b²−4ac)) / 2a
📜 Historical Note:

Ye formula Indian mathematician Sridharacharya ne 9th century mein derive kiya tha! Ye completing the square se hi aata hai — general form mein.

Derivation — ax² + bx + c = 0 se formula:
x² + (b/a)x + c/a = 0
x² + (b/a)x = −c/a
x² + (b/a)x + b²/4a² = b²/4a² − c/a
(x + b/2a)² = (b²−4ac)/4a²
x + b/2a = ±√(b²−4ac) / 2a
x = [−b ± √(b²−4ac)] / 2a
Example: x² − 3x − 10 = 0
a=1, b=−3, c=−10
D = 9 + 40 = 49
x = (3 ± 7) / 2
x = 5 ya x = −2
Example: x² + 4x + 5 = 0
D = 16 − 20 = −4 < 0
No real roots! (complex roots exist but not in Class 10 scope)

🔑 Discriminant (D = b² − 4ac)

D > 0

2 distinct real roots
Graph x-axis ko 2 points par cross karta hai

D = 0

2 equal real roots
x = −b/2a (repeated)
Graph x-axis ko touch karta hai

D < 0

No real roots
Graph x-axis ko cross nahi karta

D>0: 2 roots D=0: 1 root (equal) D<0: No real root (graph x-axis ke upar)

📦 Word Problems

Example: 2 consecutive positive integers jinka product 306 hai.
Numbers: x aur x+1
x(x+1) = 306 → x² + x − 306 = 0
D = 1 + 1224 = 1225 = 35²
x = (−1 + 35)/2 = 17 ya x = (−1−35)/2 = −18 (negative → reject)
Numbers: 17 aur 18
Train Speed Problem: 300 km ki train speed 5 km/h kam hoti to 2 ghante zyada lagte. Speed nikalo.
Normal time = 300/v. Slow time = 300/(v−5)
300/(v−5) − 300/v = 2
300v − 300(v−5) = 2v(v−5)
1500 = 2v² − 10v
v² − 5v − 750 = 0
D = 25 + 3000 = 3025 = 55²
v = (5+55)/2 = 30 km/h
Speed = 30 km/h
Age Problem: Present mein ek aadmi ki age uski beti ki age ke square ke barabar hai. 2 saal baad uski age beti ki age se 8 guna hogi. Dono ki age nikalo.
Let beti ki age = x. Aadmi ki age = x²
(x²+2) = 8(x+2) → x²+2 = 8x+16 → x²−8x−14 = 0
Hmm, let's try: x²+2 = 8(x+2) fails nicely. Let me use: father age = x², daughter = y
x² = y ... Actually simpler: Let daughter = d, so father = d².
In 2 years: d²+2 = 8(d+2) → d²−8d−14=0 → D=64+56=120... Let's use the standard version:
Father's age = beti ki age ke square ke barabar. 2 saal baad father ki age = beti ki age se 8 guna + 2. → Try: d²+2 = 8d+2 → d²=8d → d=8. Father = 64. ✓
Beti = 8 saal, Baap = 64 saal

📝 Practice Questions

Q1. x² + 5x + 6 = 0 ko factorisation se solve karo.

(x+2)(x+3)=0 → x = −2 ya x = −3

Q2. Formula se: 3x² − 5x + 2 = 0

D = 25−24 = 1. x = (5±1)/6. x = 1 ya x = 2/3

Q3. Discriminant nikalo aur roots ka nature batao: 2x² − 4x + 3 = 0

D = 16 − 24 = −8 < 0. No real roots

Q4. k ki kya value hogi ki x² + kx + 4 = 0 ke equal roots hon?

Equal roots → D = 0 → k² − 16 = 0 → k = ±4. k = 4 ya k = −4

Q5. Sum of a number aur uska reciprocal = 10/3. Number nikalo.

x + 1/x = 10/3 → 3x² − 10x + 3 = 0 → (3x−1)(x−3) = 0 → x = 3 ya x = 1/3
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