1. Definition & Standard Form:
A quadratic equation in variable x is an equation of the form ax² + bx + c = 0, where a, b, c ∈ ℜ and a ≠ 0.

2. Derivation of Quadratic Formula (Sridharacharya's Formula):
Given ax² + bx + c = 0 (a ≠ 0):
• Divide throughout by a: x² + (b/a)x + c/a = 0
• Transpose constant term: x² + (b/a)x = −c/a
• Add (b / 2a)² to both sides to complete square:
x² + 2(b/2a)x + (b/2a)² = (b/2a)² − c/a
(x + b/2a)² = (b² − 4ac) / 4a²
• Take square root on both sides:
x + b/2a = ± √(b² − 4ac) / 2a
x = (−b ± √(b² − 4ac)) / 2a ■

3. Discriminant (Δ) & Nature of Roots:
The expression Δ = b² − 4ac determines the nature of roots:

Discriminant Value (Δ)	Nature of Roots	Roots Formula
Δ > 0	Two distinct real roots	x = (−b ± √Δ) / 2a
Δ = 0	Two equal real roots	x = −b / 2a
Δ < 0	No real roots (Imaginary)	No real solution in ℜ

Step 1 — Formulate Variables:
Let speed of stream = x km/h (x > 0).
Speed upstream = (18 − x) km/h
Speed downstream = (18 + x) km/h

Step 2 — Time Equations:
Time upstream t ‹up› = ²⁴⁄_{18 − x} hours
Time downstream t ‹down› = ²⁴⁄_{18 + x} hours

Given: t ‹up› − t ‹down› = 1 hour
²⁴⁄_{18 − x} − ²⁴⁄_{18 + x} = 1

Step 3 — Simplify to Quadratic Equation:
24 [ (18 + x) − (18 − x) ] = (18 − x)(18 + x)
24 [ 2x ] = 324 − x²
48x = 324 − x² ⇒ x² + 48x − 324 = 0

Step 4 — Factorise:
(x + 54)(x − 6) = 0
x = −54 (rejected as speed cannot be negative)
x = 6

Speed of stream = 6 km/h ■

Step 1 — Identify Coefficients:
a = (k − 12), b = 2(k − 12), c = 2
Condition for quadratic: a ≠ 0 ⇒ k ≠ 12.

Step 2 — Equal Roots Condition (Δ = 0):
Δ = b² − 4ac = 0
[2(k − 12)]² − 4(k − 12)(2) = 0
4(k − 12)² − 8(k − 12) = 0

Step 3 — Factorise:
4(k − 12) [ (k − 12) − 2 ] = 0
4(k − 12)(k − 14) = 0

Either k = 12 or k = 14.
Since k = 12 makes a = 0 (not quadratic), k = 12 is rejected.

Required value of k = 14 ■

Step 1 — Define Variables:
Let smaller tap fill tank in x hours.
Larger tap fills tank in (x − 10) hours (x > 10).

Step 2 — Work Done per Hour:
Smaller tap work/hr = ¹⁄_x
Larger tap work/hr = ¹⁄_{x − 10}
Combined fill time = 9⅜ = ⁷⁵⁄₈ hours ⇒ Combined work/hr = ⁸⁄₇₅

Step 3 — Form Equation:
¹⁄_x + ¹⁄_{x − 10} = ⁸⁄₇₅
^{(x − 10) + x}⁄_{x(x − 10)} = ⁸⁄₇₅
75(2x − 10) = 8x(x − 10)
150x − 750 = 8x² − 80x
8x² − 230x + 750 = 0 ⇒ 4x² − 115x + 375 = 0

Step 4 — Solve Quadratic:
4x² − 100x − 15x + 375 = 0
4x(x − 25) − 15(x − 25) = 0
(4x − 15)(x − 25) = 0
x = 25  or  x = ¹⁵⁄₄ = 3.75.
If x = 3.75, larger tap time = 3.75 − 10 = −6.25 (impossible).

Smaller tap = 25 hours, Larger tap = 15 hours ■