source: ncert
topic: algebra
class_level: class_10
chapter: ch4
difficulty: intermediate
Class 10 | Ch4: Quadratic Equations STANDARD FORM: ax^2+bx+c=0 where a≠0. METHODS OF SOLVING: Method 1 - Factorisation: Split middle term: find p,q such that p+q=b and p×q=a×c. 2x^2-7x+3=0. a×c=6. p+q=-7, p×q=6. p=-6,q=-1. 2x^2-6x-x+3=0 → 2x(x-3)-1(x-3)=0 → (2x-1)(x-3)=0. x=1/2 or x=3. Method 2 - Completing the Square: ax^2+bx+c=0 → x^2+(b/a)x=-c/a → add (b/2a)^2 both sides. (x+b/2a)^2=(b^2-4ac)/4a^2 → x=-b/2a ± sqrt(b^2-4ac)/2a. Method 3 - Quadratic Formula: x=(-b ± sqrt(b^2-4ac))/2a. (Derived from completing the square.) DISCRIMINANT D=b^2-4ac: D>0: two distinct real roots. D=0: two equal real roots (repeated). D<0: no real roots (complex). NATURE OF ROOTS: If D is perfect square: roots are rational. If D>0 but not perfect square: roots irrational. RELATIONSHIP BETWEEN ROOTS: Sum of roots=alpha+beta=-b/a. Product of roots=alpha×beta=c/a. WORD PROBLEMS - STANDARD TYPES: Number problems, age problems, speed-distance problems, area problems. SOLVED EXAMPLES: Example 1: 2x^2-7x+3=0 by factorisation. a×c=6. Factors of 6 summing to -7: -6 and -1. 2x^2-6x-x+3=2x(x-3)-1(x-3)=(2x-1)(x-3)=0. x=1/2 or x=3. Example 2: Find nature of roots of 2x^2-6x+3=0. D=36-24=12>0. Two distinct real roots (D not perfect square, so irrational). Example 3: Sum of ages of father and son is 45. Five years ago product was 124. Let son's age=x. Father's age=45-x. (x-5)(45-x-5)=124 → (x-5)(40-x)=124. 40x-x^2-200+5x=124 → x^2-45x+324=0 → (x-9)(x-36)=0. x=9 (son) or x=36 (check: son can't be 36 if sum is 45). Son=9, Father=36. COMMON MISTAKES: Discriminant: D=b^2-4ac NOT b^2+4ac. If D=0: ONE repeated root (two equal roots, not no roots). Quadratic formula: entire (-b±sqrt(D)) is over 2a.