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School SEE Compulsory Mathematics

Quadratic Equation

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Quadratic equation → ax² + bx + c = 0 Solve by: factorisation (easy numbers) completing square (x – b/2)² formula x = [-b ± √(b² – 4ac)] / 2a Word problems → make equation from condition (area, age, sum-product, digits, triangle). Discriminant D = b² – 4ac decides roots: D>0 → 2 real, D=0 → equal, D<0 → no real roots. Memorize the 3 methods + formula → 100 % chapter done!

1. Definition

A quadratic equation is an equation of the form:
ax² + bx + c = 0 (a ≠ 0)

It always has two roots (solutions).

2. Methods to Solve Quadratic Equations

A. Factorisation Method (Most used in exams)

Steps:

Split the middle term.

Group terms and factor each group.

Solve each factor = 0.

Example:
x² – 7x + 12 = 0

Split middle term: x² – 4x – 3x + 12 = 0

Factor: x(x – 4) – 3(x – 4) = 0

(x – 4)(x – 3) = 0 → x = 4, 3

B. Completing the Square Method

Steps:

Make coefficient of x² = 1

Move constant term to RHS

Add (b/2)² to both sides

Write LHS as perfect square and solve

Example:
x² – 10x + 21 = 0

x² – 10x = –21

x² – 10x + 25 = –21 + 25 → (x – 5)² = 4

x – 5 = ±2 → x = 7, 3

C. Quadratic Formula Method (Works for all equations)

Formula:
x = [–b ± √(b² – 4ac)] / 2a

Discriminant: D = b² – 4ac

3. Nature of Roots (Very Important)

Discriminant (D)Nature of RootsD > 0Two different real rootsD = 0Two equal real rootsD < 0No real roots (imaginary)

4. Sum and Product of Roots (Without solving)

For ax² + bx + c = 0:

Sum of roots: α + β = –b / a

Product of roots: α × β = c / a

5. Word Problem Strategies

Type of ProblemHow to Form EquationExampleRectangleLet breadth = x, length = x + k → x(x + k) = areaArea = 96, length = breadth + 4 → x(x+4) = 96 → x² + 4x – 96 = 0Numbers & SquareLet number = x → x² ± x = givenNumber + reciprocal = 5 → x + 1/x = 5 → x² – 5x + 1 = 0Sum & ProductTwo numbers x, y → x + y = S, xy = P → x² – Sx + P = 0Sum of ages = 35, product = 306 → x² – 35x + 306 = 0Age ProblemsAges x and y, after/before k years → (x ± k)(y ± k) = product–Two-digit NumbersNumber = 10x + y, use conditions to form equationProduct of digits = 12, number + 27 = reverse → number = 36Right TriangleSides x, y, hypotenuse h → x² + y² = h²Hypotenuse = 25 m, difference of sides = 7 m → (x+7)² + x² = 625Perimeter/AreaPerimeter P, area A → l + b = P/2, l × b = A–Picnic/MoneyTotal students n, absent k → total/(n–k) = amount + extra–

6. Most Important Solved Examples

1. Rectangle / Area Problems

Example 1: Area = 120 m², length = breadth + 6

x(x + 6) = 120
or x² + 6x = 120
or x² + 6x – 120 = 0
or (x + 12)(x – 10) = 0
or x = 10 (ignore negative)
or Length = x + 6 = 16 m

Example 2: Area = 168 m², length = 2 × breadth – 4

x(2x – 4) = 168
or 2x² – 4x = 168
or 2x² – 4x – 168 = 0
or x² – 2x – 84 = 0
or (x – 12)(x + 7) = 0
or x = 12 (ignore negative)
or Length = 2x – 4 = 20 m

2. Sum and Product of Numbers

Example 3: Two numbers sum = 13, product = 40

x² – 13x + 40 = 0
or (x – 8)(x – 5) = 0
or x = 8, 5 (Numbers = 8 and 5)

Example 4: Two numbers sum = 12, difference of squares = 20

Let numbers x and y = 12 – x
x² – (12 – x)² = 20
or x² – (144 – 24x + x²) = 20
or x² – 144 + 24x – x² = 20
or 24x – 144 = 20
or 24x = 164
or x = 6.833 (≈7), y = 12 – 7 = 5

3. Age Problems

Example 5: Sum of ages = 27, product = 180

x² – 27x + 180 = 0
or (x – 12)(x – 15) = 0
or x = 12, 15 (Ages = 12 and 15)

Example 6: Age difference = 4 years, product = 60

(x + 4) × x = 60
or x² + 4x = 60
or x² + 4x – 60 = 0
or (x – 6)(x + 10) = 0
or x = 6 (ignore negative)
or Older age = x + 4 = 10

4. Number + Reciprocal / Square Problems

Example 7: Number + reciprocal = 7

x + 1/x = 7
or Multiply both sides by x → x² + 1 = 7x
or x² – 7x + 1 = 0
or x = [7 ± √(49 – 4)] / 2
or x = [7 ± √45] / 2

Example 8: Square of number + 3 × number = 70

x² + 3x = 70
or x² + 3x – 70 = 0
or (x + 10)(x – 7) = 0
or x = 7 (ignore negative)

5. Right Triangle / Pythagoras

Example 9: Hypotenuse = 13, difference of sides = 5

(x + 5)² + x² = 169
or x² + 10x + 25 + x² = 169
or 2x² + 10x – 144 = 0
or x² + 5x – 72 = 0
or (x – 8)(x + 9) = 0
or x = 8 (ignore negative)
or Other side = x + 5 = 13

Example 10: One side = x, other = x + 3, hypotenuse = 15

x² + (x + 3)² = 225
or x² + x² + 6x + 9 = 225
or 2x² + 6x – 216 = 0
or x² + 3x – 108 = 0
or (x – 9)(x + 12) = 0
or x = 9 (ignore negative)
or Other side = x + 3 = 12

6. Two-digit Number Problems

Example 11: Product of digits = 24, number – sum of digits = 54

Let number = 10x + y
10x + y – (x + y) = 54
or 9x = 54
or x = 6
or y = Product / x = 24 / 6 = 4
or Number = 10x + y = 64

Example 12: Number + reverse = 99, product of digits = 20

x + y = 9 (from sum of digits)
or xy = 20
or x² – 9x + 20 = 0
or (x – 5)(x – 4) = 0
or x = 5, y = 4
or Number = 54

7. Consecutive Numbers / Squares

Example 13: Sum of squares = 365

x² + (x + 1)² = 365
or x² + x² + 2x + 1 = 365
or 2x² + 2x – 364 = 0
or x² + x – 182 = 0
or (x – 13)(x + 14) = 0
or x = 13 (ignore negative)
or Numbers = 13, 14

Example 14: Product of two consecutive even numbers = 168

x(x + 2) = 168
or x² + 2x – 168 = 0
or (x – 12)(x + 14) = 0
or x = 12 (ignore negative)
or Numbers = 12, 14

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