Quadratic Equations — Class 10 IMO

Standard Form and Factorisation

What is the standard form of a quadratic equation? A quadratic equation is an equation of the form \(ax^{2} + bx + c\) = 0, where a, b, and c are real numbers, and a ≠ 0. The highest power of the variable x is 2 (hence the name "quadratic", from "quadratus" meaning square).

Why must a ≠ 0? If a = 0, the equation becomes bx + c = 0, which is linear, not quadratic.

Examples of quadratic equations:

\(x^{2} + 5x + 6\) = 0 (a=1, b=5, c=6)
\(2x^{2} - 4x\) = 0 (a=2, b=−4, c=0)
\(x^{2} - 9\) = 0 (a=1, b=0, c=−9)

Not quadratic:

\(x^{3} + 2x + 1\) = 0 (degree 3 — cubic)
2x + 5 = 0 (degree 1 — linear)

Real-life analogy: Think of a quadratic equation as a "square-shaped" problem. If you throw a ball upward, its height over time follows a quadratic pattern — that's why quadratics are used to model projectiles, bridges, and even profit maximization!

┌─────────────────────────────────────────────────────────────┐
│         STANDARD FORM OF QUADRATIC EQUATION                  │
└─────────────────────────────────────────────────────────────┘

GENERAL STRUCTURE:

        ax²  +  bx  +  c  =  0
         │       │       │
         │       │       └── Constant term
         │       └── Coefficient of x (linear term)
         └── Coefficient of x² (quadratic term)

CONDITION: a ≠ 0 (otherwise it becomes linear)


IDENTIFYING a, b, c FROM DIFFERENT FORMS:

┌─────────────────────────────┬──────┬──────┬──────┐
│        EQUATION             │  a   │  b   │  c   │
├─────────────────────────────┼──────┼──────┼──────┤
│   3x² + 5x - 2 = 0          │  3   │  5   │  -2  │
│   x² - 7x = 0               │  1   │  -7  │  0   │
│   4x² - 9 = 0               │  4   │  0   │  -9  │
│   -2x² + x + 3 = 0          │  -2  │  1   │  3   │
│   x² = 4                    │  1   │  0   │  -4  │
└─────────────────────────────┴──────┴──────┴──────┘


CONVERTING TO STANDARD FORM:

     Equation: (x + 3)(x - 2) = 0
           │
           ▼ Expand
     x² - 2x + 3x - 6 = 0
           │
           ▼ Combine like terms
     x² + x - 6 = 0  ✓ (a=1, b=1, c=-6)

     
     Equation: 2x² + 5 = 3x
           │
           ▼ Bring all terms to LHS
     2x² + 5 - 3x = 0
           │
           ▼ Rearrange in descending order
     2x² - 3x + 5 = 0  ✓ (a=2, b=-3, c=5)


WHY QUADRATIC? REAL-LIFE SHAPE:

    Path of a thrown ball:
    
        height
          ↑
        10├─────╱╲─────
          │    ╱  ╲
        5├───╱    ╲───
          │  ╱      ╲
        0└─╱────────╲──→ time
          ╱          ╲
    
    This parabolic shape is described by a quadratic equation!

Example 1: Identify a, b, and c for the quadratic equation \(3x^{2} - 2x + 7\) = 0.

Step 1: Compare with standard form \(ax^{2} + bx + c\) = 0
Step 2: Coefficient of \(x^{2}\) is 3 → a = 3
Step 3: Coefficient of x is −2 → b = −2
Step 4: Constant term is 7 → c = 7

Answer: a = 3, b = −2, c = 7

Example 2: Write the equation (x + 2)\(^{2}\) = 9 in standard quadratic form and identify a, b, c.

Step 1: Expand LHS: (x + 2)\(^{2}\) = \(x^{2} + 4x + 4
Step\) 2: Equation becomes: \(x^{2} + 4x + 4\) = 9
Step 3: Bring all terms to LHS: \(x^{2} + 4x + 4 - 9\) = 0
Step 4: Simplify: \(x^{2} + 4x - 5\) = 0
Step 5: Compare: a = 1, b = 4, c = −5

Answer: \(x^{2} + 4x - 5\) = 0; a=1, b=4, c=−5

Example 3: Determine the value of k for which the equation (k+1)\(x^{2} + 2\)(k+1)x + 2 = 0 becomes a linear equation.

Step 1: For the equation to be linear, the coefficient of \(x^{2}\) must be zero
Step 2: Set a = 0: k + 1 = 0
Step 3: Solve: k = −1
Step 4: Check: Substitute k = −1: (0)\(x^{2} + 2\)(0)x + 2 = 0 → 2 = 0 (false, so no solution actually)
Step 5: Wait — if k = −1, the equation becomes 2 = 0 which has no solution. So no value makes it a valid linear equation. The question likely expects k = −1 makes it not quadratic, but it becomes inconsistent.

Answer: k = −1 (equation becomes inconsistent, not truly linear)

Quick recap

Standard form: \(ax^{2} + bx + c\) = 0, where a ≠ 0
a, b, c are real numbers (coefficients)
a is the coefficient of \(x^{2}\), b of x, c is the constant
If a = 0, the equation is linear, not quadratic
Any quadratic equation can be rewritten in standard form
Degree of a quadratic equation is always 2

✓ Quick check

If the discriminant equals 0, the roots are:

D = 0 gives two equal real roots.

The product of the roots of 3x² − 6x + 2 = 0 is:

Product = c/a = 2/3.

Formula and Completing the Square

What are the methods to solve quadratic equations? Two important algebraic methods are:

1. Factorization Method: If \(ax^{2} + bx + c\) can be factored into (px + q)(rx + s) = 0, then either px + q = 0 or rx + s = 0. This gives two solutions.

Steps for factorization:

Write equation in standard form
Find two numbers whose product = a × c and sum = b
Split the middle term using these numbers
Factor by grouping
Set each factor = 0 and solve

2. Completing the Square Method: This method works for ALL quadratic equations. We convert \(ax^{2} + bx + c\) = 0 into the form (x + p)\(^{2}\) = q, then take square roots.

Steps for completing the square:

Make coefficient of \(x^{2}\) = 1 (divide by a if needed)
Move constant term to RHS
Add (b/2)\(^{2}\) to both sides
Write LHS as (x + b/2)\(^{2}
Take\) square root of both sides
Solve for x

┌─────────────────────────────────────────────────────────────┐
│     FACTORIZATION vs COMPLETING THE SQUARE - FLOWCHART       │
└─────────────────────────────────────────────────────────────┘

METHOD 1: FACTORIZATION

    x² + 7x + 12 = 0
           │
           ▼ Find two numbers with product=12, sum=7
       3 and 4 satisfy!
           │
           ▼ Split middle term
    x² + 3x + 4x + 12 = 0
           │
           ▼ Group terms
    x(x + 3) + 4(x + 3) = 0
           │
           ▼ Factor common binomial
    (x + 3)(x + 4) = 0
           │
           ▼ Set each factor = 0
    x + 3 = 0  or  x + 4 = 0
           │
           ▼ Solve
    x = -3  or  x = -4


METHOD 2: COMPLETING THE SQUARE

    x² + 6x + 5 = 0
           │
           ▼ Move constant to RHS
    x² + 6x = -5
           │
           ▼ Add (b/2)² = (6/2)² = 9 to both sides
    x² + 6x + 9 = -5 + 9
           │
           ▼ LHS is perfect square: (x + 3)²
    (x + 3)² = 4
           │
           ▼ Take square root
    x + 3 = ± 2
           │
           ▼ Solve
    x = -3 ± 2
    x = -1  or  x = -5


VISUAL REPRESENTATION OF COMPLETING THE SQUARE:

    x² + 6x = -5
    
    x² + 6x  →  Imagine a square of side x and a rectangle 6×x
    
        ┌─────┬────┬────┬────┬────┬────┐
        │     │    │    │    │    │    │
        │     │    │    │    │    │    │
        │  x  │  6x│    │    │    │    │
        │     │    │    │    │    │    │
        └─────┴────┴────┴────┴────┴────┘
    
    Adding 9 (which is 3²) completes a square of side (x+3):
    
        ┌─────────┬───┐
        │         │   │
        │         │ 3 │
        │  (x+3)  │   │
        │         ├───┤
        │         │   │
        └─────────┴───┘

Example 1: Solve \(x^{2} - 5x + 6\) = 0 by factorization.

Step 1: Find two numbers with product = 6 and sum = −5 → (−2) and (−3)
Step 2: Split middle term: \(x^{2} - 2x - 3x + 6\) = 0
Step 3: Group: x(x − 2) − 3(x − 2) = 0
Step 4: Factor: (x − 2)(x − 3) = 0
Step 5: Set each factor = 0: x − 2 = 0 or x − 3 = 0

Answer: x = 2, x = 3

Example 2: Solve \(x^{2} - 4x - 12\) = 0 by completing the square.

Step 1: Move constant to RHS: \(x^{2} - 4x\) = 12
Step 2: b = −4, so (b/2)\(^{2}\) = (−4/2)\(^{2}\) = (−2)\(^{2}\) = 4
Step 3: Add 4 to both sides: \(x^{2} - 4x + 4\) = 12 + 4
Step 4: LHS = (x − 2)\(^{2}\), RHS = 16
Step 5: (x − 2)\(^{2}\) = 16 → x − 2 = ±4
Step 6: x = 2 + 4 = 6 or x = 2 − 4 = −2

Answer: x = 6, x = −2

Example 3: Solve \(2x^{2} - 7x + 3\) = 0 by completing the square.

Step 1: Divide by 2 to make coefficient of \(x^{2}\) = 1: \(x^{2}\) − (7/2)x + 3/2 = 0
Step 2: Move constant: \(x^{2}\) − (7/2)x = −3/2
Step 3: b = −7/2, (b/2)\(^{2}\) = (−7/4)\(^{2}\) = 49/16
Step 4: Add 49/16 to both sides: \(x^{2}\) − (7/2)x + 49/16 = −3/2 + 49/16
Step 5: RHS = −24/16 + 49/16 = 25/16
Step 6: LHS = (x − 7/4)\(^{2}\) = 25/16
Step 7: x − 7/4 = ±5/4 → x = 7/4 ± 5/4
Step 8: x = 12/4 = 3 or x = 2/4 = 1/2

Answer: x = 3, x = 1/2

Quick recap

Factorization works only when the quadratic can be factored easily
Splitting the middle term: find p and q such that p+q = b and p×q = a×c
Completing the square works for all quadratic equations
Completing the square formula: \(x^{2} + bx\) → (x + b/2)\(^{2}\) − (b/2)\(^{2}\)
Always check your solutions by substituting back into the original equation
Both methods give the same two roots

✓ Quick check

Completing the square, x² + 6x can be written as:

x² + 6x = (x + 3)² − 9.

An equation ax² + bx + c = 0 is quadratic only if:

The coefficient of x² must be non-zero, i.e. a ≠ 0.

Discriminant and Nature of Roots

What is the discriminant? For a quadratic equation \(ax^{2} + bx + c\) = 0, the discriminant (denoted by D or Δ) is: D = \(b^{2} - 4ac\)

The discriminant tells us the nature of the roots without actually solving the equation!

Nature of roots based on discriminant:

Discriminant (D)	Nature of Roots	Roots are
D > 0	Real and distinct	Two different real numbers
D = 0	Real and equal (repeated)	One real number (double root)
D < 0	No real roots (imaginary)	Two complex conjugate roots

Quadratic formula: When D ≥ 0, the roots are given by: x = [−b ± \(\sqrt{D}\)] / (2a)

Real-life meaning:

D > 0 → The parabola (graph of quadratic) cuts the x-axis at two different points
D = 0 → The parabola touches the x-axis at exactly one point (vertex on axis)
D < 0 → The parabola never touches the x-axis (lies entirely above or below)

┌─────────────────────────────────────────────────────────────┐
│         DISCRIMINANT AND NATURE OF ROOTS - GRAPH MAP         │
└─────────────────────────────────────────────────────────────┘

CASE 1: D > 0 (Two distinct real roots)

        y
        │   \     /
        │    \   /
        │     \ /
        │      X
        │     / \
        │    /   \
        │   /     \
        └───┼───┼───► x
            r1  r2
    
    Example: x² - 5x + 6 = 0 → D = 25-24=1>0, roots: 2,3


CASE 2: D = 0 (One repeated real root)

        y
        │     ┌┐
        │    ╱  ╲
        │   ╱    ╲
        │  ╱      ╲
        │ ╱        ╲
        │╱          ╲
        └─────┼──────► x
              r
    
    Example: x² - 4x + 4 = 0 → D = 16-16=0, root: 2 (double)


CASE 3: D < 0 (No real roots)

        y
        │     ┌┐
        │    ╱  ╲
        │   ╱    ╲
        │  ╱      ╲
        │ ╱        ╲
        │╱          ╲
        └──────────────► x
        (never touches x-axis)
    
    Example: x² + x + 1 = 0 → D = 1-4 = -3 < 0, no real roots


DISCRIMINANT FLOWCHART:

        Start: ax² + bx + c = 0
                │
                ▼
        Calculate D = b² - 4ac
                │
        ┌───────┼───────┐
        ▼       ▼       ▼
       D>0     D=0     D<0
        │       │       │
        ▼       ▼       ▼
    Two real  One real  No real
    distinct  root      roots
      roots    (double)  
        │       │
        └───┬───┘
            ▼
    Use quadratic formula: x = (-b ± √D)/(2a)

Example 1: Find the discriminant of \(x^{2} - 7x + 10\) = 0 and state the nature of roots.

Step 1: Compare with \(ax^{2} + bx + c\) = 0 → a = 1, b = −7, c = 10
Step 2: D = \(b^{2} - 4ac\) = (−7)\(^{2} - 4\)(1)(10) = 49 − 40 = 9
Step 3: Since D = 9 > 0, roots are real and distinct

Answer: D = 9; roots are real and distinct

Example 2: For what value of k does \(4x^{2} - 12x + k\) = 0 have equal roots?

Step 1: For equal roots, discriminant D = 0
Step 2: a = 4, b = −12, c = k
Step 3: D = (−12)\(^{2} - 4\)(4)(k) = 144 − 16k
Step 4: Set D = 0: 144 − 16k = 0 → 16k = 144 → k = 9

Answer: k = 9

Example 3: Find the roots of \(2x^{2} - 4x + 1\) = 0 using the quadratic formula.

Step 1: a = 2, b = −4, c = 1
Step 2: D = \(b^{2} - 4ac\) = (−4)\(^{2} - 4\)(2)(1) = 16 − 8 = 8
Step 3: Since D > 0, roots are real and distinct
Step 4: \(\sqrt{D}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\)
Step 5: x = [−b ± \(\sqrt{D}\)]/(2a) = [4 ± 2\(\sqrt{2}\)]/(4)
Step 6: Simplify: x = [2(2 ± \(\sqrt{2}\))]/4 = (2 ± \(\sqrt{2}\))/2

Answer: x = (2 + \(\sqrt{2}\))/2 and x = (2 − \(\sqrt{2}\))/2

Quick recap

Discriminant D = \(b^{2} - 4ac\) determines the nature of roots
D > 0 → two distinct real roots
D = 0 → one real repeated root (double root)
D < 0 → no real roots (complex roots)
Quadratic formula: x = [−b ± \(\sqrt{\(b^{2} - 4ac\)}\)]/(2a)
The quadratic formula works for any quadratic equation

✓ Quick check

The sum of the roots of 3x² − 6x + 2 = 0 is:

Sum = −b/a = −(−6)/3 = 2.

For x² + kx + 9 = 0 to have equal roots, k equals:

Equal roots need k² − 36 = 0, so k = ±6.