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Quadratic Equations

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Quadratic Equations
Quadratic Equations - Factorization & Completing the Square

QUADRATIC EQUATIONS

Introduction

A quadratic equation is a polynomial equation in which the highest power (degree) of the unknown variable is 2. It always contains an equal sign (=) and can be written in the general form:

ax2 + bx + c = 0

where:

  • a, b, c are constants (real numbers)
  • a ≠ 0 (if a = 0, the equation becomes linear)
  • x is the unknown variable

Examples of Quadratic Equations

EquationReason it is Quadratic
2x2 – 5x – 3 = 0Highest power of x is 2
n2 + 50 = 27nCan be rearranged to n2 – 27n + 50 = 0
(4a – 9)(2a + 1) = 0When expanded gives 8a2 – 14a – 9 = 0
k2 = 49Can be rewritten as k2 – 49 = 0
💡 Class Discussion: Why is (4a – 9)(2a + 1) = 0 a quadratic equation?

Answer: When we expand the brackets:
(4a – 9)(2a + 1) = 8a2 + 4a – 18a – 9 = 8a2 – 14a – 9 = 0
The highest power of a is 2, confirming it is a quadratic equation.

Main Objective

The main goal of this topic is to learn methods of solving quadratic equations — finding the value(s) of the unknown that make the equation true. These values are called the roots or solutions of the equation. A quadratic equation generally has two solutions (which may sometimes be equal).

METHOD 1: SOLVING BY FACTORIZATION

The Zero Product Principle

This method is based on a simple but powerful idea:

If the product of two numbers equals zero, then at least one of the numbers must be zero.

Illustrations:

  • 3 × 0 = 0 ✓
  • 0 × 5 = 0 ✓
  • 0 × 0 = 0 ✓
  • 7 × 4 = 28 (NOT zero, because neither factor is zero)

In general: If a × b = 0, then either a = 0, or b = 0, or both.

Example 1: Already Factorized Equations

Solve: (x – 2)(x + 7) = 0

Solution: Since the product is zero, one of the factors must be zero:

x – 2 = 0    or    x + 7 = 0
x = 2    or    x = –7

∴ The roots are x = 2 or x = –7

Example 2: Equation with Three Factors

Solve: d(d – 4)(d + 6) = 0

Solution: Any of the three factors may equal zero:

d = 0    or    d – 4 = 0    or    d + 6 = 0
d = 0,   4,   or   –6

CLASS EVALUATION

Solve the following equations:

  1. 3d2(d – 7) = 0
  2. (6 – n)(4 + n) = 0
  3. a(2 – a)2(1 + a) = 0

Solving Quadratic Equations Using Factorization (Step-by-Step)

  1. Rearrange the equation in the form ax2 + bx + c = 0
  2. Factorize the left-hand side (LHS)
  3. Equate each factor to zero
  4. Solve for the unknown
  5. Check by substituting the values back

Example 1

Solve: 4y2 + 5y – 21 = 0

Solution — Factorize the LHS:

4y2 + 5y – 21 = 0
4y2 + 12y – 7y – 21 = 0
4y(y + 3) – 7(y + 3) = 0
(y + 3)(4y – 7) = 0

Set each factor to zero:

y + 3 = 0    or    4y – 7 = 0
y = –3    or    y = 7/4 = 1¾

Check by substitution:

When y = –3: 4(–3)2 + 5(–3) – 21 = 36 – 15 – 21 = 0 ✓

When y = 7/4: 4(7/4)2 + 5(7/4) – 21 = 49/4 + 35/4 – 21 = 21 – 21 = 0 ✓

Example 2: Difference of Two Squares

Solve: m2 = 16

Solution: Rearrange: m2 – 16 = 0

Factorize (difference of two squares):

(m – 4)(m + 4) = 0
m – 4 = 0    or    m + 4 = 0
m = 4    or    m = –4

∴ m = ±4

CLASS EVALUATION

Solve the following quadratic equations:

  1. h2 – 15h + 54 = 0
  2. 12y2 + y – 35 = 0
  3. 4a2 – 15a = 4
  4. v2 + 2v – 35 = 0

METHOD 2: COMPLETING THE SQUARE

Why Use Completing the Square?

Not all quadratic equations can be easily factorized. The completing the square method works for every quadratic equation.

Key Idea

This method transforms a quadratic expression into a perfect square trinomial plus a constant.

(x + k)2 = x2 + 2kx + k2

So, to make x2 + bx into a perfect square, we add (b/2)2:

x2 + bx + (b/2)2 = (x + b/2)2

Illustration

ExpressionAdd (b/2)2Perfect Square
x2 + 6x+ 9(x + 3)2
x2 – 10x+ 25(x – 5)2
x2 + 5x+ 25/4(x + 5/2)2

Steps for Completing the Square

  1. Arrange the equation in the form ax2 + bx + c = 0
  2. Divide every term by a (so coefficient of x2 becomes 1)
  3. Move the constant term to the right-hand side (RHS)
  4. Add (half the coefficient of x)2 to both sides
  5. Express the LHS as a perfect square
  6. Take the square root of both sides
  7. Solve for the unknown

Example 1

Solve: x2 + 6x – 7 = 0 by completing the square.

Step 1: Move the constant to the RHS:

x2 + 6x = 7

Step 2: Half of coefficient of x = 6/2 = 3; square it: 32 = 9

Step 3: Add 9 to both sides:

x2 + 6x + 9 = 7 + 9
(x + 3)2 = 16

Step 4: Take the square root of both sides:

x + 3 = ±√16 = ±4

Step 5: Solve for x:

x = –3 + 4 = 1    or    x = –3 – 4 = –7

∴ x = 1 or x = –7

Example 2

Solve: 2x2 – 8x + 5 = 0 by completing the square.

Step 1: Divide through by 2:

x2 – 4x + 5/2 = 0

Step 2: Move constant to RHS:

x2 – 4x = –5/2

Step 3: Half of –4 = –2; (–2)2 = 4. Add 4 to both sides:

x2 – 4x + 4 = –5/2 + 4
(x – 2)2 = 3/2

Step 4: Take the square root:

x – 2 = ±√(3/2) ≈ ±1.225

Step 5: Solve:

x ≈ 3.23   or   x ≈ 0.78 (to 2 d.p.)

Example 3

Solve: x2 + 5x + 2 = 0 by completing the square.

Move constant:

x2 + 5x = –2

Half of 5 = 5/2; (5/2)2 = 25/4. Add to both sides:

x2 + 5x + 25/4 = –2 + 25/4
(x + 5/2)2 = 17/4

Take square root:

x + 5/2 = ±√17 / 2
x = (–5 ± √17) / 2

Since √17 ≈ 4.123:

x ≈ –0.44   or   x ≈ –4.56

Geometric Illustration of Completing the Square

To visualize x2 + 6x, imagine a square with side x and a rectangle of length 3 on two sides. Adding a small square of area 9 completes the larger square:

x 3 ┌────────┬──────────┐ x │ x² │ 3x │ │ │ │ ├────────┼──────────┤ 3 │ 3x │ ★ 9 ★ │ ← Add this square (9) to complete! └────────┴──────────┘ The complete square has side (x + 3) and area (x + 3)²

GENERAL EVALUATION

Solve the following equations:

  1. y2(3 + y) = 0
  2. x2(x + 5)(x – 5) = 0
  3. (v – 7)(v – 5)(v – 3) = 0
  4. 9f2 + 12f + 4 = 0
  5. x2 + 8x + 3 = 0 (use completing the square)
  6. 2x2 – 5x – 3 = 0 (use completing the square)

WEEKEND ASSIGNMENT

Section A: Objective Questions

Solve and check by substitution:

  1. (4b – 12)(b – 5) = 0    A. ½, 4   B. 3, 5   C. 4, 6   D. 5, 3
  2. (11 – 4x)2 = 0    A. 11, 3   B. 23/4, 3   C. 11/4 twice   D. 24/3 twice
  3. (d – 5)(3d – 2) = 0    A. 5, 2/3   B. 4, 5   C. 5, 9   D. 2, 5
  4. u2 – 8u – 9 = 0    A. –1, 9   B. 1, 9   C. 1, 8   D. 9, –1
  5. c2 = 25    A. 5   B. –5   C. +5   D. ±5

Section B: Theory

Solve the following equations:

  1. 2x2 = 3x + 5
  2. a2 – 3a = 0
  3. p2 + 7p + 12 = 0
  4. x2 – 6x + 4 = 0 (use completing the square)
  5. 3x2 + 7x – 6 = 0 (use both methods and compare)

SUMMARY

MethodBest Used WhenAdvantage
FactorizationExpression factorizes easily into integersQuick and direct
Completing the SquareEquation does not factorize easilyWorks for all quadratics; reveals vertex form

✅ Always remember: A quadratic equation has at most two solutions (roots).

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