Quadratic Equations

Equations & Inequalities: A Comprehensive Guide

This topic is a cornerstone of O Level Mathematics, providing the essential tools for algebraic problem-solving. Mastering these techniques is crucial for tackling more advanced concepts and for applying mathematics to real-world scenarios.

1. Linear Equations

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable raised to the first power. The goal is to find the value of the unknown variable that makes the equation true.

Core Principle: The Balancing Act

Think of an equation as a balanced scale. To keep it balanced, whatever operation you perform on one side, you must perform the exact same operation on the other side.

Step-by-Step Solution:

1. Simplify: Expand any brackets and combine like terms on each side of the equation.
2. Isolate the Variable Term: Use addition or subtraction to move all terms containing the variable to one side and all constant terms to the other.
3. Solve for the Variable: Use multiplication or division to solve for the variable.

Example: Solve for `y` in `5(y - 2) = 2y + 8`

• Step 1 (Simplify): `5y - 10 = 2y + 8`
• Step 2 (Isolate): Subtract `2y` from both sides: `3y - 10 = 8`. Add `10` to both sides: `3y = 18`.
• Step 3 (Solve): Divide by 3: `y = 6`.

Common Exam Trap: When moving a term across the equals sign, students often forget to change its sign (e.g., moving `+2y` becomes `-2y` on the other side). Always use inverse operations.

2. Simultaneous Linear Equations

These are systems of two linear equations with two unknown variables. The solution is the pair of values `(x, y)` that satisfies both equations at the same time.

Method 1: Elimination

This method is best when the coefficients of one variable are the same or opposites.

1. Align: Write the equations with like terms aligned vertically.
2. Match Coefficients: Multiply one or both equations by a constant so that the coefficients of one variable are equal (or additive inverses).
3. Eliminate: Add or subtract the equations to eliminate one variable.
4. Solve: Solve the resulting single-variable equation.
5. Substitute Back: Substitute the value found back into one of the original equations to find the second variable.

Example:

Eq 1: `3x + 2y = 12`

Eq 2: `4x - y = 5`

• Multiply Eq 2 by 2: `8x - 2y = 10`
• Add this to Eq 1: `(3x + 8x) + (2y - 2y) = 12 + 10` → `11x = 22` → `x = 2`.
• Substitute `x=2` into Eq 2: `4(2) - y = 5` → `8 - y = 5` → `y = 3`.
• Solution: `(2, 3)`

Method 2: Substitution

This method is ideal when one variable can be easily expressed in terms of the other.

1. Isolate: In one equation, make one variable the subject (e.g., `y = ...`).
2. Substitute: Substitute this entire expression into the *other* equation.
3. Solve: Solve the resulting single-variable linear equation.
4. Substitute Back: Substitute your answer back into the rearranged equation from Step 1 to find the second variable.

Pakistan Application: A street vendor in Lahore sells plates of Chana Chaat and Dahi Bhallay. One customer pays Rs. 550 for 3 plates of chaat and 2 of dahi bhallay. Another pays Rs. 450 for 2 plates of chaat and 2 of dahi bhallay. Simultaneous equations can determine the price of a single plate of each item.

3. Quadratic Equations (ax² + bx + c = 0)

An equation where the highest power of the variable is 2. They can have two, one, or zero real solutions (called roots).

Method 1: Factorising

Use this for simpler quadratics.

1. Ensure the equation is in the form `ax² + bx + c = 0`.
2. Find two numbers that multiply to give `ac` and add to give `b`.
3. Rewrite the middle term using these two numbers.
4. Factor by grouping, then set each factor to zero to find the roots.

Example: `x² - 3x - 10 = 0`

• We need two numbers that multiply to -10 and add to -3. These are -5 and +2.
• `(x - 5)(x + 2) = 0`
• So, `x - 5 = 0` or `x + 2 = 0`.
• Roots: `x = 5` or `x = -2`.

Method 2: The Quadratic Formula

This formula solves *any* quadratic equation.

`x = [-b ± √(b²-4ac)] / 2a`

1. Identify `a`, `b`, and `c` from the standard form.
2. Substitute these values carefully into the formula.
3. Calculate the two possible values for `x`.

The Discriminant (b² - 4ac): This part of the formula tells you about the roots without fully solving:

• If `b² - 4ac > 0`: Two distinct real roots.
• If `b² - 4ac = 0`: One repeated real root.
• If `b² - 4ac < 0`: No real roots.

Method 3: Completing the Square

This method is useful for finding the vertex of a parabola and for solving.

1. Move the constant term `c` to the right side.
2. If `a ≠ 1`, divide the entire equation by `a`.
3. Take half of the coefficient of `x`, square it, and add it to both sides.
4. The left side is now a perfect square. Solve by taking the square root of both sides.

Application: The path of a cricket ball hit by a batsman can be modelled by a quadratic equation. We can use these methods to find how long the ball is in the air or the maximum height it reaches.

4. Linear Inequalities

Inequalities use symbols like `<`, `>`, `≤`, `≥` to show a range of possible values.

Solving Process:

Solve them just like linear equations, with one critical exception.

The Golden Rule: If you **multiply or divide both sides of an inequality by a negative number**, you **must flip the direction of the inequality sign**.

Example: Solve `-4x + 7 > 23`

• Subtract 7 from both sides: `-4x > 16`
• Divide by -4 and flip the sign: `x < -4`

Representing on a Number Line:

• For `<` or `>`, use an open circle (o) to show the endpoint is not included.
• For `≤` or `≥`, use a closed circle (•) to show the endpoint is included.

Pakistan Context: If you have a budget of Rs. 8,000 for Eid shopping at Dolmen Mall in Karachi and have already bought a kurta for Rs. 4,500, the inequality `x + 4500 ≤ 8000` helps you determine the maximum amount `x` you can spend on shoes, which is `x ≤ 3500`.