Straight Lines and Quadratic Equations

1. Straight Line Graphs

Represented on a 2D Cartesian coordinate system $(x, y)$.

The Equation of a Line

General form: $y = mx + c$

• $m$: Gradient (slope).
• $c$: $y$-intercept (where the line crosses the $y$-axis).

Gradient ($m$)

Gradient measures the steepness: $\text{gradient} = \frac{\text{vertical change}}{\text{horizontal change}} = \frac{\text{rise}}{\text{run}}$.

Calculating Gradient from Two Points (Extended): Given $(x_1, y_1)$ and $(x_2, y_2)$: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Parallel and Perpendicular Lines (Extended)

• Parallel Lines: Have the same gradient ($m_1 = m_2$).
• Perpendicular Lines: Gradients are negative reciprocals ($m_1 \cdot m_2 = -1$ or $m_2 = -\frac{1}{m_1}$).

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2. Quadratic Expressions

Expressions of the form $ax^2 + bx + c$.

Expansion

Using the distributive law (FOIL: First, Outer, Inner, Last).

• $(x + a)(x + b) = x^2 + (a+b)x + ab$.
• Difference of Two Squares: $(a - b)(a + b) = a^2 - b^2$.

Factorisation

The reverse of expansion.

• Common Factors: $3x^2 + 6x = 3x(x + 2)$.
• Simple Trinomials: $x^2 + 5x + 6 = (x + 2)(x + 3)$.
• Harder Trinomials ($ax^2 + bx + c$): Use “splitting the middle term” or trial and error.
• Difference of Two Squares: $x^2 - 16 = (x - 4)(x + 4)$.

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3. Quadratic Equations

Solving for $x$ when $ax^2 + bx + c = 0$.

Methods of Solving

1. Factorisation

   • Goal: Find the roots by breaking the quadratic expression into two linear factors.
   • Use Case: Best used when the expression is easily factorisable (e.g., simple integers for $a, b, c$).
   • Step-by-Step:
     • Step 1: Rearrange the equation to set it to zero: $ax^2 + bx + c = 0$.
     • Step 2: Factorise the quadratic expression into two linear brackets.
     • Step 3: Use the Zero Product Property: If $A \cdot B = 0$, then either $A = 0$ or $B = 0$.
     • Step 4: Solve the resulting linear equations for $x$.
       
       

2. Completing the Square

   • Goal: Transform the quadratic into a “perfect square” form to easily isolate the variable $x$ or find the vertex of the parabola. $$(x + p)^2 + q$$

     $p$ represents half the coefficient of $x$ (after $a=1$).
     In vertex form $(x + p)^2 + q$, the x-coordinate of the vertex is $-p$.

     $q$ represents the constant remaining after the square is completed.
     In vertex form $(x + p)^2 + q$, the y-coordinate of the vertex is $q$.

• Use Case: Used when factorisation is difficult, or when the vertex (turning point) of the parabola is required.
• Step-by-Step:

  • Step 1: Identify the coefficient of $x^2$ (called $a$). Ensure $a = 1$. If it isn’t, divide the entire equation by $a$.

  • Step 2: Move the constant term $c$ to the right side of the equation.

  • Step 3: Take half of the coefficient of $x$, square it, and add this value to both sides of the equation.

  • Step 4: Factorise the left side into a perfect square of the form $(x + p)^2$.

  • Step 5: Solve for $x$ by taking the square root of both sides and isolating $x$.

  • Final Form: The result should be the value(s) of $x$ (e.g., $x = 2, x = -4$). If finding the vertex, the form $(x + p)^2 + q$ is used.

    
    
    

3. Quadratic Formula
   • Goal: Directly calculate the roots using the coefficients of the equation.
   • Use Case: A universal method that works for any quadratic, especially when roots are irrational or cannot be easily factorised.
   • Step-by-Step:
     • Step 1: Write the equation in standard form: $ax^2 + bx + c = 0$.
     • Step 2: Identify the coefficients $a$, $b$, and $c$.
     • Step 3: Substitute these values into the formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
     • Step 4: Simplify the discriminant ($\sqrt{b^2 - 4ac}$) and calculate the two possible values for $x$.

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4. Graphical Solutions

The solution to a quadratic equation $ax^2 + bx + c = 0$ corresponds to the x-intercepts (roots) of the graph $y = ax^2 + bx + c$.

• Two Roots: Graph crosses $x$-axis twice ($b^2 - 4ac > 0$).
• One Root: Graph touches $x$-axis once (vertex is on axis, $b^2 - 4ac = 0$).
• No Real Roots: Graph does not touch $x$-axis ($b^2 - 4ac < 0$).