Quadratics, Coordinate Geometry, and Straight Line Graphs Study Guide

Factorising and Solving Quadratics

• Key Forms of Quadratic Expressions:

  • Standard monic quadratic: $x^2 + bx + c$ (e.g., $x^2 - 5x + 6$)
  • Non-monic quadratic: $ax^2 + bx + c$ where $a \neq 1$ (e.g., $2x^2 + 7x + 3$)
  • Difference of two squares: $a^2 - b^2 = (a - b)(a + b)$

• Core Concepts to Remember:

  • Multiply
  • Add
  • Brackets
  • Zero
  • Always check for common factors first before applying other factorisation methods.
  • For expressions of the form $ax^2 + bx + c$, identify two numbers that multiply to give $ac$ and add to give $b$

• Steps to Solve Quadratic Equations:

  1. Factorise the quadratic expression.
  2. Set each resulting factor equal to $0$
  3. Solve each linear equation for $x$
  4. Check the calculated answers (optional step for verification).
  5. Write down the final solution set.

• Other Useful Algebraic Forms:

  • $x^2 - a^2 = (x - a)(x + a)$
  • $ax^2 + bx + c = a(x - r)(x - s)$ (if factorisable)

• Worked Examples:

  • Example 1: Solve $x^2 - 5x + 6 = 0$
    • Factorise into brackets: $(x - 2)(x - 3) = 0$
    • Set each factor to zero: $x - 2 = 0$ or $x - 3 = 0$
    • Solutions: $x = 2$ or $x = 3$
  • Example 2: Solve $2x^2 - 7x + 3 = 0$
    • Factorise into brackets: $(2x - 1)(x - 3) = 0$
    • Set each factor to zero: $2x - 1 = 0$ or $x - 3 = 0$
    • Solutions: $x = \frac{1}{2}$ or $x = 3$

Forming and Solving Quadratics using the Formula or Otherwise

• Standard Form of a Quadratic Equation:

  • $ax^2 + bx + c = 0$ where $a \neq 0$

• The Quadratic Formula:

  • $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• The Discriminant ($b^2 - 4ac$):

  • If $b^2 - 4ac > 0$: there are $2$ distinct real solutions.
  • If $b^2 - 4ac = 0$: there is $1$ repeated real solution.
  • If $b^2 - 4ac < 0$: there are no real solutions.

• Procedural Steps for Using the Formula:

  1. Identify the values of $a$, $b$, and $c$ from the standard form equation.
  2. Substitute $a$, $b$, and $c$ into the quadratic formula.
  3. Simplify the expression under the square root and calculate using both the positive and negative signs ($\pm$).
  4. Write down the final numerical answers.

• Worked Example 1 (Solving via Formula):

  • Equation: $2x^2 - 4x - 6 = 0$
  • Identify coefficients: $a = 2$, $b = -4$, $c = -6$
  • Substitute into formula: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)}$
  • Simplify inside root: $x = \frac{4 \pm \sqrt{16 + 48}}{4}$
  • Evaluate root: $x = \frac{4 \pm \sqrt{64}}{4} = \frac{4 \pm 8}{4}$
  • Calculate final roots: $x = 3$ or $x = -1$

• Worked Example 2 (Forming a Quadratic Word Problem):

  • Problem statement: A number and its square add to $20$
  • Define the variable: Let $x = \text{number}$
  • Form the quadratic equation: $x^2 + x - 20 = 0$
  • Factorise: $(x + 5)(x - 4) = 0$
  • Solve for $x$: $x = 5$ or $x = -4$

• Practical Tips:

  • Always double-check the signs of $a$, $b$, and $c$
  • Use a calculator when handling complex or fractional numerical calculations.
  • If factorising cannot be performed easily, fall back on using the quadratic formula.

Completing the Square

• Fundamental Idea:

  • Transform any quadratic expression into a perfect square form.
  • For $ax^2 + bx + c$, take half of the coefficient of $x$ ($b$-value), square it, and then both add and subtract this value to preserve algebraic equality.

• General Steps:

  1. Take half of the coefficient of $x$
  2. Square this halved value.
  3. Add and subtract this squared number within the expression.
  4. Factorise the perfect square trinomial component into $(x + p)^2$

• Applications and Utility:

  • Locating the turning point (vertex) of a parabolic graph.
  • Finding the absolute minimum or maximum value of a quadratic function.
  • Solving quadratic equations algebraically without using the quadratic formula.

• Worked Example 1 (Completing the Square):

  • Expression: $x^2 + 6x + 5$
  • Half of $6$ is $3$; $3^2 = 9$. Add and subtract $9$: $x^2 + 6x + 9 - 9 + 5$
  • Form perfect square: $(x + 3)^2 - 4$
  • Vertex coordinates: $(-3, -4)$

• Worked Example 2 (Solving by Completing the Square):

  • Equation: $x^2 - 4x - 5 = 0$
  • Complete square: $(x - 2)^2 - 9 = 0$
  • Rearrange: $(x - 2)^2 = 9$
  • Take square root of both sides: $x - 2 = \pm 3$
  • Solve: $x = 5$ or $x = -1$

• Geometric Representation:

  • The vertex of the parabola is denoted as $(h, k)$.

Simultaneous Equations (Quadratic and Linear)

• Overview:

  • Involves a system of two equations where one equation is quadratic and the other equation is usually linear.

• Steps to Solve:

  1. Rearrange the linear equation to make one variable the subject (e.g., $y$).
  2. Substitute this expression into the quadratic equation.
  3. Solve the resulting single-variable quadratic equation.
  4. Substitute the calculated $x$-values back into the linear equation to determine the corresponding $y$-values.
  • Note: Systems can yield $0$, $1$, or $2$ solutions (representing the points of intersection).

• Worked Example:

  • Given system:
    • $y = x + 1$
    • $y = x^2 - 3$
  • Equate expressions for $y$: $x + 1 = x^2 - 3$
  • Rearrange into standard form: $x^2 - x - 4 = 0$
  • Factorise: $(x - 2)(x + 2) = 0$
  • Solve for $x$: $x = 2$ or $x = -2$
  • Find $y$ when $x = 2$: $y = 2 + 1 = 3$
  • Find $y$ when $x = -2$: $y = -2 + 1 = -1$
  • Final intersection coordinates: $(2, 3)$ and $(-2, -1)$

• Visual Interpretation:

  • The straight line and parabola intersect at exactly $2$ distinct points: $(2, 3)$ and $(-2, -1)$.

• Key Reminders:

  • Substitute algebraic terms carefully.
  • Always check both sets of solutions in both original equations.
  • The points of intersection on a graph represent the algebraic solutions to the simultaneous system.

Distance Between Two Points and Mid-Points

• Distance Formula:

  • $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Midpoint Formula:

  • $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$

• Worked Example (Distance):

  • Given points $A(1, 2)$ and $B(4, 6)$:
  • Substitute coordinates: $d = \sqrt{(4 - 1)^2 + (6 - 2)^2}$
  • Simplify: $d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

• Worked Example (Midpoint):

  • Given points $A(1, 2)$ and $B(4, 6)$:
  • Substitute coordinates: $M = \left(\frac{1 + 4}{2}, \frac{2 + 6}{2}\right)$
  • Simplify fractions: $M = \left(\frac{5}{2}, \frac{8}{2}\right) = (2.5, 4)$

• Key Concepts:

  • Distance: Derived directly from Pythagoras' Theorem ($a^2 + b^2 = c^2$).
  • Midpoint: Calculated as the average of the $x$-coordinates and the average of the $y$-coordinates.
  • Direction Independence: The order of points does not affect the calculation (calculating from $A$ to $B$ gives the same result as $B$ to $A$).

Straight Line Graphs

• Slope-Intercept Form:

  • $y = mx + c$
  • $m = \text{gradient (slope)}$
  • $c = y\text{-intercept}$

• Gradient Formula:

  • $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$

• Key Geometric Properties:

  • Parallel lines: Possess equal gradients ($m_1 = m_2$).
  • Perpendicular lines: Gradients multiply to $-1$ ($m_1 \times m_2 = -1$).
  • $x$\text{-intercept}: Found by setting $y = 0$
  • $y$\text{-intercept}: Found by setting $x = 0$

• Worked Example:

  • Find the equation of a line passing through point $(2, 3)$ with a gradient of $4$:
  • Use point-slope form: $y - 3 = 4(x - 2)$
  • Expand and solve for $y$: $y - 3 = 4x - 8 \implies y = 4x - 5$
  • The $y$\text{-intercept} is $-5$ (or coordinate $(0, -5)$).

Regions on Graphs

• Inequality Representation Rules:

  • Strict inequalities ($> , <$) are drawn using a dashed line.
  • Non-strict inequalities ($\ge , \le$) are drawn using a solid line.
  • Shade the region containing points that satisfy the inequality.

• Procedural Steps:

  1. Draw the boundary line (solid or dashed depending on inequality operator).
  2. Test a sample point not on the line (e.g., $(0, 0)$).
  3. Shade the region that satisfies the inequality condition.
  4. Confirm whether the boundary line is included or excluded from the solution set.

• Worked Example:

  • Inequality: Shade the region $y > 2x + 1$
  • Boundary line equation: $y = 2x + 1$ (drawn as a dashed line because of strict inequality $>$).
  • Test point $(0, 0)$:
    • Substitute into inequality: $0 > 2(0) + 1 \implies 0 > 1$ (False).
  • Conclusion: The origin $(0, 0)$ is not included in the solution set.
  • Action: Shade the correct side of the boundary line above/left of $(0, 0)$.