IB MYP 4-5 Maths- Quadratic equations and its Solving - Study Notes - New Syllabus
IB MYP 4-5 Maths- Quadratic equations and its Solving – Study Notes
Standard
- Quadratic equations and its Solving
IB MYP 4-5 Maths- Quadratic equations and its Solving – Study Notes – All topics
Quadratic Equations
Quadratic Equations
A quadratic equation is an equation of the form:
\( ax^2 + bx + c = 0 \), where \( a \ne 0 \).
Why Study Quadratics?
- They model real-world phenomena: projectile motion, area, optimization, etc.
- Key to understanding polynomial behavior.
- Appear in MYP exams and later IBDP curriculum.
Nature of Roots (Using Discriminant)
The discriminant \( D \) of a quadratic equation is given by:
\( D = b^2 – 4ac \)
| Discriminant \( D \) | Nature of Roots |
|---|---|
| \( D > 0 \) | Two distinct real roots |
| \( D = 0 \) | One repeated real root |
| \( D < 0 \) | No real roots (complex roots) |
Solving by Factorisation
Used when the quadratic can be written as a product of two brackets:
\( ax^2 + bx + c = (dx + e)(fx + g) = 0 \)
Set each bracket = 0 and solve.
Example:
Solve: \( x^2 + 5x + 6 = 0 \)
▶️ Answer/Explanation
Factor: \( (x + 2)(x + 3) = 0 \)
So, \( x = -2 \) or \( x = -3 \)
Example:
Solve: \( 2x^2 – 5x – 3 = 0 \)
▶️ Answer/Explanation
Split middle term: \( 2x^2 – 6x + x – 3 = 0 \)
Group: \( 2x(x – 3) + 1(x – 3) = (2x + 1)(x – 3) = 0 \)
So, \( x = -\frac{1}{2} \) or \( x = 3 \)
Solving by Quadratic Formula
Use when factorisation is difficult or impossible.
\( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
Example:
Solve: \( x^2 – 4x – 1 = 0 \)
▶️ Answer/Explanation
Use formula: \( a = 1,\ b = -4,\ c = -1 \)
\( x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(1)(-1)}}{2(1)} = \frac{4 \pm \sqrt{20}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5} \)
Example:
Solve: \( 3x^2 + 2x + 5 = 0 \)
▶️ Answer/Explanation
Discriminant \( D = 2^2 – 4(3)(5) = 4 – 60 = -56 \)
\( D < 0 \), so no real solutions (complex roots)
Solving by Completing the Square
Used to convert a quadratic into vertex form: \( y = a(x – h)^2 + k \)
Steps:
- Make coefficient of \( x^2 \) equal to 1 (if needed)
- Write as: \( x^2 + bx = (x + \frac{b}{2})^2 – (\frac{b}{2})^2 \)
- Add/subtract the constant term to complete the square
Convert into the form: \( (x + p)^2 = q \)
Example:
Solve: \( x^2 + 6x + 5 = 0 \)
▶️ Answer/Explanation
\( x^2 + 6x = -5 \)
Half of 6 is 3. Add and subtract 9 on LHS:
\( x^2 + 6x + 9 = 4 \Rightarrow (x + 3)^2 = 4 \Rightarrow x + 3 = \pm 2 \Rightarrow x = -1,\ -5 \)
Example:
Solve: \( x^2 + 4x + 8 = 0 \)
▶️ Answer/Explanation
\( x^2 + 4x = -8 \)
Half of 4 is 2 → add and subtract 4:
\( (x + 2)^2 = -4 \) → no real solutions
Solving by Graphing
Plot the function \( y = ax^2 + bx + c \). The x-intercepts (where \( y = 0 \)) give the solutions.
Tips:
- Use a table of values to sketch the parabola.
- If the parabola touches x-axis once → 1 root
- If it crosses → 2 roots; if it doesn’t → no real roots
Example:
Solve graphically: \( y = x^2 – 3x – 4 \)
▶️ Answer/Explanation
Step 1: Create a table of values
| \( x \) | \( y = x^2 – 3x – 4 \) |
|---|---|
| -2 | 6 |
| -1 | 0 |
| 0 | -4 |
| 1 | -6 |
| 2 | -6 |
| 3 | -4 |
| 4 | 0 |
| 5 | 6 |
Step 2: Plot the points on a graph
Join the points with a smooth curve to form a parabola.
Step 3: Interpret the graph
The parabola cuts the x-axis at \( x = -1 \) and \( x = 4 \).
Final Answer: The solutions are \( \boxed{x = -1 \text{ and } x = 4} \)
Example:
Solve graphically: \( y = x^2 + 2x + 5 \)
▶️ Answer/Explanation
Step 1: Create a table of values
| \( x \) | \( y = x^2 + 2x + 5 \) |
|---|---|
| -4 | 13 |
| -3 | 8 |
| -2 | 5 |
| -1 | 4 |
| 0 | 5 |
| 1 | 8 |
Step 2: Plot the points and sketch the curve
The curve forms a parabola that opens upward.
Step 3: Check for x-axis intersections
The graph does not intersect the x-axis, so there are no real roots.
Final Answer: \( \boxed{\text{No real solutions}} \)
Graph of a Quadratic Equation
The graph of a quadratic equation is a parabola.
- If \( a > 0 \), the parabola opens upward (smiley)
- If \( a < 0 \), it opens downward (frowny)
Key Features of a Parabola:
- Vertex: The turning point (minimum or maximum value)
- Axis of Symmetry: Vertical line through the vertex: \( x = -\frac{b}{2a} \)
- x-intercepts (roots): Solutions to \( ax^2 + bx + c = 0 \)
- y-intercept: Value of \( y \) when \( x = 0 \) → \( y = c \)
Standard form: \( y = ax^2 + bx + c \)
Example:
Find the vertex and x-intercepts of: \( y = x^2 – 6x + 5 \)
▶️ Answer/Explanation
Vertex: Use \( x = -\frac{b}{2a} = \frac{6}{2} = 3 \)
\( y = 3^2 – 6(3) + 5 = 9 – 18 + 5 = -4 \)
Vertex: \( (3, -4) \)
x-intercepts: Solve \( x^2 – 6x + 5 = 0 \) → \( (x – 1)(x – 5) = 0 \)
Roots: \( x = 1,\ 5 \)
