Quadratic Equations Cambridge (CIE) IGCSE Maths Extended Study Guide

Solving Quadratics by Factorising

• General Form of a Quadratic Equation: A quadratic equation must be rearranged into the form $ax^2 + bx + c = 0$.     - It is essential that zero is on one side of the equation.     - Factorisation is generally easier if the equation is rearranged so that the coefficient $a$ is positive.

• The Factorisation Process: To solve using factorisation, you must factorise the quadratic expression and then solve by setting each resulting bracket equal to zero.     - Logic: If two things multiply together to give zero, then at least one of them must be equal to zero. For example, if $(x + 4)(x - 1) = 0$, then either $x + 4 = 0$ or $x - 1 = 0$.

• Case Study: Solving $(x - 3)(x + 7) = 0$:     - Solve the first bracket: $x - 3 = 0 \rightarrow x = 3$.     - Solve the second bracket: $x + 7 = 0 \rightarrow x = -7$.     - The two solutions are $x = 3$ or $x = -7$.     - Note: In simple cases, the solutions are the numbers in the brackets with opposite signs.

• Solving with Coefficients in Front of $x$ within Brackets:     - The process remains identical, but more work is required to isolate $x$. You cannot simply change the signs.     - Example: Solve $(2x - 3)(3x + 5) = 0$:         - First bracket: $2x - 3 = 0 \rightarrow 2x = 3 \rightarrow x = \frac{3}{2}$.         - Second bracket: $3x + 5 = 0 \rightarrow 3x = -5 \rightarrow x = -\frac{5}{3}$.         - The solutions are $x = \frac{3}{2}$ or $x = -\frac{5}{3}$.

• Handling $x$ as a Factor:     - If an expression is factorised as $x(x - 4) = 0$, the standard process applies.     - It may help to think of the single $x$ as $(x - 0)$ or $(x)$.     - Solve the first "bracket": $(x) = 0$, so $x = 0$.     - Solve the second bracket: $x - 4 = 0 \rightarrow x = 4$.     - The solutions are $x = 0$ or $x = 4$.     - Common Mistake: Dividing both sides by $x$ at the beginning will result in losing the $x = 0$ solution. Never cancel $x$ from both sides.

• Examiner Tips and Tricks: If your calculator can solve quadratics, use the solutions to work backward to find factors. For instance, if a calculator gives solutions for $6x^2 + x - 2 = 0$ as $x = -\frac{2}{3}$ and $x = \frac{1}{2}$, rearrange these to $3x + 2 = 0$ and $2x - 1 = 0$. This identifies the factors as $(3x + 2)(2x - 1)$.

• Worked Examples:     - (a) Solve $x^2 + 3x - 10 = 0$: Factorises to $(x - 2)(x + 5) = 0$. Solutions: $x = 2$ or $x = -5$.     - (b) Solve $5x^2 - x = 0$: Factorises to $x(5x - 1) = 0$. Solutions: $x = 0$ or $x = \frac{1}{5}$.

The Quadratic Formula

• Formula Definition: For a quadratic in the form $ax^2 + bx + c = 0$ (where $a \neq 0$), the quadratic formula is:     - $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

• Application Steps:     1. Ensure the equation equals zero on the right-hand side.     2. Identify the values for $a$, $b$, and $c$.     3. Substitute values into the formula using brackets around any negative numbers.     4. Calculate the result twice: once using the $+$ sign and once using the $-$ sign.

• Worked Example: Solve $2x^2 - 8x - 3 = 0$:     - Identify: $a = 2, b = -8, c = -3$.     - Substitute: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 2 \times (-3)}}{2 \times 2}$.     - Solutions: $x = 4.3452078...$ or $x = -0.34520787...$.     - Correct to 3 d.p.: $x = 4.345$ or $x = -0.345$.     - Correct to 3 s.f.: $x = 4.35$ or $x = -0.345$.

• Exact (Surd) Form: You may be asked for exact answers. This involves simplifying the expression under the square root (the discriminant).     - Example: For $x = \frac{8 \pm \sqrt{88}}{4}$.     - Simplify $\sqrt{88} = \sqrt{4 \times 22} = \sqrt{4} \times \sqrt{22} = 2\sqrt{22}$.     - Substitute back: $x = \frac{8 \pm 2\sqrt{22}}{4}$.     - Simplify the fraction: $x = \frac{2(4 \pm \sqrt{22})}{4} = \frac{4 \pm \sqrt{22}}{2}$.     - Exact solutions: $x = \frac{4 + \sqrt{22}}{2}$ or $x = \frac{4 - \sqrt{22}}{2}$.

• The Discriminant ($b^2 - 4ac$):     - If $b^2 - 4ac > 0$: There are 2 different real solutions.     - If $b^2 - 4ac = 0$: There is exactly 1 real solution.     - If $b^2 - 4ac < 0$: There are no real solutions.     - If the discriminant is a perfect square ($0, 1, 4, 9, 16,...$), the quadratic can be factorised using integers.

• Examiner Tips and Tricks: Always check what accuracy the question demands (e.g., 2 decimal places). Use a calculator to check answers, but always show the substitution step in the formula manually.

Completing the Square

• Basic Rule: The first two terms of $x^2 + bx + c$ can be written as the difference of two squares:     - $x^2 + bx = (x + p)^2 - p^2$, where $p = \frac{b}{2}$.     - Example: $x^2 + 20x = (x + 10)^2 - 10^2 = (x + 10)^2 - 100$.     - Note: A negative $b$ value does not change the subtraction of $p^2$ at the end.

• The General Process of Completing the Square:     - To complete the square for $x^2 + 10x + 9$:         - Replace $x^2 + 10x$ with $(x + 5)^2 - 5^2$.         - Add the constant: $(x + 5)^2 - 25 + 9$.         - Simplify: $(x + 5)^2 - 16$.

• Handling Coefficients ($a \neq 1$):     - Form: $ax^2 + bx + c$.     - First, factorise $a$ out of the $x^2$ and $x$ terms only using square brackets: $a[x^2 + \frac{b}{a}x] + c$.     - Complete the square on the expression inside the brackets: $a[(x + p)^2 - p^2] + c$, where $p = \frac{b}{2a}$.     - Multiply through by $a$: $a(x + p)^2 - ap^2 + c$.     - Example: $y = 4x^2 + 16x + 5$:         - Factor out $4$ within square brackets: $y = 4[x^2 + 4x] + 5$.         - Replace $x^2 + 4x$ with $(x + 2)^2 - 2^2$ (since $p = \frac{4}{2} = 2$): $y = 4[(x + 2)^2 - 4] + 5$.         - Expand the square brackets: $y = 4(x + 2)^2 - 16 + 5$.         - Final form: $y = 4(x + 2)^2 - 11$.

• Turning Points and Graphs:     - Completing the square identifies the turning point of a quadratic graph.     - If $y = a(x + p)^2 + q$, the turning point is at $(-p, q)$. Note the sign change for the $x$-coordinate.     - Transformation: This represents a translation of $y = x^2$ by $p$ units left and $q$ units up.     - Minimum vs. Maximum:         - If $a > 0$, the turning point is a minimum.         - If $a < 0$, the turning point is a maximum.     - Squared brackets $(x \pm p)^2$ are always $\ge 0$. The smallest a squared term can be is $0$.

• Working Example: Turning Point:     - For $y = x^2 + 6x - 11$:         - $p = \frac{6}{2} = 3$.         - $y = (x + 3)^2 - 3^2 - 11 = (x + 3)^2 - 20$.         - Turning point is at $(-3, -20)$.

• Solving by Completing the Square:     - Replace $x^2 + bx$ with $(x + p)^2 - p^2$.     - Rearrange the equation to make $x$ the subject using the $\pm\sqrt{}$ operation.     - Example: Solve $x^2 + 10x + 9 = 0$:         - Completion: $(x + 5)^2 - 25 + 9 = 0 \rightarrow (x + 5)^2 = 16$.         - Root: $x + 5 = \pm\sqrt{16} \rightarrow x + 5 = \pm 4$.         - Solve: $x = -5 + 4 = -1$ or $x = -5 - 4 = -9$.     - Trick for Equations: If you have $ax^2 + bx + c = 0$, you can divide the entire equation by $a$ before completing the square. Note: This only works for equations with "$= 0$", not for rewriting expressions.

Deciding the Quadratic Method

• When to Solve by Factorisation:     - When the question explicitly asks for it.     - For two-term quadratics (e.g., $x^2 - 4x = 0$ which becomes $x(x - 4) = 0$).     - For "Difference of Two Squares" (e.g., $x^2 - 9 = 0$ which becomes $(x + 3)(x - 3) = 0$).     - Often the quickest method if factors are recognizable.

• When to Use the Quadratic Formula:     - When requested to leave solutions to specific accuracy (decimal places or significant figures).     - When the equation is difficult to factorise (e.g., $36x^2 + 33x - 20 = 0$).     - If in doubt, as this method always works.

• When to Solve by Completing the Square:     - When requested in a multi-part question (e.g., Part A asks for completing the square, Part B asks for the solution).     - When making $x$ the subject of more complex formulas containing both $x^2$ and $x$ terms.     - Example: Rearrange $x^2 + 6x = y$ for $x$.         - $(x + 3)^2 - 9 = y \rightarrow (x + 3)^2 = y + 9 \rightarrow x + 3 = \pm\sqrt{y + 9} \rightarrow x = -3 \pm\sqrt{y + 9}$.

• Examiner Tips and Tricks:     - Use your calculator to verify solutions. If solutions are integers or simple fractions, the quadratic can be factorised.     - When solving by completing the square, never expand the squared bracket back out once it is formed.

Questions & Discussion

• How do I solve a quadratic equation using factorisation?     - Rearrange to $ax^2 + bx + c = 0$, factorise the left side, and set each bracket to zero. Example: In $(x - 3)(x + 7) = 0$, solutions are $3$ and $-7$.

• What if $x$ is a factor?     - Treat it as a bracket $(x - 0) = 0$, which gives the solution $x = 0$. Do not divide by $x$.

• How do I use the quadratic formula?     - Substitute identified $a$, $b$, and $c$ into $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Ensure you use the formula for both the addition and subtraction parts to get two solutions.

• What is the discriminant?     - It is the value $b^2 - 4ac$. Its sign determines the number of real solutions (Positive = 2, Zero = 1, Negative = 0).

• How do I find the turning point by completing the square?     - Once in the form $y = a(x + p)^2 + q$, the turning point is always at $(-p, q)$. This relates directly to graph transformations of $y = x^2$.

• How does completing the square link to the quadratic formula?     - The formula is derived by completing the square on the general equation $ax^2 + bx + c = 0$. They share the same underlying structure.