# Discriminant, Roots and Quadratic Inequalities

> CIE IGCSE Additional Mathematics · 0606 2025-2027
> Source: https://www.owlsprep.com/study/cie-0606-u2-discriminant-roots-and-quadratic-inequalities/

This guide covers core quadratic function skills for CIE IGCSE Additional Mathematics 0606, including discriminant root classification, line-curve intersection rules, and solving quadratic inequalities with correct solution set notation.

**Prerequisites:** Ability to rearrange algebraic equations and solve linear inequalities; Knowledge of quadratic factorisation and the quadratic formula

## Learning objectives

- Use the discriminant to classify the number of real roots of quadratic equations
- Apply discriminant conditions to line and quadratic curve intersection problems
- Solve quadratic inequalities using factorisation or the quadratic formula
- Represent quadratic inequality solution sets correctly using standard notation

## 1. The Discriminant and Classification of Real Roots

**Discriminant** — For a quadratic equation of the form $ax^2 + bx + c = 0$ where $a \neq 0$, the discriminant is the value that determines the number of real roots of the equation.

*Notation:* \Delta = b^2 - 4ac

*Example:* For $2x^2 + 5x + 2 = 0$, $\Delta = 5^2 - 4(2)(2) = 9$

The discriminant maps to three possible cases for real roots (complex roots are not assessed in 0606):
- $\Delta > 0$: Two distinct real roots
- $\Delta = 0$: One repeated (equal) real root
- $\Delta < 0$: No real roots

**Worked example:** Find the values of $k$ for which the quadratic equation $3x^2 + kx + 12 = 0$ has equal real roots.

1. Identify coefficients: $a = 3$, $b = k$, $c = 12$
2. For equal roots, discriminant $\Delta = 0$
3. $$k^2 - 4(3)(12) = 0$$
4. $$k^2 - 144 = 0 \implies k = \pm 12$$

> **tip**
>
> Always confirm $a \neq 0$ before calculating the discriminant, as the formula only applies to quadratic equations.

## 2. Discriminant Applications: Line and Curve Intersections

When analyzing if a straight line intersects a quadratic parabola, substitute the line equation into the curve equation to form a single quadratic equation in one variable. The discriminant of this resulting equation tells you the number of intersection points:
- $\Delta > 0$: Two distinct intersection points
- $\Delta = 0$: One intersection point (the line is tangent to the curve)
- $\Delta < 0$: No intersection points

**Worked example:** Find the value of $m$ for which the line $y = mx + 3$ is tangent to the curve $y = 2x^2 + 5x + 4$.

1. Equate the two expressions for $y$:
2. $$mx + 3 = 2x^2 + 5x + 4$$
3. Rearrange into standard quadratic form:
4. $$2x^2 + (5 - m)x + 1 = 0$$
5. For tangent condition, $\Delta = 0$:
6. $$(5 - m)^2 - 4(2)(1) = 0$$
7. $$m^2 - 10m + 17 = 0$$
8. Solve using quadratic formula:
9. $$m = 5 \pm 2\sqrt{2}$$

> **Exam tip:** This is a frequent 3-4 mark exam question: always follow the sequence substitute → rearrange → apply discriminant rule.

## 3. Solving Quadratic Inequalities

Quadratic inequalities are expressions of the form $ax^2 + bx + c > 0$, $ax^2 + bx + c < 0$, $ax^2 + bx + c \geq 0$, or $ax^2 + bx + c \leq 0$. Follow these steps to solve them:
1. Rearrange the inequality so all terms are on one side, with the coefficient of $x^2$ positive.
2. Solve the corresponding quadratic equation to find the roots.
3. Sketch a quick upward-opening parabola through the roots to identify the region that satisfies the inequality.
4. Write the solution set using correct notation.

**Worked example:** Solve the inequality $2x^2 - 5x - 3 < 0$.

1. Coefficient of $x^2$ is already positive, so no sign change is needed.
2. Solve the corresponding quadratic equation by factorisation:
3. $$(2x + 1)(x - 3) = 0 \implies \text{roots at } x = -0.5, x = 3$$
4. Sketch upward opening parabola crossing x-axis at $-0.5$ and $3$. The region where the parabola is below the x-axis (for $<0$) is between the two roots.
5. Solution set: $-0.5 < x < 3$

> **warning**
>
> If you multiply or divide both sides of an inequality by a negative number, reverse the direction of the inequality sign. Always make the $x^2$ coefficient positive before solving to avoid sign errors.

## 4. Correct Notation for Quadratic Inequality Solutions

Quadratic inequalities produce two types of solution sets for positive $x^2$ coefficients:
1. For $< 0$ or $\leq 0$: Solution is a single interval between the two roots, written as $a < x < b$.
2. For $> 0$ or $\geq 0$: Solution is two separate intervals, written as $x < a$ or $x > b$.
Never write $a > x > b$ for two intervals, as this implies $x$ is simultaneously less than $a$ and greater than $b$, which is impossible.

**Check your understanding**

1. What is the solution set for $(x - 2)(x + 4) > 0$?

   - $-4 < x < 2$
   - $x < -4$ or $x > 2$
   - $x < -2$ or $x > 4$

   *Why:* For $>0$ with positive leading coefficient, values are outside the roots $x=-4$ and $x=2$.

2. What is the solution set for $x^2 - 6x + 9 \leq 0$?

   *Why:* The quadratic factorises to $(x-3)^2 = 0$, which has a repeated root at $x=3$, so only $x=3$ satisfies the inequality.

## Common pitfalls

- **Wrong:** Forgetting to make the coefficient of $x^2$ positive before solving a quadratic inequality
  - Why it fails: This flips the direction of the parabola, leading to the wrong region being selected for the solution set
  - Correct: Multiply both sides of the inequality by -1, reversing the inequality sign, to make the $x^2$ coefficient positive before finding roots
- **Wrong:** Writing a solution set for a $>0$ inequality as $a < x < b$
  - Why it fails: This incorrectly selects the region between the roots, which is the region where the quadratic is $<0$ for positive leading coefficient
  - Correct: For positive leading coefficient and $>0$ inequality, select the regions outside the two roots, written as $x < a$ or $x > b$
- **Wrong:** Using the discriminant for non-quadratic equations where $a = 0$
  - Why it fails: The discriminant formula only applies to equations of the form $ax^2 + bx + c = 0$ where $a \neq 0$
  - Correct: First confirm the coefficient of $x^2$ is non-zero before applying discriminant rules
- **Wrong:** Rounding discriminant values too early when solving line-curve intersection problems
  - Why it fails: This leads to incorrect root values, losing accuracy marks
  - Correct: Keep values in exact surd form until the final step, unless told otherwise by the question

## Cheatsheet

| Concept | Condition | Result |
| --- | --- | --- |
| Discriminant for $ax^2+bx+c=0$ | $\Delta > 0$ | Two distinct real roots |
|  | $\Delta = 0$ | One repeated real root |
|  | $\Delta < 0$ | No real roots |
| Line-curve intersection | $\Delta > 0$ | Two intersection points |
|  | $\Delta = 0$ | Line is tangent to curve |
|  | $\Delta < 0$ | No intersection points |
| Quadratic inequality ($a>0$) | $ax^2+bx+c < 0$ | Solution: between roots ($a < x < b$) |
|  | $ax^2+bx+c > 0$ | Solution: outside roots ($x < a$ or $x > b$) |

## What's next

Now that you have mastered discriminant analysis and quadratic inequalities, you are ready to apply these skills to more advanced quadratic function topics in CIE IGCSE Additional Mathematics 0606. These concepts are foundational for upcoming units including quadratic graph sketching, maximum and minimum value problems, and kinematics applications where quadratic motion equations are used. You will also encounter discriminant rules again when solving simultaneous equations involving one linear and one quadratic equation, a common 5-6 mark exam question. Practice these skills regularly using past paper questions to build speed and avoid common sign errors, especially when rearranging inequalities and applying discriminant conditions.

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