Quadratic Inequalities (AQA A-Level Mathematics): Revision Notes

1. Write the Inequality in Standard Form:

• Ensure the inequality is in the form $\ ax^2 + bx + c$ with one side equal to 0.

• Example: For $x^2 - 3x - 4 = 0$, factor it into $\ (x - 4)(x + 1) = 0$, so the roots are $\ x = :highlight[4]$ and $\ x = :highlight[-1].$

📑Example:

📑Example: The solution for $\ x^2 - 3x - 4 > 0 \ is \ x \in (-\infty, -1) \cup (4, \infty).$

📑Example: Solve $x^2 - 4x - 5 < 0$

1. Solve for = 0 and sketch the curve:

• $(x + 1)(x - 5) < 0$
• Roots are $x = :highlight[-1]$ and $x = :highlight[5]$

2. Divide the graph vertically into strips through the roots and label these strips + or -:

• The graph will have three regions:
• Left of $x = -1$ (positive)
• Between $x = -1$ and $x = 5$ (negative)
• Right of$\ x = 5$ (positive)

6. Decide which strips to use based on the inequality:

• Need negative strip(s) since $< 0$
• Solution: -1 < x < 5

7. Calculator Instructions:

• Go to the inequality solving mode.
• Select polynomial degree ( 2 for quadratic).
• Input the coefficients.

📑Example: Solve $8 + 6x - x^2 \leq 0$

8. Rewrite the inequality:

• $-x^2 + 6x + 8 \leq 0$
• Doesn't factorise easily, so use the quadratic formula.
• $a = -1, b = 6, c = 8$
• $\frac {-6 \pm \sqrt {6^2-4(-1)(8)}}{2(-1)}$
• Solutions: $x = :highlight[3 \pm \sqrt{17}]$

9. Sketch the curve and determine the regions:

• Since it's a downward-facing parabola, regions are $x \leq 3 - \sqrt{17} and x \geq 3 + \sqrt{17}$.

10. Solution:

• x ≤ 3 - √17 or x ≥ 3 + √17

📝Q5 (Jun 2014, Q6)

Example: Solve the inequality $3x^2 + 10x + 3 > 0$

3. Factor the quadratic equation:

• Given equation: $3x^2 + 10x + 3$
• Factor the quadratic:
• Identify pairs of factors for $3 \times 3 = 9$ that add up to 10:
• Pairs: (1, 9)
• Rewrite the quadratic:
• $3x^2 + x + 9x + 3 = 0$
• Factor by grouping:
• $x(3x + 1) + 3(3x + 1) = 0$
• $(3x + 1)(x + 3) = 0$

11. Find the roots of the quadratic:

• Set each factor to zero:
• $3x + 1 = 0$
• $x = :highlight[-\frac{1}{3}]$
• $x + 3 = 0$
• $x = :highlight[-3]$

12. Sketch the graph:

• The parabola opens upwards because the coefficient of $x^2$ is positive.
• The roots divide the $x$-axis into three intervals:
• $x < -3$ (positive region)
• $-3 < x < -\frac{1}{3}$ (negative region)
• $x > -\frac{1}{3}$ (positive region)

13. Determine the solution intervals:

• The inequality $3x^2 + 10x + 3 > 0$ is satisfied in the positive regions.
• Solution: $x < -3 \ or\ x > -\frac{1}{3}$

Final Solution:

• x < -3 or x > -1/3

📑Example: $x^2 + 3x + 4k = 0$