Solving Linear and Quadratic Inequalities for the ESAT

Rules for Manipulating Inequalities

The most important principle to grasp when dealing with inequalities is that they do not behave exactly like equations. While you can add or subtract the same value from both sides without restriction, you cannot multiply or divide both sides, raise both sides to an even power, or apply specific functions without first verifying if the operation preserves the inequality.

Consider these cases where standard algebraic logic might fail if applied blindly:

1. Multiplication by a negative: $-7 < 5$ is true, but multiplying by $-1$ gives $7 < -5$, which is false.
2. Squaring both sides: $-2 < -1$ is true, but squaring gives $4 < 1$, which is false.
3. Cubing both sides: $-2 < -1$ is true, and cubing gives $-8 < -1$, which remains true.
4. Division by a negative: $-8 < -4$ is true, but dividing by $-2$ gives $4 < 1$, which is false.
5. Trigonometric functions: $30 < 60$ is true, but $\cos 30 < \cos 60$ is false, whereas $\sin 30 < \sin 60$ is true.

When solving inequalities, ensure that your manipulations do not turn a true statement into a false one or generate incorrect solutions.

Linear Inequalities

Linear inequalities involve $x$ to the power of 1 and can generally be solved by rearranging, provided you remember the negative multiplication rule.

Example 1: Solve $3x + 2 \leq x + 5$.

1. Subtract $x$ from both sides: $2x + 2 \leq 5$.
2. Subtract 2 from both sides: $2x \leq 3$.
3. Divide by 2: $x \leq \frac{3}{2}$.

Quadratic Inequalities

Solving a quadratic inequality typically requires two steps: algebraic factorisation and graphical sketching. The sketch helps you visualise which regions of the $x$-axis satisfy the inequality.

Example 2: Solve $x^2 + 5x + 6 \geq 0$.

First, factorise the expression: $(x + 2)(x + 3) \geq 0$. This tells us the quadratic crosses the $x$-axis at $x = -2$ and $x = -3$. Next, sketch the graph to see where the curve sits above the $x$-axis (the region where $y \geq 0$):

From the diagram, the solution consists of the two outer regions: $x \leq -3$ or $x \geq -2$.

Rational Inequalities

When an inequality has a variable in the denominator, you must be careful. You cannot simply multiply by the denominator because its sign changes depending on the value of $x$. To avoid this, multiply by the square of the denominator, as a squared term is always positive for real numbers.

Example 3: Solve $\frac{2x + 5}{x + 3} > 1$.

1. Multiply both sides by $(x + 3)^2$: $\frac{2x + 5}{x + 3} \times (x + 3)^2 > 1 \times (x + 3)^2$.
2. This simplifies to: $(2x + 5)(x + 3) > (x + 3)^2$.
3. Rearrange to set to zero: $(2x + 5)(x + 3) - (x + 3)^2 > 0$.
4. Factorise out the common term $(x + 3)$: $(x + 3)((2x + 5) - (x + 3)) > 0$.
5. Simplify: $(x + 3)(x + 2) > 0$.

As seen in Example 2, this is a U-shaped quadratic with roots at $-3$ and $-2$. The solution for $(x + 3)(x + 2) > 0$ is $x < -3$ or $x > -2$.

Correct Range Notation

Avoid common errors in writing solution ranges. A string like $1 < x < -5$ is mathematically impossible as $x$ cannot be simultaneously greater than 1 and less than $-5$. Similarly, using "and" incorrectly, such as $-2 < x$ and $x < -4$, creates an empty set. Always use "or" for separate regions: $x < -4$ or $x > -2$.

Modulus Inequalities

The modulus function $|x - a|$ represents the positive distance of $x$ from $a$ on the number line. This interpretation is often the fastest way to solve modulus inequalities.

Example 4: Solve $|x - 3| < |x - 5|$.

This asks: for what values of $x$ is the distance from 3 less than the distance from 5? The midpoint between 3 and 5 is 4. Any value less than 4 is closer to 3. Thus, the solution is $x < 4$. Graphically, this is where the V-shape of $y = |x - 3|$ sits below $y = |x - 5|$:

Example 5: Solve $|2x - 4| < |x + 2|$, which is $2|x - 2| < |x + 2|$.

We solve this by finding the points where the graphs intersect:

To find the intersection points, we solve the equations for the relevant branches of the modulus functions:

1. For the first intersection: $x + 2 = -(2x - 4) \implies x + 2 = -2x + 4 \implies 3x = 2 \implies x = \frac{2}{3}$.
2. For the second intersection: $x + 2 = 2x - 4 \implies x = 6$.

The graph shows $|2x - 4|$ is below $|x + 2|$ between these points. So, the solution is $\frac{2}{3} < x < 6$.