Linear and Quadratic Inequalities for the TMUA

The Fundamental Behaviour of Inequalities

When dealing with inequalities, it is vital to remember that they do not behave exactly like equations. While you can add and subtract terms on both sides freely, you cannot multiply or divide both sides, raise both sides to an even power, or apply certain functions without first checking if the operation preserves the inequality.

Consider these examples where standard algebraic moves fail if applied carelessly:

1. The statement $-7 < 5$ is true, but multiplying both sides by $-1$ yields $7 < -5$, which is clearly false.
2. The statement $-2 < -1$ is true, but squaring both sides yields $4 < 1$, which is false.
3. The statement $-8 < -4$ is true, but dividing both sides by $-2$ yields $4 < 1$, which is false.

In general, you must ensure your manipulations do not turn a true statement into a false one or generate incorrect solution sets.

Solving Linear Inequalities

Linear inequalities can often be solved by simple rearrangement, provided you remember to flip the sign if you multiply or divide by a negative number.

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

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

Solving Quadratic Inequalities

Quadratic inequalities are best solved using a combination of factorisation and graphical sketching. Simply finding the roots is not enough: you must determine which regions of the $x$ axis satisfy the inequality.

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

First, factorise the expression: $(x + 2)(x + 3) \geq 0$

Next, produce a rough sketch of the quadratic $y = (x + 2)(x + 3)$. The roots (x-intercepts) are at $x = -2$ and $x = -3$.

From the diagram, we identify the $x$ values where the graph sits on or above the $x$ axis (where $y \geq 0$). The solution is $x \leq -3$ or $x \geq -2$.

Rational Inequalities and the Square Method

A common mistake when solving rational inequalities is multiplying by a denominator that could be negative, which would flip the sign in ways the algebraist might not account for.

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

It is tempting to multiply by $x + 3$, but since $x + 3$ is negative for $x < -3$, this is a poor strategy. Instead, multiply both sides by $(x + 3)^2$, which is guaranteed to be positive for all $x$ in the domain.

$\frac{2x + 5}{x + 3} \times (x + 3)^2 > 1 \times (x + 3)^2$

Rearrange the terms into a single side to find a common factor: $(2x + 5)(x + 3) - (x + 3)^2 > 0$

Extract the common factor $(x + 3)$: $(x + 3)((2x + 5) - (x + 3)) > 0$ $(x + 3)(x + 2) > 0$

By sketching this quadratic, we see it is positive outside of the roots. Thus, the solution is $x < -3$ or $x > -2$.

Writing Ranges and Correct Notation

Correct logical notation is essential. Avoid 'howlers' such as $1 < x < -5$, which is a logical impossibility. If a solution set consists of two disjoint regions, use the word 'or'. For example, if $x$ must be less than $-4$ or greater than $-2$, write $x < -4$ or $x > -2$.

Inequalities Involving the Modulus Sign

The modulus function $|f(x)|$ makes the value of $f(x)$ positive. The expression $|x - a|$ can be interpreted as the positive distance of $x$ from $a$ on a number line.

Consider the inequality $|x - 3| < |x - 5|$. Graphically, we sketch $y = |x - 3|$ and $y = |x - 5|$ and see where the first graph sits below the second.

From the diagram, the solution is $x < 4$. Conceptually, this asks: which $x$ values are closer to 3 than they are to 5? The midpoint between 3 and 5 is 4, so everything to the left of 4 is closer to 3.

For more complex cases like $|2x - 4| < |x + 2|$, which is $2|x - 2| < |x + 2|$, we use a sketch:

To find the intersection points $A$ and $B$, we solve the equations where the lines meet. For $A$, we solve $x + 2 = -(2x - 4)$, giving $x = \frac{2}{3}$. For $B$, we solve $2x - 4 = x + 2$, giving $x = 6$. The range where the first graph is below the second is $\frac{2}{3} < x < 6$.