Algebra

Linear & Quadratic Inequalities

Introduction

A mathematical statement which says that two expressions do not name the same number is called an inequality. When two expressions are not equal, they can either be:

Inequality symbols

\(\gt\)	greater than
\(\lt\)	less than
\(\geq\)	greater than or equal to
\(\leq\)	less than or equal to

the other.

To solve an inequality in the single variable \(x\) means to find the numbers which make the inequality true when they are used as replacements for the variable. Unlike an equation, an inequality usually has an infinite number of solutions.

Inequalities, like equations, are solved by finding a series of successively simpler inequalities. We need the following axioms of inequalities to simplify an inequality:

Addition Axiom of Inequality – For any mathematical expressions \(P\), \(Q\), and \(R\), the inequalities \(P \lt Q\) and \(P + R \lt Q + R\) have the same solutions.

Multiplication Axiom of Inequality – For any mathematical expressions \(P\), \(Q\), and \(R\), where \(R \gt 0\), the inequalities \(P \lt Q\) and \(P \times R \lt Q \times R\) have the same solutions. For \(R \lt 0\), the inequalities \(P \lt Q\) and \(P \times R \gt Q \times R\).

It is important to note that multiplication or division of both sides of an inequality by a negative number reverses the direction of the inequality from \(\lt\) to \(\gt\).

Example

Example 1

Question

Find the set of values of \(x\) for which \(5(2x + 1) - 3x + 4 \geq 2(x - 2) - 2x - 1\)

Solution

1. Step 1 — Multiply \(5\) and \((2x + 1)\); \(2\) and \((x - 2)\)
   \[ 5(2x + 1) - 3x + 4 \geq 2(x - 2) - 2x - 1 \]
   \[ 10x + 5 - 3x + 4 \geq 2x - 4 - 2x - 1 \]
2. Step 2 — Combine similar terms
   \[ 7x + 9 \geq -5 \]
3. Step 3 — Subtract \(9\) from both sides
   \[ 7x + 9 - 9 \geq -5 - 9 \]
4. Step 4 — Combine similar terms
   \[ 7x \geq -14 \]
5. Step 5 — Divide both sides by \(7\)
   \[ x \geq -2 \]
6. Step 6 — In graph

   On the number line, a filled circle at \(-2\) with a ray extending to the right shows the solution set: every value of \(x\) greater than or equal to \(-2\). The circle is filled because \(-2\) itself is included.

7. Step 7 — State the solution set

   \(\therefore\) the solution set of the inequality \(5(2x + 1) - 3x + 4 \geq 2(x - 2) - 2x - 1\) is \(x \geq -2\).

Inequalities involving the quadratic form \(ax^2 + bx + c\)

To solve inequalities involving the quadratic form \(ax^2 + bx + c\), we need to consider the following:

A. The product of two quantities is positive if both are positive or both are negative quantities, i.e.

\(ab \gt 0\) iff i. \(a \gt 0\) and \(b \gt 0\), or ii. \(a \lt 0\) and \(b \lt 0\).

\(ab \geq 0\) iff i. \(a \geq 0\) and \(b \geq 0\), or ii. \(a \leq 0\) and \(b \leq 0\).

B. The product of two quantities is negative if one quantity is positive and the other is negative, i.e.

\(ab \lt 0\) iff i. \(a \gt 0\) and \(b \lt 0\), or ii. \(a \lt 0\) and \(b \gt 0\).

\(ab \leq 0\) iff i. \(a \geq 0\) and \(b \leq 0\), or ii. \(a \leq 0\) and \(b \geq 0\).

Example

Example 2

Question

Find the set of values of \(x\) for which \(4x(3x + 2) - x \lt 3(1 + 3x^2) - 7\)

Solution

1. Step 1 — Multiply \(4x\) and \((3x + 2)\); \(3\) and \((1 + 3x^2)\)

   Use B since the inequality is \(\lt\).

   \[ 4x(3x + 2) - x \lt 3(1 + 3x^2) - 7 \]
   \[ 12x^2 + 8x - x \lt 3 + 9x^2 - 7 \]
2. Step 2 — Combine similar terms
   \[ 12x^2 + 7x \lt 9x^2 - 4 \]
3. Step 3 — Subtract \(9x^2\) from and add \(4\) to both sides
   \[ 12x^2 + 7x - 9x^2 + 4 \lt 9x^2 - 4 - 9x^2 + 4 \]
4. Step 4 — Combine similar terms
   \[ 3x^2 + 7x + 4 \lt 0 \]
5. Step 5 — Factorise
   \[ (3x + 4)(x + 1) \lt 0 \]
6. Case 1 — 1st factor is positive, 2nd is negative

   Solve using both cases.

   \[ 3x + 4 \gt 0 \quad \text{and} \quad x + 1 \lt 0 \]

   Subtract \(4\) from both sides of the 1st inequality and subtract \(1\) from both sides of the 2nd inequality:

   \[ 3x + 4 - 4 \gt 0 - 4 \quad \text{and} \quad x + 1 - 1 \lt 0 - 1 \]

   Combine similar terms:

   \[ 3x \gt -4 \quad \text{and} \quad x \lt -1 \]

   Divide both sides of the 1st inequality by \(3\):

   \[ x \gt -\frac{4}{3} \quad \text{and} \quad x \lt -1 \]

   Graph to see the solution set — it is the intersection of the 2 graphs. On the number line, an open circle at \(-\dfrac{4}{3}\) carries a ray extending to the right, and an open circle at \(-1\) carries a ray extending to the left. The two rays overlap between \(-\dfrac{4}{3}\) and \(-1\); the overlap, marked by the thick segment on the number line, is the intersection.

   Solution set of case 1: \(-\dfrac{4}{3} \lt x \lt -1\)

7. Case 2 — 1st factor is negative, 2nd is positive

   Solve using both cases.

   \[ 3x + 4 \lt 0 \quad \text{and} \quad x + 1 \gt 0 \]

   Subtract \(4\) from both sides of the 1st inequality and subtract \(1\) from both sides of the 2nd inequality:

   \[ 3x + 4 - 4 \lt 0 - 4 \quad \text{and} \quad x + 1 - 1 \gt 0 - 1 \]

   Combine similar terms:

   \[ 3x \lt -4 \quad \text{and} \quad x \gt -1 \]

   Divide both sides of the 1st inequality by \(3\):

   \[ x \lt -\frac{4}{3} \quad \text{and} \quad x \gt -1 \]

   Graph to see the solution set — it is the intersection of the 2 graphs. On the number line, an open circle at \(-\dfrac{4}{3}\) carries a ray extending to the left, and an open circle at \(-1\) carries a ray extending to the right. The two rays point away from each other and do not overlap, so there is no intersection.

   Solution set of case 2: \(\varnothing\) or null set

8. State the solution set

   \(\therefore\) the solution set of the inequality \(4x(3x + 2) - x \lt 3(1 + 3x^2) - 7\) is \(-\dfrac{4}{3} \lt x \lt -1\).