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Algebra

Question 8 of 8

Linear & Quadratic Inequalities

Question

Find the set of values of \(x\) for which \(4x^2 + 5 \le 21x\)

Show solution Hide solution Fully worked — 6 steps
  1. Subtract 21x from both sides
    \[ 4x^2 + 5 \le 21x \]
    \[ 4x^2 + 5 - 21x \le 21x - 21x \]
  2. Combine similar terms
    \[ 4x^2 - 21x + 5 \le 0 \]
  3. Factorise
    \[ (4x - 1)(x - 5) \le 0 \]
  4. Case 1 — 1st factor is positive, 2nd is negative
    \[ 4x - 1 \ge 0 \quad \text{and} \quad x - 5 \le 0 \]

    Add 1 to both sides of the 1st inequality and add 5 to both sides of the 2nd inequality:

    \[ 4x - 1 + 1 \ge 0 + 1 \quad \text{and} \quad x - 5 + 5 \le 0 + 5 \]

    Combine similar terms:

    \[ 4x \ge 1 \quad \text{and} \quad x \le 5 \]

    Divide both sides of the 1st inequality by 4:

    \[ x \ge \frac{1}{4} \quad \text{and} \quad x \le 5 \]

    Graph to see the solution set: on the number line, a filled circle at \(5\) with a ray extending left (\(x \le 5\)) and a filled circle at \(\dfrac{1}{4}\) with a ray extending right \(\left(x \ge \dfrac{1}{4}\right)\). The rays overlap between \(\dfrac{1}{4}\) and \(5\), endpoints included — the intersection, marked by the thick segment on the number line.

    Number line graph of case 1: filled circles at 1/4 and 5 with rays towards each other; the overlap between them gives the solution set 1/4 ≤ x ≤ 5

    Solution set of case 1: \(\dfrac{1}{4} \le x \le 5\)

  5. Case 2 — 1st factor is negative, 2nd is positive
    \[ 4x - 1 \le 0 \quad \text{and} \quad x - 5 \ge 0 \]

    Add 1 to both sides of the 1st inequality and add 5 to both sides of the 2nd inequality:

    \[ 4x - 1 + 1 \le 0 + 1 \quad \text{and} \quad x - 5 + 5 \ge 0 + 5 \]

    Combine similar terms:

    \[ 4x \le 1 \quad \text{and} \quad x \ge 5 \]

    Divide both sides of the 1st inequality by 4:

    \[ x \le \frac{1}{4} \quad \text{and} \quad x \ge 5 \]

    Graph to see the solution set: on the number line, a filled circle at \(\dfrac{1}{4}\) with a ray extending left \(\left(x \le \dfrac{1}{4}\right)\) and a filled circle at \(5\) with a ray extending right (\(x \ge 5\)). The rays point in opposite directions and share no values, so the inequalities have no common solution.

    Number line graph of case 2: filled circles at 1/4 and 5 with rays in opposite directions — no overlap, so the solution set is ∅

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

  6. Solution set

    \(\therefore\) the solution set of the inequality \(4x^2 + 5 \le 21x\) is \(\dfrac{1}{4} \le x \le 5\). That is, the solution set is every value of \(x\) from one quarter to 5 inclusive.