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Algebra

Question 7 of 8

Linear & Quadratic Inequalities

Question

Find the set of values of \(x\) for which \(10x^2 + 9x \ge 9\)

Show solution Hide solution Fully worked — 6 steps
  1. Subtract 9 from both sides of the inequality
    \[ 10x^2 + 9x \ge 9 \]
    \[ 10x^2 + 9x - 9 \ge 9 - 9 \]
  2. Combine similar terms
    \[ 10x^2 + 9x - 9 \ge 0 \]
  3. Factorise
    \[ (2x + 3)(5x - 3) \ge 0 \]
  4. Case 1 — both factors are positive
    \[ 2x + 3 \ge 0 \quad \text{and} \quad 5x - 3 \ge 0 \]

    Subtract 3 from both sides of the 1st inequality and add 3 to both sides of the 2nd inequality:

    \[ 2x + 3 - 3 \ge 0 - 3 \quad \text{and} \quad 5x - 3 + 3 \ge 0 + 3 \]

    Combine similar terms:

    \[ 2x \ge -3 \quad \text{and} \quad 5x \ge 3 \]

    Divide both sides of the 1st inequality by 2; divide both sides of the 2nd inequality by 5:

    \[ x \ge -\frac{3}{2} \quad \text{and} \quad x \ge \frac{3}{5} \]

    Graph to see the solution set: on the number line, a filled circle at \(-\dfrac{3}{2}\) with a ray extending right \(\left(x \ge -\dfrac{3}{2}\right)\) and a filled circle at \(\dfrac{3}{5}\) with a ray extending right \(\left(x \ge \dfrac{3}{5}\right)\). Both inequalities hold from \(\dfrac{3}{5}\) rightwards, endpoint included — the intersection, marked by the thick segment on the number line.

    Number line graph of case 1: filled circles at −3/2 and 3/5 with rays right; the overlap from 3/5 rightwards gives the solution set x ≥ 3/5

    Solution set of case 1: \(x \ge \dfrac{3}{5}\)

  5. Case 2 — both factors are negative
    \[ 2x + 3 \le 0 \quad \text{and} \quad 5x - 3 \le 0 \]

    Subtract 3 from both sides of the 1st inequality and add 3 to both sides of the 2nd inequality:

    \[ 2x + 3 - 3 \le 0 - 3 \quad \text{and} \quad 5x - 3 + 3 \le 0 + 3 \]

    Combine similar terms:

    \[ 2x \le -3 \quad \text{and} \quad 5x \le 3 \]

    Divide both sides of the 1st inequality by 2; divide both sides of the 2nd inequality by 5:

    \[ x \le -\frac{3}{2} \quad \text{and} \quad x \le \frac{3}{5} \]

    Graph to see the solution set: on the number line, a filled circle at \(\dfrac{3}{5}\) with a ray extending left \(\left(x \le \dfrac{3}{5}\right)\) and a filled circle at \(-\dfrac{3}{2}\) with a ray extending left \(\left(x \le -\dfrac{3}{2}\right)\). Both inequalities hold from \(-\dfrac{3}{2}\) leftwards, endpoint included — the intersection, marked by the thick segment on the number line.

    Number line graph of case 2: filled circles at −3/2 and 3/5 with rays left; the overlap from −3/2 leftwards gives the solution set x ≤ −3/2

    Solution set of case 2: \(x \le -\dfrac{3}{2}\)

  6. Solution set

    \(\therefore\) the solution set of the inequality \(10x^2 + 9x \ge 9\) is \(x \le -\dfrac{3}{2}\) or \(x \ge \dfrac{3}{5}\). That is, the solution set is every value of \(x\) less than or equal to minus three halves, together with every value of \(x\) greater than or equal to three fifths.