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O-Level Additional Mathematics · Secondary 3 and 4

Quadratic Functions and Inequalities

Quadratic functions are expressions of the form y = ax² + bx + c, where a ≠ 0. In A-Math you go beyond solving quadratic equations: you complete the square to find maximum and minimum values, use the discriminant to decide how many roots there are, find when a line meets a curve, and solve quadratic inequalities. These skills appear throughout the rest of the course, from coordinate geometry of circles to stationary points in calculus, so it pays to make them automatic early.

What the syllabus expects

  • Complete the square to write ax² + bx + c in the form a(x - h)² + k, and state the maximum or minimum value and the turning point
  • Sketch the graph of a quadratic function, showing the turning point and the intercepts with the axes
  • Use the discriminant b² - 4ac to determine the nature of the roots of a quadratic equation
  • Find the conditions for ax² + bx + c to be always positive or always negative
  • Determine whether a straight line cuts, touches or does not meet a curve
  • Solve quadratic inequalities and represent the solution on a number line
  • Solve simultaneous equations where one equation is linear and the other is quadratic

Completing the square and turning points

Any quadratic y = ax² + bx + c can be rewritten as y = a(x - h)² + k. Because a squared term is never negative, (x - h)² ≥ 0 for every x. If a > 0, the smallest value of y is k, reached when x = h, so the curve is a U shape with minimum point (h, k). If a < 0, the curve is an inverted U and k is the maximum value. To complete the square, first take out the coefficient of x² from the x terms, then halve the coefficient of x inside the bracket. For example, 2x² - 12x + 5 = 2(x² - 6x) + 5 = 2[(x - 3)² - 9] + 5 = 2(x - 3)² - 13. So the minimum value is -13 when x = 3, and the line of symmetry is x = 3. The y-intercept comes from putting x = 0, which gives 5 here.

The discriminant and the nature of roots

For ax² + bx + c = 0, the discriminant is b² - 4ac. If b² - 4ac > 0, there are two real and distinct roots and the curve cuts the x-axis twice. If b² - 4ac = 0, there are two real and equal roots, so the curve touches the x-axis. If b² - 4ac < 0, there are no real roots and the curve does not meet the x-axis. The phrase 'real roots' with no other word means b² - 4ac ≥ 0, since equal roots are still real. A quadratic is always positive (positive for all real x) when a > 0 and b² - 4ac < 0, because the U-shaped curve lies entirely above the x-axis. It is always negative when a < 0 and b² - 4ac < 0. For example, x² + 4x + 7 has discriminant 16 - 28 = -12 < 0 and a = 1 > 0, so it is positive for all real x.

Lines meeting curves

To find where the line y = mx + c meets a curve, substitute the line into the curve to get a single quadratic in x, then look at the discriminant of that new quadratic. Two distinct real roots means the line cuts the curve at two points. Equal roots means the line is a tangent to the curve. No real roots means the line does not meet the curve. For example, to find k such that y = x + k is a tangent to y = x² - 3x + 6, set x² - 3x + 6 = x + k, so x² - 4x + (6 - k) = 0. For a tangent, 16 - 4(6 - k) = 0, so 16 - 24 + 4k = 0 and k = 2. The same substitution solves simultaneous equations where one equation is linear: make one variable the subject of the linear equation, substitute, solve the quadratic, then find the other variable for each root.

Solving quadratic inequalities

To solve a quadratic inequality, bring every term to one side, factorise, find the roots and sketch the curve. For (x - a)(x - b) < 0 with a < b, the U-shaped curve is below the x-axis between the roots, so a < x < b. For (x - a)(x - b) > 0, the curve is above the axis outside the roots, so x < a or x > b. For example, x² - x - 6 ≤ 0 factorises to (x - 3)(x + 2) ≤ 0, giving -2 ≤ x ≤ 3. Never divide both sides by an expression containing x, since you do not know its sign. If the coefficient of x² is negative, multiply through by -1 and reverse the inequality sign first. Show the answer on a number line with filled circles for ≤ or ≥ and open circles for < or >.

Worked example

Find the range of values of k for which the line y = kx - 1 does not meet the curve y = x² + x + 3.

  1. Substitute the line into the curve: x² + x + 3 = kx - 1.
  2. Collect the terms on one side: x² + (1 - k)x + 4 = 0.
  3. The line does not meet the curve, so this equation has no real roots and b² - 4ac < 0.
  4. Here a = 1, b = 1 - k and c = 4, so (1 - k)² - 16 < 0.
  5. Expand: 1 - 2k + k² - 16 < 0, which gives k² - 2k - 15 < 0.
  6. Factorise: (k - 5)(k + 3) < 0, so the roots are k = -3 and k = 5.
  7. The U-shaped graph of (k - 5)(k + 3) is below the axis between its roots, so -3 < k < 5.

Answer: -3 < k < 5

Common mistakes

  • Writing the condition for 'real roots' as b² - 4ac > 0. Real roots include equal roots, so the condition is b² - 4ac ≥ 0.
  • Forgetting to take out the coefficient of x² before completing the square, for example writing 2x² - 12x + 5 as 2(x - 6)² plus a constant. The 2 multiplies everything in the bracket, so take it out of the x terms first.
  • Using the discriminant of the curve alone when the question asks about a line meeting the curve. Substitute the line first to form a new quadratic, and use the discriminant of that.
  • Writing the solution of (x - 3)(x + 2) > 0 as -2 > x > 3, which no number satisfies. Two separate regions must be written as x < -2 or x > 3.
  • Claiming a quadratic is always positive from b² - 4ac < 0 alone. You must also check a > 0, since a curve with a < 0 and no roots is always negative.
  • Getting the sign of the turning point wrong. The curve y = (x + 4)² - 1 has its minimum point at (-4, -1), not at x = 4.

Exam tips

  • Underline the key word before writing the condition: 'meets' or 'intersects' means b² - 4ac ≥ 0, 'cuts at two distinct points' means > 0, 'tangent' means = 0 and 'does not meet' means < 0.
  • For any inequality in k, finish with a quick sketch of the quadratic in k. It shows at a glance whether the answer lies between the roots or outside them.
  • After completing the square, expand your answer in your head to check that it gives back the original expression.

Practice questions

1. Express x² - 8x + 21 in the form (x - h)² + k. State the minimum value of the expression and the value of x at which it occurs.

Show answer

x² - 8x + 21 = (x - 4)² - 16 + 21 = (x - 4)² + 5. The minimum value is 5, when x = 4.

2. Find the range of values of p for which the equation x² + px + 9 = 0 has two real distinct roots.

Show answer

p < -6 or p > 6. Two real distinct roots need b² - 4ac > 0, so p² - 36 > 0, which is (p - 6)(p + 6) > 0. The solution lies outside the roots -6 and 6.

3. Find the value of c for which the line y = 2x + c is a tangent to the curve y² = 8x, and find the coordinates of the point of contact.

Show answer

c = 1, and the point of contact is (1/2, 2). Substituting gives (2x + c)² = 8x, so 4x² + (4c - 8)x + c² = 0. For a tangent, (4c - 8)² - 16c² = 0, which simplifies to 64 - 64c = 0, so c = 1. Then 4x² - 4x + 1 = 0, so (2x - 1)² = 0, x = 1/2 and y = 2(1/2) + 1 = 2.

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