Quadratic equations and inequalities; discriminant practice questions.

Solve the quadratic equation x² − 5x + 6 = 0.

• x = 1 or x = 6
• x = 2 or x = 3
• x = −2 or x = −3
• x = 1 or x = 5

Answer: x = 2 or x = 3

Factorising: (x − 2)(x − 3) = 0, so x = 2 or x = 3 (they sum to 5 and multiply to 6).

The equation 2x² + kx + 8 = 0 has two equal real roots. Determine the possible values of k.

• k = 8 only
• k = −8 only
• k = ±8
• k = ±4

Answer: k = ±8

Equal roots require the discriminant to be zero: k² − 4(2)(8) = k² − 64 = 0, so k = 8 or k = −8. Forgetting the negative square root gives the incomplete answer k = 8 only.

Determine the values of k for which x² + 6x + k = 0 has two distinct real roots.

• k > 9
• k < 9
• k ≤ 9
• k < −9

Answer: k < 9

Two distinct real roots require b² − 4ac > 0, so 36 − 4k > 0, giving k < 9. The strict inequality matters: k = 9 gives a repeated root, not two distinct roots.

The function f(x) = x² − 4x + 7 is written in the form (x − h)² + k. Which expression is correct?

• (x − 2)² + 3
• (x + 2)² + 3
• (x − 2)² − 3
• (x + 2)² − 3

Answer: (x − 2)² + 3

Completing the square: x² − 4x + 7 = (x − 2)² − 4 + 7 = (x − 2)² + 3. A common error is adding 4 instead of subtracting it, or flipping the sign inside the bracket.

The equation 3x² − 12x + 5 = 0 has roots α and β. Calculate α + β.

• −4
• 4
• 12
• 5/3

Answer: 4

For ax² + bx + c = 0, the sum of the roots is −b/a = −(−12)/3 = 4. Sign slips give −4, and forgetting to divide by a gives 12.

Solve the inequality x² − x − 6 < 0.

• −2 < x < 3
• x < −2 or x > 3
• −3 < x < 2
• x < −3 or x > 2

Answer: −2 < x < 3

Factorising gives (x − 3)(x + 2) < 0, so the expression is negative between the roots: −2 < x < 3. Choosing the outside regions is the classic error for a positive leading coefficient.