Leaving Cert Mathematics Higher Paper 1 free practice questions with answers

Question 1 (30 marks) [Algebra and equations]

(a) Solve the following equation in x. Give each answer correct to 2 decimal places. 5x² + 14x − 6 = 0

(a) Answer: x ≈ −3.18 or x ≈ 0.38

• Compare with ax² + bx + c = 0: a = 5, b = 14, c = −6.
• Substitute into x = (−b ± √(b² − 4ac)) / 2a: x = (−14 ± √((14)² − 4(5)(−6))) / 10.
• The discriminant is b² − 4ac = 316, so √316 ≈ 17.7764.
• x = (−14 + 17.7764) / 10 ≈ 0.38 and x = (−14 − 17.7764) / 10 ≈ −3.18.

Quadratic formula: write a, b, c carefully with their signs, check the discriminant is positive, keep four or more decimals until the final rounding to 2 d.p.

(b) Find the range of values of p ∈ ℝ for which: |3p + 1| ≥ 9

(b) Answer: p ≤

• |A| ≥ k means A ≥ k or A ≤ −k, so 3p + 1 ≥ 9 or 3p + 1 ≤ −9.
• Solve each linear inequality: p ≥
• Combine the two outer ranges: p ≤

Modulus inequalities: |A| ≥ k splits into two outer ranges (or), |A| ≤ k gives one inner range (and); divide by a positive number without flipping the sign.

(c) Solve the equation for x ∈ ℝ: log₃(x + 20) + log₃(x − 6) = 3

(c) Answer: x = 7

• Both arguments must be positive, so x > 6 (the larger of −20 and 6).
• Add the logs as a single log of a product: log₃((x + 20)(x − 6)) = 3, so (x + 20)(x − 6) = 3³ = 27.
• Expand and collect: x² + 14x − 147 = 0.
• Factorise: (x − 7)(x + 21) = 0, so x = 7 or x = −21.
• x = 7 is greater than 6, so it is valid. x = −21 is less than 6, so it is rejected. Therefore x = 7.

Adding logs of the same base multiplies the arguments; after solving the quadratic, test each root against the domain of every log.

−10

3 or p ≥ 8

3

8

3 and p ≤ −10

3 .

−10

3 or p ≥ 8

3 .