n is an integer such that 3n + 2 ≤ 14, and 6n/(n^2 + 5) >1. Find all possible values of n.

Step 1: Simplify 3n + 2 ≤ 14 3n ≤ 12 n ≤ 4 and 6n > n^2 + 5 0 > n^2 -6n + 5 Factorise (n-5)(n-1) < 0
Step 2: Let (n-5)(n-1) = 0, so n=5 or n=1 If (n-5)(n-1) < 0, then 1<n<5 (use graph/substitution)
Step 3: Combine, n can take values 2,3,4.

Answered by Catriona M. Maths tutor

10051 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Find the roots of 2x^2-5x-6=0 to 3dps


x – 7x + 10 = 0


Given the equations: x + 3y = 1 and 2x - y = -5, solve for x and y.


What is (x-5)^2


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences