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

First of all, solving the equation 3n + 2 < 14 to find n. 3n < 14 -2 = 3n < 12. n < 12/3 = n < 4Secondly solve the equation 6n/(n^2+5) > 1 to find n. Collect all terms on one side of the to solve the equation as a quadratic. Therefore, multiplying both sides by (n^2 + 5) will give: 6n > n^2 +5. Then minus both sides by 6n to give n^2 -6n +5 < 0. Solve the quadratic n^2 -6n +5, which gives (n-5)(n-1)<0. Therefore n = 5 and n=1, thus 1<n<5Finally, finding all n that satisfy both equation so the answer is 2 and 3.

KP

Related Maths GCSE answers

All answers ▸

How do you complete the square?


What is the gradient of the graph y = 3x^2 + 5x + 2, when x is 4?


The population of sheep on an island is 170. The population of the sheep is expected to increase by 3% each year, what will the population of sheep be in 5 years time? [3 marks]


Find the solutions to the quadratic equation x^2-9x+20=0