Given a^2 < 4 and a+2b = 8. Work out the range of possible values of b. Give your answer as an inequality.

For this question, we want to start by finding the possible values of a from the first equation and then using that to give us information about b from the second equation.

Now, the condition that a^2 < 4 actually gives us two pieces of information (a way to remember this is that for inequalities, squared gives us two things). We have that a < 2 but also that a >  -2 since for example, (-3)^2 = 9 which is larger than 4. Therefore, we have -2 < a < 2.

Rearranging the second equation for b, we get b = (8-a)/2. Therefore, putting in our values for a, we get that the maximum value of b is (8-(-2))/2 = 10/2 = 5 and the minimum value of b is (8-2)/2 = 6/2 = 3. Since the inequality for a does not include 2 and -2, we do not include 3 and 5 for b and so we get the answer 3 < b < 5.

Related Further Mathematics GCSE answers

All answers ▸

Find and describe the stationary points of the curve y = x^2 + 2x - 8


find the stationary point of the curve for the equation y=x^2 + 3x + 4


Solve the following simultanious equations: zy=28 and 2z-3y=13


Can you explain rationalising surds?


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 2013–2024

Terms & Conditions|Privacy Policy