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.

FS
Answered by Felix S. Further Mathematics tutor

8340 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Given f(x)= 8 − x^2, solve f(3x) = -28


Given y=x^3-x^2+6x-1, use diffferentiation to find the gradient of the normal at (1,5).


How can I find the equation of a straight line on a graph?


Let Curve C be f(x)=(1/3)(x^2)+8 and line L be y=3x+k where k is a positive constant. Given that L is tangent to C, find the value of k. (6 marks approx)


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