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 ▸

x^3 + 2x^2 - 9x - 18 = (x^2 - a^2)(x + b) where a,b are integers. Work out the three linear factors of x^3 + 2x^2 - 9x - 18. (Note: x^3 indicates x cubed and x^2 indicates x squared).


How do you use derivatives to categorise stationary points?


Find the gradient of the line x^2 + 3x - 6 at the point (5,34)


Rationalise and simplify (root(3) - 7)/(root(3) + 1) . Give your answer in the form a + b*root(3) where a, b are integers.


We're here to help

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