(a) Find the set of values of k that satisfy the inequality k^2 - k - 12 < 0. (b) We have a triangle ABC, of lengths AC = 4 and BC = 2. Given that cos B < 1/4 , find the range of possible values for AB:
Solution: Let us consider the function f(k) = k2 - k -12. Firstly, we want to find out the roots of this function. We calculate the discriminant: Δ = b2 – 4ac = 49 hence we get the roots: k1 = (1-7)/(21) and k2 = (1+7)/(21) hence k1 = -3 and k2 = 4. We know that in ak2 + bk + c = 0 quadratic equation, a>0, so we have the following table sign:For k < -3 ===> f(k) > 0 For k = -3 ===> f(k) = 0 For -3 < k < 4 ===> f(k) < 0 For k = 4 ===> f(k) = 0 For k > 4 ===> f(k) > 0For f(k) < 0, we get that -3 < k < 4For the next part, we know that cos B < ¼, so we have to use the cosine theorem on the line opposite to the angle B:AC2 = AB2 + BC2 – 2ABBCcos B; Let AB = k and let us plug in the known values:16 = k2 + 4 -2k2cos B => cos B = (-12+k2)/(4*k) < ¼ => we get k2 - k -12 < 0 with solutions -3 < k < 4
