Suppose r and s are the roots of the equation 3x^2 + 9x + 15 = 0. Find (r+1)(s+1)
By the quadratic formula, the roots are -1/2*i*(sqrt(11) +- 3i). Plugging those in, we get (r + 1)(s + 1) = -6.
+2666 We want to find\((r+1)(s+1) = rs + s + r + 1\)
From Vieta's, we know that \(rs = {c \over a} = {15 \over 3} = 5\) and that \(r + s = -{b \over a} = -{9 \over 3} = -3\)
So, we have \(5 - 3 + 1= \color{brown}\boxed{3}\)
