Suppose r and s are the roots of the equation 3x^2 + 9x + 15 = 2x^2 + 8x - 17. Find (r+1)(s+1)
+394 We have:
\(3x^2+9x+15=2x^2+8x-17\)
Start off by simplifying.
\(x^2+x+32=0\)
We want to find (r+1)(s+1) = rs+r+s+1
We apply Vieta's formula:
rs = 32
r+s = -1
32-1+1 = 32
