2nd VMO 1963 Problem 2 For what values of m does the equation x 2 + (2m + 6)x + 4m + 12 = 0 has two real roots, both of them greater than -2.
m ≤ -3 Solution For real roots we must have (m+3) 2 ≥ 4m+12 or (m-1)(m+3) ≥ 0, so m ≥ 1 or m ≤ -3. If m ≥ 1, then - (2m+6) ≤ -8, so at least one of the roots is  -2. So we need - (m+3) - √(m 2 +2m-3) > -2, or √(m 2 +2m-3) < -m-1 = √(m 2 +2m+1), which is always true. 2nd VMO 1963 © John Scholes jscholes@kalva.demon.co.uk 7 March 2004 Last corrected 7 Mar 04