For what values of m does the equation x2 + (2m + 6)x + 4m + 12 = 0 has two real roots, both of them greater than -2.
Answer
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 must have m ≤ -3.
The roots are -(m+3) ±√(m2+2m-3). Now -(m+3) ≥ 0, so -(m+3) + √(m2+2m-3) ≥ 0 > -2. So we need -(m+3) - √(m2+2m-3) > -2, or √(m2+2m-3) < -m-1 = √(m2+2m+1), which is always true.
© John Scholes
jscholes@kalva.demon.co.uk
7 March 2004
Last corrected 7 Mar 04