Problem: If r1 and r2 are the distinct real roots of x2+px+8=0; then it must follow that:
Answer Choices:
A. ∣r1+r2∣>42
B. ∣r1∣>3 or ∣r2∣>3
C. ∣r1∣>2 and ∣r2∣>2
D. r1<0 and r2<0
E. ∣r1+r2∣<42
Solution:
Since r1 and r2 are real and distinct, p2−32>0∴∣p∣>42. But r1+r2=−p ∴∣r1+r2∣=∣p∣∴∣r1+r2∣>42.