Problem: The roots of 64x3−144x2+92x−15=0 are in arithmetic progression. The difference between the largest and smallest roots is:
Answer Choices:
A. 2
B. 1
C. 1/2
D. 3/8
E. 1/4
Solution:
Let the roots be a−d, a, and a+d. Then [x−(a−d)](x−a)[x−a+d)]=x3−3ax2+x(3a2−d2) +(a3−ad2)=0. But 64x3−144x2+92x−15=64(x3−49​x2+1623​x−6415​)=0∴x3−49​x2+1623​x−6415​=0 =x3−3ax2+x(3a2−d2)−(a3−ad2)∴−3a=−49​,a=43​ and 3a2−d2=1623​,d2=164​,d=+21​ or −21​. ∴ the roots are 45​,43​, and 41​, and the required difference is 1 .