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(x3−49x2+1623x−6415)=0∴x3−49x2+1623x−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 .