Problem:
If a,b,c,d are the solutions of the equation x4−bx−3=0, then an equation whose solutions are d2a+b+c,c2a+b+d,b2a+c+d,a2b+c+d is
Answer Choices:
A. 3x4+bx+1=0
B. 3x4−bx+1=0
C. 3x4+bx3−1=0
D. 3x4−bx3−1=0
E. none of these
Solution:
Since the coefficient of x3 in the polynomial function f(x)=x4−bx−3 is zero, the sum of the roots of f(x) is zero and therefore,
d2a+b+c=d2a+b+c+d−d=d−1
Similarly,
b2a+c+d=b−1,c2a+b+d=c−1,a2b+c+d=a−1
Hence the equation f(−x1)=0 has the specified solutions:
x41+xb−3=01+bx3−3x4=03x4−bx3−1=0