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​